Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

This week marks the third of four CrossProduct™ puzzles. This time, there are *seven* three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.

280 | |||

168 | |||

162 | |||

360 | |||

60 | |||

256 | |||

126 | |||

183,708 | 245,760 | 117,600 |

Can you find all seven three-digit numbers and complete the table?

In the game of Jenga, you build a tower and then remove its blocks, one at a time, until the tower collapses. But in *Riddler Jenga*, you start with one block and then place more blocks on top of it, one at a time.

All the blocks have the same alignment (e.g., east-west). Importantly, whenever you place a block, its center is picked randomly along the block directly beneath it. For example, the following animation shows Riddler Jenga towers that were randomly constructed before ultimately collapsing when the fifth, 10th and 15th blocks were placed. The block highlighted in red is the one above which the blocks were no longer balanced.

On average, how many blocks must you place so that your tower collapses — that is, until at least one block falls off?

(Note: This problem is *not* asking for the average height of the tower after any unbalanced blocks have fallen off. It is asking for the average number of blocks added in order to make the tower collapse in the first place.)

Congratulations to 👏David Mercado 👏 of Fairfield, CT, winner of last week’s Riddler Express.

In last week’s CrossProduct™ puzzle, there were six three-digit numbers, with products shown in the following table:

210 | |||

144 | |||

54 | |||

135 | |||

4 | |||

49 | |||

6,615 | 15,552 | 420 |

As always, a good place to start was to write out the possible ways each row’s product could be the product of three digits:

- 210 was 5×6×7.
- 144 was 2×8×9, 3×6×8, 4×4×9 or 4×6×6.
- 54 was 1×6×9, 2×3×9 or 3×3×6.
- 135 was 3×5×9.
- 4 was 1×2×2 or 1×1×4.
- 49 was 1×7×7.

Next, many solvers worked out the prime factorization of the three columns’ products, so they could figure out which digit went into which column.

- 6,615 was 3
^{3}×5×7^{2}. - 15,552 was 2
^{6}×3^{5}. - 420 was 2
^{2}×3×5×7.

Since the middle column’s product didn’t have a factor of 7, that meant the last row’s number had to be **717**. There was just one 7 left in the puzzle: in the first column and the first row. Also, since there were no factors of 5 in the second column, that meant the first row’s number was **765**. Now there was just one 5 left in the puzzle: in the first column and the fourth row. And because the last column had just one factor of 3 (i.e., not two factors of 3), the number in the fourth row was **593**.

Now that all the 5s and 7s were accounted for, all that remained were factors of 2 and 3. To make the numbers work out, the three remaining digits of the middle column somehow had to multiply to 288. The only two ways this was possible with three digits were 4×8×9 or 6×6×8. Since there was no way to get 6, 6 and 8 from the three remaining rows, those digits had to be 4, 8 and 9. Moreover, an 8 was only possible in the second row, meaning the 9 came in the third row and the 4 in the fifth row, which meant the number in the fifth row had to be **141**. Finally, because the first column’s product had only factors of 3 remaining and the third column had only factors of 2, the number in the second row was **982** and the number in the third row was **392**.

Here was an alternative animated approach, courtesy of solver Andrew Heairet:

Congratulations to 👏 Justin Ahmann 👏 of Bloomington, Indiana, 👏 Jenny Mitchell 👏 of Nashville, Tennessee and 👏 Peter Exterkate 👏 of Sydney, Australia, winners of last week’s Riddler Classic.

Last week, you looked at *self-intersecting* polygons, whose sides cross over each other. (These are distinct from the simple polygons you learned about in school, whose sides do not intersect each other.)

For example, it’s possible to draw a polygon for which every side intersects exactly two other sides (not counting the vertices, of course). The polygon with the fewest sides that still meets this criterion is the pentagram:

But was it possible to draw a polygon where each side intersected exactly *three* other sides? And if so, what was the minimum number of sides this polygon can have?

Despite the disbelief of several readers, it was indeed possible to draw such a polygon! The solutions and drawings you submitted were endlessly creative, replete with zig, zags and symmetries.

Many solvers found different 18-gons where each side intersected three other sides. These figures varied quite a bit, and included David Devore’s three concentric triangles, each of whose sides is crossed by inward and outward zags:

But it was possible to do even better! Jenny Mitchell was among this week’s winners for her submission of this 14-sided polygon:

Sure enough, 14 was the least number of sides anyone was able to find. Unless, of course, you look beyond Euclidean geometry, as Steve Curry did. Steve found that you could draw a 6-sided polygon on a sphere in which each side intersected three other sides, shown below. Meanwhile, Jenny found a 2-sided polygon (yes, a “bi-gon”) in a cylindrical geometry.

Finally, let’s return to two dimensions for a moment. Solvers Rohan Lewis and James Anderson both explored the minimum number of sides for polygons in which each side intersected exactly *k* other sides for different values of *k*. Starting from *k *= 1, it appeared that this sequence was 6, 5, 14 (this week’s solution), 7, 10, etc. At the time of this writing, no such sequence exists on OEIS. Knowing Riddler Nation, a new sequence will surely be authored there sooner rather than later.

And if this sort of riddle wasn’t for you, I hope you’ll let bi-gons be bi-gons.

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{2} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

It’s the second week in our four weeks of CrossProduct™ puzzles!

This time around, there are *six* three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.

210 | |||

144 | |||

54 | |||

135 | |||

4 | |||

49 | |||

6,615 | 15,552 | 420 |

Can you find all six three-digit numbers and complete the table?

The solution to this Riddler Express can be found in the following column.

From James Bach comes a doozy on doodles:

James likes to draw doodles in the shapes of different polygons. He especially likes to doodle *self-intersecting* polygons, where the sides cross over each other. (These are distinct from the simple polygons you might have learned about in school, whose sides do not intersect each other.)

The other day, James was able to draw a self-intersecting polygon, each of whose sides intersected with exactly two other sides. Not only that, he drew a polygon with the fewest possible sides that met these criteria.

What was it, you ask? It was a pentagram, which has just five sides.

Lovely. But this got James really thinking — can you draw a polygon where each side intersects exactly *three* other sides? And if so, what is the minimum number of sides this polygon can have?

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Greg 👏 of Brooklyn, New York, winner of last week’s Riddler Express.

Last week you had to solve a CrossProduct™ with five three-digit numbers. The products of the three digits of each number were shown in the rightmost column below, while the products of the digits in the hundreds, tens and ones places were shown in the bottom row.

135 | |||

45 | |||

64 | |||

280 | |||

70 | |||

3,000 | 3,969 | 640 |

Once again, a good way to start was to write out the possible ways each row’s product could be the product of three digits:

- 135 was 3×5×9.
- 45 was 1×5×9 or 3×3×5.
- 64 was 1×8×8, 2×4×8 or 4×4×4.
- 280 was 5×7×8.
- 70 was 2×5×7.

Next, many solvers worked out the prime factorization of the three columns’ products, so they could figure out which digit went into which column.

- 3,000 was 2
^{3}×3×5^{3}. - 3,969 was 3
^{4}×7^{2}. - 640 was 2
^{7}×5.

First off, the middle column had an odd product, which meant every middle digit had to be odd. This was only possible if the number in the third row was **818**.

Meanwhile, only the middle column had 7s in its prime factorization, which meant the middle digits in the fourth and fifth rows were both 7. As for the first column, since the 8 in the third row accounted for all three factors of 2, that meant the first digits in the fourth and fifth rows had to be odd as well. The number in the fourth row was therefore **578**, while the number in the fifth row was **572**.

In the middle column of the first and second rows, you still somehow needed to account for four factors of 3. The only way this was possible was if both of these middle digits were 9. To make the remaining products work, that meant the number in the first row had to be **395**, while the number in the second row was **591**.

There was certainly more than one way to solve this puzzle. For another approach, check out Andrew Heairet’s animated solution:

It looks like this will be the next Twitch sensation!

Congratulations to 👏 Stephen Berg 👏 of Troy, New York, winner of last week’s Riddler Classic.

Last week, Cassius the ape was trying to solve Lucas’ Tower puzzle with three disks, all of which start on the same pole. The disks had different diameters, with the biggest disk at the bottom and the smallest disk on top. The goal was to move all three disks from one pole to any other pole, one at a time — but at no point could a larger disk sit atop a smaller disk.

The *minimum* number of moves to solve the puzzle was not in question — that result is well known: *N* disks can be solved in exactly 2^{N}−1 moves.

It turned out that Cassius couldn’t care less about actually solving the puzzle, but he was very good at following directions and understood that a larger disk could never sit atop a smaller disk. With each move, he randomly chose from among the set of valid moves.

On average, how many moves would it take for Cassius to solve this puzzle with three disks?

Given the complexity of this problem, several solvers, like Kenneth W and Kei Nishimura-Gasparian, turned to their computers for assistance. After 1 million simulations, Kenneth found that it took Cassius approximately 70.7 moves on average to solve the puzzle.

Many solvers found an exact answer. One way to do this was to map out all 27 possible states of the disks and poles, as Dean Ballard did:

Based on Dean’s diagram, the question then became: If you started at A and randomly walked your way back and forth between adjacent states, what was the average number of steps it took to reach either Z1 or Z2?

“Random walk” problems like these are commonly solved using systems of equations or Markov chains, which was what solver Michael Ringel did:

Either way, you found that the average number of moves to go from A to Z1 or Z2 was **637/9**, or about 70.78 — very close to Kenneth’s numerical approximation. If you alternatively interpreted the problem as going from A to Z1 *and only* Z1 (or to Z2 and only Z2) — an answer I accepted — you had double the result, or **1274/9**, due to the bilateral symmetry in the state diagram.

For extra credit, you had to find the average number of moves in the general case of *N* disks and three poles. The challenge here was that the number of possible states scaled exponentially with *N*, which made this rather difficult.

Puzzle submitter Toby Berger, along with coauthor Max A. Alekseyev, solved this exact problem in a 2014 paper. It may have been difficult to see from Dean’s or Michael’s drawings, but there was a recursive nature to the diagram of states. Here’s an illustration from Toby’s paper that shows this nature a little more clearly when there were one, two and three disks:

By leveraging the recursive nature of these states, the authors were able to show the general formula for the average number of steps to randomly move the disks from one pole to another: (3^{N}−1)(5^{N}−3^{N})/(2·3^{N−1}). The average number of steps to randomly move the disks to *either* of the other poles was then half of that, or **(3 ^{N}−1)(5^{N}−3^{N})/(4·3^{N−1})**.

Finally, Eric Thompson-Martin further studied the shape of the probability distribution for the number of steps it took Cassius to solve the puzzle. According to Eric, this distribution appeared to be a log-normal curve as the number of disks increased.

In any case, I hope no one blew a (Sierpiński) gasket solving this!

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{3} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

By all accounts, Riddler Nation had a lot of fun hunting for the mysterious numbers a few weeks back. So here’s what we’re going to do: For the next four weeks, the Riddler Express will feature a similar puzzle that combines multiplication and logic. We’ll be calling these *CrossProducts*.

For your first weekly CrossProduct, there are five three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.

135 | |||

45 | |||

64 | |||

280 | |||

70 | |||

3,000 | 3,969 | 640 |

Can you find all five three-digit numbers and complete the table?

The solution to this Riddler Express can be found in the following column.

From Toby Berger comes a towering challenge:

Cassius the ape (a friend of Caesar’s) has gotten his hands on a Lucas’ Tower puzzle (also commonly referred to as the “Tower of Hanoi”). This particular puzzle consists of three poles and three disks, all of which start on the same pole. The three disks have different diameters — the biggest disk is at the bottom and the smallest disk is at the top. The goal is to move all three disks from one pole to any other pole, one at a time, but there’s a catch. At no point can a larger disk ever sit atop a smaller disk.

For *N* disks, the minimum number of moves is 2^{N}−1. (Spoiler alert! If you haven’t proven this before, give it a shot. It’s an excellent exercise in mathematical induction.)

But this week, the *minimum* number of moves is not in question. It turns out that Cassius couldn’t care less about solving the puzzle, but he is very good at following directions and understands a larger disk can never sit atop a smaller disk. With each move, he randomly chooses one among the set of valid moves.

On average, how many moves will it take for Cassius to solve this puzzle with three disks?

*Extra credit:* On average, how many moves will it take for Cassius to solve this puzzle in the general case of *N* disks?

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Edouard Duriez 👏 of Toulouse, France, winner of last week’s Riddler Express.

Last week, I had found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. I even had photo evidence:

If each of the four cubes had side length 1, then how far above the ground was the bottommost corner of the cube on top?

One approach was to look at the portion of the cube on top that dipped below the surface of the other three cubes. This shape was a tetrahedron, whose base (on top) was an equilateral triangle with side length 1. (These sorts of connections between cubes and equilateral triangles are nothing new to The Riddler.) The remaining three faces of the tetrahedron were all congruent isosceles right triangles whose hypotenuse had length 1.

If only you could figure out the height of this tetrahedron, then you could subtract it from 1 to determine how high the bottommost corner was from the ground. The height — along with one of the legs of the isosceles right triangles and a line connecting the equilateral base’s centroid with one of its corners — formed yet another right triangle.

Putting the Pythagorean theorem into action, you found that the tetrahedron’s height was 1/√6. And that meant the bottommost corner was a height of **1−1/√6**, or about 0.592, off the ground.

Meanwhile, solver Thomas Stone picked out a different right triangle within the tetrahedron to calculate its height, ultimately arriving at the same answer:

Blocks are one thing. But if any architects are reading this, please make a building with this geometric arrangement of four cubes. Most of the math has already been done for you!

Congratulations to 👏 Mark Bradwin 👏 of Seattle, Washington and 👏 Nic Tamburello 👏of Gaithersburg, Maryland, winners of last week’s Riddler Classic.

Last week, you were a contestant on the game show Lingo, where your objective was to determine a five-letter mystery word. You were told this word’s first letter, after which you had five attempts to guess the word. You were allowed to guess any five-letter word, even one that had a different first letter.

After each of your guesses, you were told which letters of your guess were also in the mystery word and whether any of the letters were in the correct position. In the example below, T was in the correct position (remember, the first letter was provided to you), while A and C were in the mystery word but not in the correct positions.

For this example, here was how you might have figured out the mystery word (TACOS) using all five guesses:

The mystery word and guesses could have also contained multiple instances of a letter. For example, the mystery word MISOS contains one O, so a guess with more than one O (like MOSSO) would only have had the first O marked as correct (but in this case, in the wrong position).

As a contestant, your plan was to make a mockery of the game show by adopting a bold strategy: No matter what, before you were even told what the first letter of the mystery word was, you had decided what your first four guesses would be. Then, with your fifth guess, you would use the results of your first four guesses (and your encyclopedic knowledge of five-letter words!) to determine all remaining possibilities for the mystery word. If multiple mystery words were still possible, you would pick one of these at random. You had to assume that the mystery word was selected randomly from this word list, which was also the list your guesses had to be chosen from.

Which four five-letter words would you have chosen to maximize your chances of victory?

Coming up with any old five-letter words wasn’t particularly challenging. But to pick four *good* words that also worked well together, you pretty much had to tackle last week’s extra credit, which asked you to determine your chances of victory given your choice of four words.

This turned out to be a coding exercise. The most challenging part was getting all those edge cases right, particularly when there were repeated letters in one of your guesses or in the mystery word itself. A neat result that came out of all this was that if there were *N* five-letter words in the dictionary (here, *N* happened to be 8,636) and there were *K* distinct states that the Lingo board took on in terms of correct letters and the given first letter, your chances of guessing the correct word were exactly *K*/*N*.

In any case, the puzzle’s creator, Vince Vatter, and myself independently wrote code to score sequences of five-letter words and verified that our code always returned the same results. (Translation: Vince helped me debug my code.) So if you think you were scored in error, it’s far more likely that you were in error than both Vince and myself. (Translation: Vince didn’t screw up, you did.)

Prior to submitting the puzzle, Vince himself found a sequence of words (WIDTH, BARES, CLOMP, GUNKY) that won a whopping 95.79 percent of the time. But this week’s winners did even better! Without further ado, here were all the submitters who did better than 95 percent:

Top performers at Riddler Lingo by their win percentage, plus their winning five-letter words

Rank | Name | Words | Win % |
---|---|---|---|

1 | Mark Bradwin | BINTS, CLOAK, GYRED, WHUMP | 95.96 |

1 | Nic Tamburello | BINTS, GYRED, WHUMP, CLOAK | 95.96 |

3 | Jenny Mitchell | BLINK, CHOMP, GUDES, WARTY | 95.90 |

4 | James Anderson | BARES, GUNKY, WIDTH, CLOMP | 95.79 |

5 | Q P Liu | CARES, BUMPY, KLONG, WIDTH | 95.74 |

5 | Nikhil Mahajan | CARES, KLONG, BUMPY, WIDTH | 95.74 |

7 | Michael Engen | CLIPT, GYBED, KHOUM, WARNS | 95.48 |

8 | David Devore | CHOMP, FURAN, GYBED, KILTS | 95.40 |

9 | Dallas Trinkle | BLOTS, CAGER, DINKY, WHUMP | 95.39 |

10 | James Bach | FUMED, BLOCS, PINKY, GARTH | 95.08 |

As part of his strategy, Mark only considered word combinations with 20 unique letters — a trend you can see up and down the leaderboard. Nic, meanwhile, automated his search by using simulated annealing.

Finally, since writing a scoring algorithm was the majority of the work required for this puzzle, I wanted to give a special shoutout to all submitters who correctly scored their own submission in the extra credit. In addition to most of those on the leaderboard, this list included: David Ding, Mike Onigman, Laurynas Navidauskas, Bryce Wargin, Peter Ji and Paulina Leperi.

Even without knowing the first letter in advance, Riddler Nation has truly put Lingo’s contestants to shame. Well done, all!

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{4} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter

If you have young children (or if you’re still a child at heart), you probably have small blocks somewhere in your home.

I recently found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. Here’s a photo I took of this arrangement:

If you too have blocks at home (I mean, of course you do), see if you can make the same arrangement.

Now, if each of the four cubes has side length 1, then how far above the ground is the bottommost corner of the cube on top?

The solution to this Riddler Express can be found in the following column.

From Vince Vatter, one of the two reigning Wordsmiths Extraordinaire of Riddler Nation, comes a puzzle that is sure to please anyone who fondly remembers that time Spelling Bee appeared in this column:

You are a contestant on the game show Lingo, where your objective is to determine a five-letter mystery word. You are told this word’s first letter, after which you have five attempts to guess the word. You can guess any five-letter word, even one that has a different first letter.

After each of your guesses, you are told which letters of your guess are also in the mystery word and whether any of the letters are in the correct position. In the example below, T is in the correct position (remember, the first letter is provided to you), while A and C are in the mystery word but not in the correct positions.

For this example, here’s how you might have figured out the mystery word (TACOS) using all five guesses:

The mystery word and guesses can contain multiple instances of a letter. For example, the mystery word MISOS contains one O, so a guess with more than one O (like MOSSO) will only have the first O marked as correct (but in this case, in the wrong position).

As a contestant, your plan is to make a mockery of the game show by adopting a bold strategy: No matter what, before you are even told what the first letter of the mystery word is, you have decided what your first four guesses will be. Then, with your fifth guess, you will use the results of your first four guesses (and your encyclopedic knowledge of five-letter words!) to determine all remaining possibilities for the mystery word. If multiple mystery words are still possible, you will pick one of these at random.

Which four five-letter words would you choose to maximize your chances of victory? Assume that the mystery word is selected randomly from this word list, which is also the list your guesses must be chosen from.

*Extra credit:* For the four five-letter words you chose, what are your chances of victory?

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Martin Miller 👏 of Mill Valley, California, winner of last week’s Riddler Express.

Last week, you reviewed some survey data that was randomly collected from the residents of Riddler City, which had a very large population.

Ten randomly selected residents were asked how many people (including them) lived in their household. As it so happened, their answers were 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

Your job was to use this (admittedly limited) data to estimate the average household size in Riddler City. Your co-worker suggested averaging the 10 numbers, giving an answer of about 5.5 people. But you weren’t so sure.

Would your best estimate have been exactly 5.5, less than 5.5 or greater than 5.5?

On its face, it seemed like the answer should have been exactly 5.5 — the average of the 10 numbers from 1 through 10.

But Riddler Nation was not fooled. Among the 836 submissions I received before the midnight deadline on Monday, only 8.6 percent thought the best estimate was exactly 5.5, while another 19.5 percent thought the best estimate was greater than 5.5. Meanwhile, the overwhelming majority (71.9 percent) thought the best estimate was less than 5.5.

These solvers read the puzzle carefully, noting that it was *residents* — rather than *households* — that were randomly selected for the survey. If households had been randomly selected instead, then yes, your best estimate would have been 5.5 people per household. But because it was residents that were selected, solvers like Penelope Ackerman realized that a household with 10 people was 10 times more likely to have been chosen than a household with just one person. In other words, the sampling was *biased* toward larger households. For a household with one person and a household with 10 people to be selected with equal likelihood, there had to have been 10 times as many households with one person.

As it turned out, applying this reasoning to all 10 household sizes meant 2,520/7,381 of the households in Riddler City had one person, 1,260/7,381 had two people, 840/7,381 had three people, and so on, with 252/7,381 of the households having 10 people. The denominator was 7,381 because that was the sum of the smallest set of integers satisfying the ratios in the problem. These ratios were nicely illustrated by solver Reece J. Goiffon:

All this meant the average household size in Riddler City — calculated as 2,520/7,381·1 + 1,260/7,381·2 + 840/7,381·3 + … + 252/7,381·10 — was about 3.4 people. That is, **less than 5.5**.

Any readers who also happen to be pollsters in Riddler City should take notice. You have a pollster rating to maintain!

Congratulations to 👏 Ali Farhat 👏 of Dearborn, Michigan, winner of last week’s Riddler Classic.

Last week, you were competing in the finals of the Riddler Ski Federation’s winter championship! There was just one opponent left to beat, and then the gold medal would be yours.

Both of you were completing *two* runs down the mountain, and the times of your runs would be added together. Whoever skied in the least overall time would be the winner. Also, you knew that you and your opponent were evenly matched, and you both had the same normal probability distribution of finishing times for each run. And each skiing run was independent of all the others.

For the first run, your opponent went first. Then, as you crossed the finish line on your own first run, your coach excitedly signaled to you that you were faster than your opponent. Without knowing either exact time, what was the probability that you would still be ahead after the second run and earn your gold medal?

To no one’s surprise, many solvers simulated the skiing championship. Paulina Leperi and Harold Doran both wrote some code that simulated hundreds of thousands of pairs of skiing runs for both you and your opponent, finding that you emerged the winner just about 75 percent of the time. Could the answer have been 75 percent?

A few solvers, like Stergios Athanasoglou and David Ding, tackled the puzzle head on, working through the mathematics (and calculus) of the normal distribution.

But in a cruel (or clever?) twist of mathematics, it turned out that probability distribution of finishing times was *irrelevant* for this problem. As long as it was the same for both you and your opponent, it didn’t matter if this distribution was normal, uniform or even a Laplace distribution.

As John from Washington, D.C. observed, there was a 50 percent chance that you would be faster than your opponent on the second run, in which case you were guaranteed to be the overall winner. As for the other 50 percent of the time, when your opponent had the faster second run, it all came down to whether the time gap was greater in the first run or the second run — two cases that were equally likely, thanks to the symmetry in the problem. The 25 percent of the time the gap was greater in the first run, you won, while you lost the other 25 percent of the time. Putting it all together, your chance of victory was 50 percent plus 25 percent, or exactly **75 percent**.

While it turned out that the normal probability distribution didn’t matter for the puzzle, the same could not be said for the extra credit, where you were asked to repeat the exercise in the case of 30 snowboarders (including you). Again, you were the last to complete the first run, and your coach signaled that you were in the lead at that point. What was the probability that you would win gold in snowboarding?

Even by Riddler standards, this problem was surprisingly *hard* to calculate. Forget 30 snowboarders with a normal probability distribution — even just three snowboarders with a uniform probability distribution was a challenge, requiring some hefty calculus and order statistics.

Everyone who solved this did so via computation, finding that you had **between a 31.4 and 31.5 percent** chance of winning the snowboarding championship. Josh Silverman took it one step further, looking at how your chances of winning depended on the number of competitors *N*. He found that this probability (the lower blue points in the graph below) appeared to be inversely proportional to the cube root of *N* (modeled by the higher orange points).

Of course, an even more general version of this puzzle was if there were *N* competitors and *R* runs in the competition, rather than just two runs. But I’ll save that riddle for our computer overlords to solve.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{5} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Ernie Cohen comes a scintillating stumper of a survey:

You’re reviewing some of the survey data that was randomly collected from the residents of Riddler City. As you’ll recall, the city is quite large.

Ten randomly selected residents were asked how many people (including them) lived in their household. As it so happened, their answers were 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

It’s your job to use this (admittedly limited) data to estimate the average household size in Riddler City. Your co-worker suggests averaging the 10 numbers, which would give you an answer of about 5.5 people. But you’re not so sure.

Would your best estimate be exactly 5.5, less than 5.5 or greater than 5.5?

The solution to this Riddler Express can be found in the following column.

Congratulations, you’ve made it to the finals of the Riddler Ski Federation’s winter championship! There’s just one opponent left to beat, and then the gold medal will be yours.

Each of you will complete *two* runs down the mountain, and the times of your runs will be added together. Whoever skis in the least overall time is the winner. Also, this being the *Riddler* Ski Federation, you have been presented detailed data on both you and your opponent. You are evenly matched, and both have the same normal probability distribution of finishing times for each run. And for both of you, your time on the first run is completely independent of your time on the second run.

For the first runs, your opponent goes first. Then, it’s your turn. As you cross the finish line, your coach excitedly signals to you that you were faster than your opponent. Without knowing either exact time, what’s the probability that you will still be ahead after the second run and earn your gold medal?

*Extra credit:* Over in the snowboarding championship, there are 30 finalists, including you (apparently, you’re a dual-sport threat!). Again, you are the last one to complete the first run, and your coach signals that you are in the lead. What is the probability that you’ll win gold in snowboarding?

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Daniel Baker 👏 of Salt Lake City, Utah, winner of last week’s Riddler Express.

Last week, you were a contestant on the hit new game show, “You Bet Your Fife.” On the show, a random real number was chosen between 0 and 100. Your job was to guess a value that was *less than* this randomly chosen number. Your reward for winning was a novelty fife that was valued precisely at your guess. For example, if the number was 75 and you guessed 5, you would have won a $5 fife, but if you guessed 60, you would have won a $60 fife. Meanwhile, a guess of 80 would have won you nothing.

What number should you have guessed to maximize the average value of your fifing winnings?

If you guessed a smaller number, you were more likely to win a fife. But if you guessed a higher number — and you happened to win — you got a *more valuable* fife. With these competing priorities, some solvers reasoned that the answer should be right in the middle, with a guess of 50. But how could we show this mathematically?

One way was to determine the average winnings as a function of the number you guessed, which we’ll call *x*. This average was the product of your winnings (which would be *x*) and your probability of winning, which was (100−*x*)/100. In other words, this probability started at 100 percent when you guessed a value of zero, and linearly decreased to zero percent when you guessed a value of 100.

Putting these expressions together, a guess of *x* returned an average value of *x*(100−*x*)/100. To maximize your average winnings, you had to find the maximum of this quadratic function, which you could do with calculus (setting its derivative equal to zero), graphing or reasoning with symmetry. Any which way you did it, this maximum occurred when *x* was **50**.

Plugging in a value of 50 for *x* in the expression *x*(100−*x*)/100 meant that your average winnings were $25. Not too shabby. But if you didn’t win anything on “You Bet Your Fife,” you can always bid on a $100,000 flute.

Congratulations to 👏 Reuven 👏 of Ottawa, Ontario, Canada, winner of last week’s Riddler Classic.

Last week, the mysterious Barbara Yew offered a mysterious number puzzle, where you had to find eight three-digit numbers. The products of the three digits of each number were shown in the rightmost column of the table below. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, were shown in the bottom row.

First off, I’m delighted to report that this riddle was part of the annual (and this year, virtual!) MIT Mystery Hunt, puzzles from which have previously appeared in this column. The puzzle’s submitter, “Barbara Yew,” was a fictional character within the hunt, which was created and organized by last year’s winning team, the ✈✈✈ Galactic Trendsetters ✈✈✈. Indeed, this week’s winner, Reuven, participated in the hunt as part of team Control Group.

Your first step was to list out the different ways to multiply three digits from 1 to 9 to achieve each row’s product:

- 294 was 6×7×7.
- 216 was 3×8×9, 4×6×9 or 6×6×6.
- 135 was 3×5×9.
- 98 was 2×7×7.
- 112 was 2×7×8 or 4×4×7.
- 84 was 2×6×7 or 3×4×7.
- 245 was 5×7×7.
- 40 was 1×5×8 or 2×4×5.

For a few solvers, that last row — 40 — was tricky. You had to realize that 1 was an allowable digit, even though it isn’t a prime number.

From here, there were a total of 2,729,376 unique ways to place these digits in the table. That wasn’t a prohibitively large number, and so some solvers, like Siddhartha Srivastava of Patna, India, powered forth with computational brute force to find the unique solution. But this puzzle was indeed solvable with pen (okay, pencil), paper and logic.

Your next step was to find the prime factorization of the products of each column:

- 8,890,560 was 2
^{6}×3^{4}×5^{1}×7^{3}. - 156,800 was 2
^{7}×5^{2}×7^{2}. - 55,566 was 2
^{1}×3^{4}×7^{3}.

Solver Sara McArdle took a closer look at the first column, whose prime factorization implied that four of the eight digits had to be 5, 7, 7 and 7. That meant the remaining factorization, 2^{6}×3^{4}, or 5,184, had to be the product of the remaining four digits. There was only one way to break down 5,184 in this way: 8, 8, 9 and 9. The two factors of 9 had to come from the second (216) and third (135) rows. Also, because the third column did not have a factor of 5, that meant the three-digit number in the third row had to be 953.

Because all of the first column’s factors of 3 were accounted for, that meant the first row’s 6 couldn’t be in the first column. Also, the second column did not include a factor of 3, which meant the last column had the 6. In other words, the three-digit number in the first row was 776.

That 6 from the first row was responsible for the third column’s single factor of 2. Since the first digit of the second row was 9, the only way that row’s third digit could be odd was for the three-digit number in the second row to be 983.

But what about those two 8s in the first column? At this point, they had to be in the fifth and eighth rows. Because the last column had no factors of 5, the three-digit number in the eighth row had to be 851. And because the third column wasn’t allowed any more factors of 2, the three-digit number in the fifth row had to be 827.

The first column still needs a factor of 5, and the only remaining place it can get it is from the seventh row, whose three-digit number had to be 577. Meanwhile, the first column had all its factors of 2 accounted for, so the three-digit number of the fourth row had to be 727. Last but not least, the three-digit number of the sixth row was 743.

When all was said and done, here was the completed table:

Finally, a big congratulations to the winning team of this year’s incredible MIT Mystery Hunt, Palindrome!

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>For the final FiveThirtyEight Politics podcast episode of the Trump presidency, the crew is joined by ABC News White House Correspondent Karen Travers to discuss Trump’s legacy, how he changed politics and what the lasting effects will be. To help, they look back at an article editor-in-chief Nate Silver wrote shortly after Trump was inaugurated, “14 Versions Of Trump’s Presidency, From #MAGA To Impeachment,” to see what came true (and what Nate didn’t predict).

*You can listen to the episode by clicking the “play” button in the audio player above or by **downloading it in iTunes**, the **ESPN App** or your favorite podcast platform. If you are new to podcasts, **learn how to listen**.*

*The FiveThirtyEight Politics podcast is recorded Mondays and Thursdays. Help new listeners discover the show by **leaving us a rating and review on iTunes**. Have a comment, question or suggestion for “good polling vs. bad polling”? Get in touch by email, **on Twitter** or in the comments.*

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{6} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Lucas Jaeger comes a “flute-iful” challenge:

You’re a contestant on the hit new game show, “You Bet Your Fife.” On the show, a random real number (i.e., decimals are allowed) is chosen between 0 and 100. Your job is to guess a value that is *less than* this randomly chosen number. Your reward for winning is a novelty fife that is valued precisely at your guess. For example, if the number is 75 and you guess 5, you’d win a $5 fife, but if you’d guessed 60, you’d win a $60 fife. Meanwhile, a guess of 80 would win you nothing.

What number should you guess to maximize the average value of your fifing winnings?

The solution to this Riddler Express can be found in the following column.

In a world of sudokus, KenKens and kakuros, Barbara Yew offers a different sort of number puzzle:

There are eight three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens, and ones places, respectively, are shown in the bottom row.

Can you find all eight three-digit numbers and complete the table? It’s a bit of a mystery, but I’m sure you have it within you to hunt down the answer!

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Jacob Kopczynski 👏 of San Francisco, California, winner of last week’s Riddler Express.

Last week, you were slicing one big square into smaller squares (not necessarily of equal size), so that the smaller squares didn’t overlap, while still making up the entire area of the big square.

What whole numbers of squares could you *not* have sliced the big square into?

Many solvers began by trying out small numbers of squares. Of course, it was possible to slice the big square into one square by simply leaving it alone. The next smallest number of squares you could get was 4, each a quarter of the big square. That meant getting either two squares or three squares was impossible. (A rigorous proof of this is left as an exercise for you, the reader — hah!)

Another possible number was 9, since that would be a 3-by-3 array of equally sized squares. Similarly, any square number was possible. But what about numbers *between* the perfect squares?

As noted by solver America Masaros, if it was possible to create N squares, it was also possible to create *N*+3 squares by taking any undivided square and slicing it into four equally sized squares. This effectively replaced one square with four new squares — a net gain of three squares. That meant any number that was 4 or 9 plus a multiple of 3 (i.e., of the form 4+3*k* or 9+3*k*, where *k* is a whole number) was also possible, ruling out a whole bunch of numbers.

But there were still a few numbers less than 9 that had to be checked: 5, 6 and 8 (7 was 3 more than 4, so it was possible). While 5 was not possible (again, you are welcome to prove this!), both 6 and 8 *were* possible using unequally sized squares, as shown below:

Indeed, this strategy with one larger square and an elbow of smaller squares around it worked for any even number greater than 2.

Finally, since 8 was possible, any number that was a multiple of 3 greater than 8 was also possible, again by picking one of the squares and slicing it into four equal squares. That meant the only numbers that were *not* possible were **2, 3 and 5**.

Meanwhile, Vince Vatter (of Battle for Riddler Nation fame) recognized this problem from one of his colleague’s books:^{7}

It remains unclear why a value of 14 was chosen for this exercise, rather than the actual lower bound of 5.

At least one solver extended the puzzle further, looking at how a cube could be partitioned into smaller cubes. But let’s save that for a future riddle!

Congratulations to 👏 Eilon 👏 of Chicago, Illinois, winner of last week’s Riddler Classic.

Last week, Robin of Foxley entered the FiveThirtyEight archery tournament. She was guaranteed to hit the circular target, which had no subdivisions — it was just one big circle. However, her arrows were equally likely to hit each location within the target.

Her true love, Marian, had issued a challenge. Robin had to fire as many arrows as she could, such that each arrow was closer to the center of the target than the previous arrow. For example, if Robin fired three arrows, each closer to the center than the previous, but the fourth arrow was farther than the third, then she was done with the challenge and her score was *4*.

On average, what score could Robin have expected to achieve in this archery challenge?

First off, a surprising and subtle fact about this riddle was that the geometry (in this case, a circle) didn’t matter. What did matter was how close each arrow was to the center *relative to the other arrows*. That meant the target could have been a square, a line segment or even a sphere — the answer would be the same.

A good first step was then to restate the problem and forget about the geometry: If you pick random values between 0 and 1 uniformly — each representing the relative distance of an arrow to the center — how many consecutive decreasing values would you expect (plus one, for the arrow that broke the streak)?

Solver Balthazar Potet approached this by thinking about the values for the first *N* arrows Robin fired and the probability they’d result in a score of *N*. With any *N* values, there were *N*! ways to order them. For Robin to have a score of *N*, the smallest value couldn’t have been in the *N*th position, since it had to be greater than the previous value. And when each of the other *N*−1 values occurred in the *N*th position, there was exactly one way to order the remaining values so that they formed a decreasing sequence. So of the *N*! orderings, *N*−1 resulted in a score of *N*, meaning the probability was (*N*−1)/*N*!

From there, you had to use these probabilities to compute an average score, which you could find by multiplying each score by its probability and then adding up all those products. The probability Robin scored 2 was (2−1)/2!, or 1/2, which meant a score of 2 contributed 2·1/2, or 1, to her average score. The probability Robin scored 3 was (3−1)/3!, or 1/3, which meant a score of 3 contributed 3·1/3, or 1 (again!), to her average score. In general, the probability Robin scored *N* was (*N*−1)/*N*!, which meant a score of *N* contributed *N*·(*N*−1)/*N*!, or 1/(*N*−2)!, to her average score. Since *N* was at least 2 — meaning Robin fired at least two arrows — her average score was 1/0! + 1/1! + 1/2! + 1/3! + …, a sum that converges to ** e**, which is

For extra credit, you had to calculate Robin’s average score when the target had 10 concentric circles, whose radii were 1, 2, 3, etc., all the way up to 10 (the radius of the entire target). This time, Robin had to fire as many arrows as she could, such that each arrow fell within a smaller concentric circle than the previous arrow.

Here, the geometry of the target was relevant, since there was now a nonzero probability that consecutive arrows could fall within the same ring. The chances that any given arrow landed in one of the rings (from the smallest ring to the largest) were 1 percent, 3 percent, 5 percent, 7 percent, 9 percent, 11 percent, 13 percent, 15 percent, 17 percent and 19 percent.

Having the discrete rings (perhaps counterintuitively) made the problem more complex, but several solvers persisted. Emma Knight was able to set up and solve a system of 10 equations, finding that the average number of arrows was **approximately 2.5585**. Josh Silverman was further able to come up with a closed formula for the solution and a precise rational result.

Meanwhile, solvers like Paulina Leperi and Angelos Tzelepis (whose results are shown below) approximated the answer by simulating many arrows.

That was the answer when there were 10 rings. As the number of rings increased, the average score also increased, approaching *e* in the limit of infinitely many rings.

By now, I bet you’re curious how Robin of Foxley actually performed at Marian’s archery challenge. Robin was such a good archer that she was able to fire off two arrows that were *exactly* the same distance from the target’s center. Marian’s mind was blown by this probability-zero event, and they lived happily ever after.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{8} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Rob Peacock comes a matter of squaring a multitude of squares:

There are many ways to slice a big square into smaller squares (not necessarily of equal size), so that the smaller squares don’t overlap, while still making up the entire area of the big square.

For example, you can slice the big square into four smaller squares, each a quarter of the area of the big square. Or you could slice it into seven squares, if you take one of those four squares and slice it into four yet smaller squares.

What whole numbers of squares can you *not* slice the big square into?

The solution to this Riddler Express can be found in the following column.

Robin of Foxley has entered the FiveThirtyEight archery tournament. Her aim is excellent (relatively speaking), as she is guaranteed to hit the circular target, which has no subdivisions — it’s just one big circle. However, her arrows are equally likely to hit each location within the target.

Her true love, Marian, has issued a challenge. Robin must fire as many arrows as she can, such that each arrow is closer to the center of the target than the previous arrow. For example, if Robin fires three arrows, each closer to the center than the previous, but the fourth arrow is farther than the third, then she is done with the challenge and her score is *four*.

On average, what score can Robin expect to achieve in this archery challenge?

*Extra credit:* Marian now uses a target with 10 concentric circles, whose radii are 1, 2, 3, etc., all the way up to 10 — the radius of the entire target. This time, Robin must fire as many arrows as she can, such that each arrow falls within a smaller concentric circle than the previous arrow. On average, what score can Robin expect to achieve in *this* version of the archery challenge?

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Eric Thompson-Martin 👏 of Amherst, Massachusetts, winner of the previous Riddler Express.

Last time, I was playing around in my kitchen with a tall glass and a smaller disk, when, to my surprise, I was able to balance the disk neatly atop the rim of the glass:

Suppose you had a disk of radius *R* and a glass with a circular rim of radius 2*R*. If you randomly placed the disk so that its center lay within the glass’s rim, what was the probability that the disk would balance atop the glass? (Assume the distribution was uniformly spread across the circular area inside the rim.)

At first, this might have sounded like a physics question. However, the physics gave way to mathematics rather quickly. For the disk to balance atop the rim, the disk’s center (i.e., its center of mass) had to lie between the arc of the glass’s rim and the straight line connecting the endpoints. Otherwise, the disk would either fall into or out of the glass. This region was illustrated by Tom Epp, for the case when the disk was on the right edge of the rim:

But the disk didn’t have to lie on the right edge of the rim — it could have been anywhere around the circumference. That meant the region in which the center would balance was a ring, or annulus. Solver Alex Vornsand approximated the size of this region by picking random points in the circle and seeing how many would have resulted in a balanced disk.

Of the 200 random points, exactly 50 (i.e., one-quarter of them — a nice round number) were in the annulus.

At this point, solvers like Thanh Nguyen used trigonometry to calculate the inner radius of the annulus, which turned out to be *R*√3. Thar meant the area of the ring was the area of the entire circle — 𝜋(2*R*)^{2}, or 4𝜋*R*^{2} — minus the area of the inner circle — 𝜋(*R*√3)^{2}, or 3𝜋*R*^{2}. This difference was simply 𝜋*R*^{2}.

Finally, the question was asking for the *probability* that picking a random point would result in a balanced disk. That meant you had to find the ratio of the area of the annulus (where the disk would balance) to the area of the entire rim. This was 𝜋*R*^{2}/(4𝜋*R*^{2}), or **one-quarter**.

When it’s safe to return to bars and restaurants, I’m sure many readers will apply what they’ve learned here by balancing as many coins as they can on the rim of a pint glass. Your friends will be impressed! (And if they’re not, get better friends.)

Congratulations to 👏 Peter Flynn 👏 of Thunder Bay, Ontario, Canada, winner of the previous Riddler Classic.

Last time, you were introduced to the two-player Game of Attrition, where each player started with a whole number of “power points.” Players took turns “attacking” each other, which involved subtracting their own number of power points from their opponent’s until one of the players was out of points.

For example, suppose Player A (who went first) started with 5 points, and Player B started with 7 points. After A’s first attack, A still had 5 points, while B had been reduced to 2 points (i.e., 7 minus 5). Now it was B’s turn, who reduced A to 5 minus 2, or 3 points. Finally, on A’s second turn, B was reduced from 2 points to nothing (since 2 minus 3 is −1). Despite starting with fewer points, A won!

Now suppose A went first and started with *N* points. In terms of *N*, what was the greatest number of points B could start with so that A still emerged victorious?

First off, if B started with *N* or fewer points, then A would win with a single attack. So you immediately knew that the answer was greater than *N*. But beyond this simplest case, the math got hairier.

One way to gain some intuition for what was happening was to make a plot of who won for each combination of points. The following graph shows the results when each player started with anywhere from 1 to 100 points:

The black region on the top left shows where A lost, while the colorful bands indicate where A won. The different colors show how many turns it took for A to win.

As we said before, when A started with at least as many points as B (the blue triangle in the bottom right), then A won in a single turn. Looking closely at the graph, you can see that A won in two turns when B started with at most 3/2·*N*, in three turns when B started with at most 8/5·*N*, in four turns when B started with at most 21/13·*N* and in five turns when B started with at most 55/34·*N*. If these fractions look familiar, that’s because they contain consecutive Fibonacci numbers!

There are countless patterns contained within the Fibonacci sequence, one of which is that the ratio of consecutive numbers approaches the golden ratio, also written as 𝜑, which equals (1+√5)/2, or approximately 1.618. If we kept going, we’d find that A won whenever B started with at most 𝜑·*N* points. Technically, since B had to have a whole number of points, the solution was **⌊𝜑· N⌋** — that is, the “floor” of 𝜑·

Another way to arrive as this solution was to look for edge cases, where A and B were headed for a tie — that is, where the ratio of A’s points to B’s points didn’t fluctuate from one round to the next. Suppose B started with *k*·*N* points, where *k* is a constant we’ll be solving for in a minute. After A’s first turn, B had *k*·*N*−*N*, or (*k*−1)·*N* points. Then, after B’s turn, A had *N*−(*k*−1)·*N*, or (2−*k*)·*N* points. For the ratio to remain the same, we need *k*·*N*/*N* to equal (*k*−1)·*N*/((2−*k*)·*N*). Some rearrangement and simplification (like canceling the *N*’s) led to the quadratic equation *k*2−*k*−1=0, whose positive solution was — wait for it — 𝜑. Once again, as long as B started with at most **⌊𝜑· N⌋** points, A was destined to win.

So before you play the Game of Attrition, make sure you know a thing or two (or three or five) about the Fibonacci sequence.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Over the course of 2020, FiveThirtyEight’s visual journalists covered a historic election, an unprecedented year in sports, a raging pandemic and an economy in free fall. So to cap off this long, strange, difficult year, we’re continuing our tradition of celebrating the best — and weirdest — charts we’ve published in the last 12 months. Charts are grouped by topic, but they’re not listed in any particular order beyond that. Click any of them to read the story where they originally ran. Enjoy!

Did you enjoy this long list of weird charts? Then boy do we have content in the archives for you! Check out our lists from 2019, 2018, 2016, 2015 and 2014.

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{9} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Due to the holidays, the next column will appear on Jan. 8, 2021. See you in the new year!

The other day I was playing around in my kitchen with a tall glass and a smaller disk, when, to my surprise, I was able to balance the disk neatly atop the rim of the glass:

First of all, try this at home!

Now suppose you have a disk of radius *R* and a glass with a circular rim of radius 2*R*. If you randomly place the disk so that its center lies within the glass’s rim, what is the probability that the disk will balance atop the glass? (Assume the distribution is uniformly spread across the circular area inside the rim.)

From Keith Wynroe comes an epic power battle that is sure to wear you down:

The Game of Attrition has two players, each of whom starts with a whole number of “power points.” Players take turns “attacking” each other, which involves subtracting their own number of power points from their opponent’s until one of the players is out of points.

For example, suppose Player A (who goes first) starts with 5 points and Player B starts with 7 points. After A’s first attack, A still has 5 points, while B has been reduced to 2 points (i.e., 7 minus 5). Now it’s B’s turn, who reduces A to 5 minus 2, or 3 points. Finally, on A’s second turn, B is reduced from 2 points to nothing (since 2 minus 3 is −1). Despite starting with fewer points, A wins!

Now suppose A goes first and starts with *N* points. In terms of *N*, what is the greatest number of points B can start with so that A will still emerge victorious?

Congratulations to 👏 Andrew Young 👏 of Sarasota, Florida, winner of last week’s Riddler Express.

Tonight is the last night of Hanukkah, which means millions of Jews have been lighting their menorahs over the last eight nights. It is traditional to start with one candle in the rightmost position, which is lit using a central candle called a shamash. On the second night, a second candle is added to the left of the first. On the third night, a third candle is added to the left of the second, and so on, until all eight candles have been added on the final night.

With this system for adding candles, whether you were looking at the menorah from the front (i.e., in your home) or back (i.e., through a window), you could easily determine which of the eight nights of Hanukkah it was by counting the number of candles (other than the shamash). But what if you wanted to make that same menorah capable of tracking *more* than eight nights?

To do that, you had to devise a new system of adding or moving candles (other than the shamash in the middle, which is always lit) around the menorah so that it could uniquely indicate as many nights as possible. Your system had to read the same forward and backward, so that regardless of whether you were looking at the menorah from the front or the back you’d still know what night it was. What was the greatest number of nights that could be indicated using your system?

Without resorting to methods *not* stated in the problem (like playing with the heights or colors of the candles), your best bet was to use a binary code. Lit candles represented a one, and unlit positions on the menorah were zeros. With this strategy, the menorah could theoretically indicate up to 256 nights. However, since some of these 256 states read the same forward and backward — an ambiguity that was specifically to be avoided — the answer had to be less than 256.

Realizing that a simple binary code was insufficient, some solvers turned to ternary code. Alex Vornsand took this approach, looking at pairs of candles on either side of the central shamash — either neither candle was lit, one candle was lit, or both candles were lit. With four pairs of candles, this meant there were 3^{4}, or 81 distinct states, no two of which read the same forward and backward:

But it was possible to do even better! As solver Francesca Maroney noted, there were 16 symmetric configurations that read the same forward and backward — there were 2^{4} different ways to light the four candles on one side of the shamash, and for each of those there was just one way to light the other four to make the menorah symmetric.

Since 16 of the 256 states were symmetric, that meant the remaining 240 were asymmetric — they read differently forward and backward. These could be paired up as reflections about the shamash, so that you could see either of the two states depending on which side of the menorah you were on. In the end, there were 120 paired states and 16 symmetric states, for a total of **136** distinguishable states — the maximum number of nights your menorah could track.

Sign me up for 136 nights of Hanukkah — and the 136 presents that go with it!

Congratulations to 👏 Jack Stade 👏 of Lafayette, Colorado, winner of last week’s Riddler Classic.

Last week, the Potentate of Puzzles challenged five lucky citizens with a test. The potentate had countless red hats, green hats and blue hats. She said to the citizens, “Tomorrow, you will all be blindfolded as I place one of these hats on each of your heads. Once all the hats are placed, your blindfolds will be removed. At this point, there will be no communication between any of you! As soon as I give a signal, everyone must guess — at the same time — the color of the hat atop their own head. To win this challenge, *at least one* of you must guess correctly!”

The potentate continued: “The good news is that there’s a little more information you’ll have. I will be arranging you into two rows, with two of you in one row and three of you in the other. Citizens in the same row cannot see each other, but they can see all the citizens in the other row. Finally, each citizen will know their placement *within* their own row — that is, whether they are seated on the left or right or in the middle.”

What strategy could the citizens have devised beforehand so that at least one of them guessed correctly?

At first blush, this seemed hopeless. With five hats, each of which could be three colors, there were a total of 3^{5}, or 243, cases to consider. How could the five citizens possibly cobble together enough information to ensure that at least one of them guessed correctly?

Virtually all solvers treated this as the logic problem it was intended to be, experimenting with different rules for the citizens. But before proceeding down this road, I want to give a tip of the hat (see what I did there?) to the creative approach of solver Josh Silverman, who used a genetic algorithm to computationally find a solution. But enough with the computers!

The puzzle’s submitter, Alex van den Brandhof, originally penned this puzzle with fellow Dutchman Dion Gijswijt for Pythagoras, a Dutch magazine for high school math students. One solution, courtesy of Alex, is as follows:

First, let’s slap some labels on the positions and the hat colors. Let’s call the citizens in the row of two *A* and *B*, and the citizens in the row of three *C*, *D* and *E*. Let’s also convert those colors into numbers — a red hat has a value of 0, a green hat has a value of 1 and a blue hat has a value of 2. We then call the values (again, representing the colors) of the respective positions *a*, *b*, *c*, *d* and *e*.

First off, *C* guessed *a *(the color of A’s hat), and *D* guessed *b* (the color of B’s hat). Meanwhile, *E* guessed the sum of *a* and *b*, modulo 3. For example, if *A* and *B* both had blue hats — each with a value of 2, according to the aforementioned scheme — then *E* would have guessed green (which has a value of 1), since 2 + 2 equals 4, which is congruent to 1 (mod 3).

Should any one of those three (*C*, *D* and *E*) have guessed their own hat color correctly, then the citizens would be victorious. Therefore, when *A* and *B* made their guesses, they might as well have assumed that *C*, *D* and *E* were all *wrong*. That meant *A* shouldn’t have guessed *c*, because if *A* and *C* had the same color hat, then C would already have been correct. Similarly, *B* shouldn’t have guessed *d*, because if *B* and *D* had the same color hat, then D would already have been correct. And finally, *both A* and *B* should have assumed that the sum of their hat values (mod 3) was different from *E*’s guess — otherwise *E* would have been correct.

At this point, no matter what colors *A* and *B* saw on the hats of *C*, *D* and *E*, it was *always* possible for at least one of them to be correct — in the event that *C*, *D* and *E* were all wrong.

For example, suppose the color values are as follows: *a* = 0, *b* = 2, *c* = 2, *d* = 1 and *e* = 1. Here’s how this scenario would play out for *C*, *D* and *E*:

- Seeing that
*a*= 0,*C*guesses 0 — incorrectly. - Seeing that
*b*= 2,*D*guesses 2 — incorrectly. - Seeing that
*a*= 0 and*b*= 2,*E*guesses 0 + 2 (mod 3), or 2 — incorrectly.

With three incorrect guesses, the pressure is then on *A* and *B*. Looking at the hats of *C*, *D* and *E*, *A* and *B* both know that *A* shouldn’t guess 2, *B* shouldn’t guess 1, and that the sum of their guesses (mod 3) shouldn’t be 1. That left just three possible cases to consider:

*A*guesses 0 and*B*guesses 0, for a sum of 0.*A*guesses 0 and*B*guesses 2, for a sum of 2.*A*guesses 1 and*B*guesses 2, for a sum of 0.

If *A* guesses 0 and *B* guesses 2, at least one of them is guaranteed to be correct! This sort of result played out for all 243 combinations of *a*, *b*, *c*, *d* and *e* — **it was always possible for at least one citizen to guess correctly**.

Next time, the potentate will be sure to make use of her countless *yellow* hats as well, in which case it will surely take more than five citizens working together to complete the challenge …

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{10} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Yesterday was the first night of Hanukkah, which means millions of Jews began lighting their menorahs. It is traditional to start with one candle in the rightmost position, which is lit using a central candle called a shamash. On the second night, a second candle is added to the left of the first. On the third night, a third candle is added to the left of the second, and so on, until all eight candles have been added on the final night.

With this system for adding candles, whether you are looking at the menorah from the front (i.e., in your home) or back (i.e., through a window), you can easily determine which of the eight nights of Hanukkah it is by counting the number of candles (other than the shamash). But what if you wanted to make that same menorah capable of tracking *more* than eight nights?

To do that, you’d have to devise a new system of adding or moving candles (other than the shamash in the middle, which is always lit) around the menorah so that it can uniquely indicate as many nights as possible. Your system must read the same forward and backward, so that regardless of whether you’re looking at the menorah from the front or the back you will still know what night it is. What is the greatest number of nights that could be indicated using your system?

The solution to this Riddler Express can be found in the following column.

From Alex van den Brandhof comes a matter of life and death:

The Potentate of Puzzles decides to give five unlucky citizens a test. The potentate has countless red hats, green hats and blue hats. She says to the citizens, “Tomorrow, you will all be blindfolded as I place one of these hats on each of your heads. Once all the hats are placed, your blindfolds will be removed. At this point, there will be no communication between any of you! As soon as I give a signal, everyone must guess — at the same time — the color of the hat atop their own head. If *at least one* of you guesses correctly, all of you will survive! Otherwise …”

The potentate continues: “The good news is that there’s a little more information you’ll have. I will be arranging you into two rows facing each other, with two of you in one row and three of you in the other. Citizens in the same row cannot see each other, but they can see all the citizens in the other row. Finally, each citizen will know their placement *within* their own row — that is, whether they are seated on the left or right or in the middle.”

Can the citizens devise a strategy beforehand that ensures their survival? If so, what is the strategy?

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Helen Jannke 👏 of Acton, Massachusetts, winner of last week’s Riddler Express.

Last week, a family of five had a book exchange for Christmas. First, each person put their name in a hat. The hat was shaken, and then each person drew a name from the hat and gave that person a book. However, if anyone drew their own name, they all put their names back into the hat and started over.

What is the probability that no one drew their own name?

There were five possible names the first person could have drawn, four names the second person could draw, three for the third, two for the fourth and just one for the fifth. That meant the total number of scenarios to consider here was 5·4·3·2·1 — that is, 5!, or 120. For many solvers, this number was small enough that it was possible to list out all 120 possibilities and count up how many of them resulted in no one drawing their own name.

But with some clever counting, you could add up these cases more efficiently. One way to do this was to represent the five people as five points on a directed graph. Then, if person A drew the name of person B, you drew a line from point A to point B on the graph. After you drew all five lines (one for each drawing), there were three possibilities:

- The lines formed
*one*big cycle, looping around all five points before returning to the start. - The lines formed
*two*cycles, one with three points and another with two points. - One of the lines connected a point to itself.

It was that third case — when a line connected a point to itself — that was problematic. It meant that someone had drawn their own name from the hat. Fortunately, counting up the first two cases was manageable.

To make one big cycle, you could pick any point to start with. There were four remaining points to which you could draw a line. From that selected point, there were three points to visit next — then two, and then one, before returning back to your starting point. All told, that accounted for 4!, or 24, unique cycles.

Counting up the graphs with two cycles (one with three points and the other with two) was a little trickier. First you had to choose which two points belonged in the cycle of two, while the remaining three points were then relegated to the cycle of three. There were 5 choose 2, or 10, ways to do this. Then, for each of these 10 ways, there was just one way to link up two points to make a cycle (each point went to the other), but *two* ways to link up three points (i.e., they could loop clockwise or counterclockwise). That made for 20 graphs with two cycles.

And so, among the 120 possible graphs, 44 of them (24 + 20) had no points connected to themselves. That meant the probability that no one picked their own name was 44/120, or **11/30**, or about 0.37. On average, it would take the family of five about three drawings to get a successful one in which no one chose their own name.

This problem was equivalent to finding the probability that a random permutation of five people was also a derangement, meaning no one wound up in their own starting position. The number of derangements for *N* objects is commonly written as !*N*. That meant the probability that no one picked their own name from a family of *N* could be written as !*N*/*N*! — a delightful notational palindrome.

As the number of family members continued to increase, !*N*/*N*! approached a rather remarkable value, for which 11/30 (this week’s answer) was a decent approximation. To learn more, I recommend David Ding’s writeup of this puzzle.

Congratulations to 👏 Angela Zhou 👏 of New York City, winner of last week’s Riddler Classic.

Last week, three friends were baking holiday cookies together. They had a flat layer of cookie dough in the shape of an isosceles right triangle (shown below). They wanted to design a cookie cutter that would cut out three identical (i.e., congruent) cookies. The cookies had to be as large as possible while staying within the triangle and without overlapping each other.

Had there been two friends or four friends, they could have used all the cookie dough, but with three friends, some dough went to waste.

What was the greatest percentage of cookie dough that could be used up by the three identical cookies?

This was a delicious challenge. Full disclosure: According to the puzzle’s submitter, Dean Ballard, this puzzle was proposed by Karl Scherer of Auckland, New Zealand, in the 2002-03 volume of the Journal of Recreational Mathematics. Along with the problem, Sherer submitted a solution that used exactly 90 percent of the cookie dough. Two issues later, Dean had his own solution published, which we’ll return to in just a moment.

For now, here are a few of the top submissions from the past week:

Bram Carlson used approximately 86.6 percent of the cookie dough with a cyclic quadrilateral cookie cutter:

Ignas from London, England, used approximately 90.3 percent of the cookie dough with a a truncated parallelogram cutter:

Meanwhile, Thomas Stone did a shade better than Ignas, using approximately 91.3 percent of the cookie dough with a funky-looking hexagonal cookie cutter:

Angela Zhou, this week’s winner, used approximately 94.1 percent of the cookie dough with a trapezoidal cookie cutter. This was exactly the same result that Dean achieved almost 20 years ago.

But the story didn’t end there. According to Dean, in the very same issue in which his solution appeared, an even better solution had been identified by Robert Wainwright and Richard Hess, which used up almost **95.5 percent** of the cookie dough:

Dean and I believe this remains an open problem. If you can do even better than 95.5 percent, let us know!

For extra credit, you had to make three identical cookies from a sphinx of cookie dough. Keen-eyed solvers noted that there was no restriction on the convexity of the cookie cutter. That made it possible to get arbitrarily close to using **100 percent** of the cookie dough.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{11} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Christopher “CJ” Halverson comes a bibliophilic game of Secret Santa:

Every year, CJ’s family of five (including CJ) does a book exchange for Christmas. First, each person puts their name in a hat. The hat is shaken, and then each person draws a random name from the hat and gifts that person a book. However, if anyone draws their own name, they all put their names back into the hat and start over.

What is the probability that no one will draw their own name?

From Dean Ballard come trials and tribulations of troublesome triangles:

Three friends are baking holiday cookies together. They have a flat layer of cookie dough in the shape of an isosceles right triangle (shown below). They want to design a cookie cutter that will cut out three identical (i.e., congruent) cookies. The cookies should be as large as possible while staying within the triangle and without overlapping each other.

Had there been two friends or four friends, they would have been able to use all of the cookie dough. But with three friends, there will unfortunately be some cookie dough that goes to waste.

What is the greatest percentage of cookie dough that can be made up by the three identical cookies? When explaining your work, please be as detailed as you can about the arrangement of your cookies so I can verify your work!

*Extra credit:* Instead of a right isosceles triangle, the three friends now want to make identical cookies from a sphinx of cookie dough. Again, what is the greatest percentage of cookie dough that can go into the three identical cookies?

Congratulations to 👏 Piotr L. 👏 of Newbridge, Ireland, winner of the most recent Riddler Express.

Depending on the year, there can be one, two or three Friday the 13ths. Several weeks ago happened to be the second Friday the 13th of 2020.

What was the *greatest* number of Friday the 13ths that could occur over the course of four consecutive calendar years?

The “four” years in this problem was intentional — it guaranteed that there was at most one leap year in the mix. It was also possible for there to be no leap years, since some years that are multiples of four — like 1900 and 2100 — are not leap years.

Most solvers turned to spreadsheets so they could efficiently track all the possibilities. And to simplify matters a bit, rather than look for months whose 13th days were Fridays, you could have equivalently found months whose first days were Sundays.

From here, many solvers, like Amy Leblang, used modular arithmetic. Amy renamed the days of the week A-day, B-day, C-day, D-day, E-day, F-day and G-day. We haven’t decided yet which of these days is Sunday — but if A-day were Sunday, then B-day would be Monday, C-day would be Tuesday, etc. Suppose the first year of our four-year interval was a leap year. Then if Jan. 1 was an A-day, that would mean Feb. 1 was a D-day, since January has 31 days (*three* more than a multiple of seven). And then March would be an E-day, because February has 29 days in a leap year (*one* more than a multiple of seven).

Continuing month by month over the four years, there were nine months that started with an A-day, six months that started with a B-day, seven months that started with a C-day, seven months that started with a D-day, six months that started with an E-day, six months that started with an F-day and seven months that started with an G-day. So when the first year was a leap year, A-day was the most frequent first day of the month, occurring nine times. And when A-day happened to be a Sunday, you’d have nine Friday the 13ths over four years.

Had the leap year been the second, third or fourth year in the interval — or if there had been no leap years — then none of the seven lettered days started a month more than eight times. That meant you could have at most **nine** Friday the 13ths over four years. Indeed, the last time this occurred was (checks notes) from 2012 to 2015. Sure enough, 2012 was a leap year that started on a Sunday.

For extra credit, you were asked about four-year intervals that could start on any day of the year. As it so happened, this maximum was again **nine**.

To be honest, riddles about leap years always make my head spin a little. We’ll have no more of these for at least another four years.

Congratulations to 👏 Daniel Silva-Inclan 👏 of Chicago, Illinois, winner of the most recent Riddler Classic.

To celebrate Thanksgiving, you and 19 of your family members were seated at a circular table (socially distanced, of course). Everyone at the table wanted a helping of cranberry sauce, which happened to be in front of you at the moment.

Instead of passing the sauce around in a circle, you passed it randomly to the person seated directly to your left or to your right. They then did the same, passing it randomly either to the person to *their* left or right. This continued until everyone had, at some point, received the cranberry sauce.

Of the 20 people in the circle, who had the greatest chance of being the *last* to receive the cranberry sauce?

First off, this was essentially a repeat of a Riddler Express from several years ago. That puzzle was submitted by Chris Thornett of Brooklyn, New York, who — surprise, surprise — got the right answer once again!

Chris explained his solution by looking at a specific family member (other than you), whom we might as well call … Chris. Now for Chris to be the last person to receive the cranberry sauce, exactly one of two things had to happen:

- At some point, the person directly to Chris’s
*left*received the sauce. However, the sauce was*not*passed to Chris, instead making its way around the entire table until Chris received it from the person on his*right*. - At some point, the person directly to Chris’s
*right*received the sauce. However, the sauce was*not*passed to Chris, instead making its way around the entire table until Chris received it from the person on his*left*.

None of this depended on where the sauce had already been, which meant everyone at the table had the *same* chances of being last: **1/19**.

This was a rather surprising result! One might have expected that those sitting far away from you were more likely to have received the cranberry sauce last. A few solvers, like David Robinson and Quoc Tran, reached for their nearest computer and verified Chris’s results with thousands or even millions of simulations.

The animation below shows one simulation at first, with the cranberry sauce randomly bouncing its way around the table. It then speeds up, showing a grand total of 100,000 simulations. The pie chart (how appropriate for Thanksgiving!) in the middle reveals the relative frequencies with which each person is last to receive the sauce. Sure enough, the distribution appears to be uniform — exactly what Chris’s solution had predicted!

As you’d expect, this result was generalizable to any number of people in the circle. For any value of *N*, each person (other than you) had a 1/(*N*−1) chance of being the last to receive the cranberry sauce.

Solver Rajeev Pakalapati took the problem a step further and solved the general case when your family was right-leaning or left-leaning — that is, in which way the cranberry sauce was passed around the table. (Apologies for the attempt at political humor.) If each member of your family passed the sauce to the right with probability *p* > 0.5, then the person to your left was most likely to receive it last. According to Rajeev’s calculations, that maximum probability was *x*^{18}·(*x*−1)/(*x*^{19}−1), where *x* was defined as *p*/(1−*p*).

This nonlinear result was very sensitive to changes in *p*. Even when *p* was only slightly greater than 0.5 (as in the second animation above, where *p* was 0.51), there were tremendous shifts in the probability distribution of who was the last to receive the sauce.

Finally, I wanted to acknowledge that for most of us, the value of *N* for our Thanksgiving dinners last week was much smaller than 20. Here’s hoping this puzzle will be more realistic for Thanksgiving in 2021.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>There are few truly great football movies. The things that draw fans to the sport — a strange mix of athleticism, schematic complexity, courage and violence — seem to defy easy capture on the big screen.

Perhaps it’s because our rooting interests change when we watch movies. We cheer for dramatic character arcs rather than rainbow Hail Marys caught as time runs out. Good football movies aren’t about big moments on the field but the big characters who shape them. And there’s no bigger personality than a movie football coach.

On the big screen, a head coach is typically part father figure, part disciplinarian extracting the utmost from his team. Some serve as a foil — a common enemy players can rally around to come together and ultimately overcome insurmountable odds. Others are inspirational lords of the locker room, dispensing life advice. But which movie coach is the best? We watched nine football movies — from “Air Bud” to “Varsity Blues” — and tried to capture what makes each film’s head football coach tick.

Football movies watched by rating on Rotten Tomatoes, with year released and competition level portrayed in the film

Movie | Year | Competition Level | Tomato Rating |
---|---|---|---|

“North Dallas Forty” | 1979 | Pro | 84% |

“Friday Night Lights” | 2004 | High school | 82 |

“The Longest Yard” | 1974 | Semi-pro | 79 |

“Remember the Titans” | 2000 | High school | 73 |

“Any Given Sunday” | 1999 | Pro | 52 |

“We Are Marshall” | 2006 | College | 48 |

“Varsity Blues” | 1999 | High school | 43 |

“Air Bud: Golden Receiver” | 1998 | Junior high | 21 |

“Johnny Be Good” | 1988 | College | 0 |

Where possible, we calculated the expected points added of each coach’s play calls, and we evaluated coaches’ decision making — both on the field and in the locker room. Finally, we ranked the leadership qualities of each coach along an alignment chart, defining each of the nine as lawful good, lawful evil, chaotic good and chaotic evil.

“Lawful” means adhering to the law generally. In cases where a coach deviates from established norms, they will at least remain consistent with respect to their worldviews. Conversely, a “chaotic” coach would be an individual willing to buck norms and traditions — including those he’s established — in pursuit of his goals. A “good” coach is selfless as a leader and puts the needs of others ahead of his own interests, while an “evil” coach is willing to actively harm others to get what he wants.

And yes, we will spoil the movies here … but the most recent one came out in 2006, so if you haven’t seen them by now, that’s on you.

In a cartoonish family drama about a dog suiting up and helping protagonist Josh Framm’s junior high school football team turn its program around, it’s perhaps unsurprising that Fanelli’s only flaw appears to be an unhealthy love of junk food. Fanelli’s job is in danger because of the Timberwolves’ persistent losing, but his perspective on coaching never wavers: It’s about the kids. “If winning’s all they’re interested in, I’m not the right guy anyway,” Fanelli says.

And while Fanelli is open to the unorthodox — he allows Josh to bring Buddy onto the team — he does so in service to his ethos that “football is about having fun.” Most of “Air Bud: Golden Receiver” is undeniably corny, but Robert Costanzo as Fanelli is able to transcend the kitsch in one touching moment when he convinces runaway Josh that opening his heart to his mother’s new boyfriend doesn’t mean he has to stop loving his deceased father. That, coupled with his ability to adapt to the strengths of his personnel, is enough to place the coach firmly in the upper right corner of our alignment chart.

Denzel Washington’s Herman Boone is firm but fair. He’s a mean cuss to everyone, Black or white. He’s authoritarian and perhaps not quite the coach the players deserved, but maybe the one they needed. In the situation the movie’s writers put him in — lose one game and you’re fired, in an environment of dusty racism and heightened tensions — Boone’s sometimes excessive drive to win is at least somewhat understandable. “We will be perfect” is a laughable statement from any football coach, but it somehow manages to be stoic and endearing from Washington. It also helps, of course, that the Titans do end up going undefeated.

If winning is the best measure of a coach’s ability, the title of best movie coach surely goes to Boone. In fact, his Titans were so dominant in real life — the 1971 T.C. Williams High School team outscored its opponents 357-45 with nine shutouts — that screenwriter Gregory Allen Howard had to manufacture drama to make the movie more compelling.

1971 schedule for T.C. Williams High School, the basis of the film “Remember the Titans”

Opponent | Result | Score |
---|---|---|

Herndon | W | 19-0 |

Yorktown | W | 25-0 |

Hayfield | W | 26-7 |

Jefferson | W | 25-0 |

Marshall | W | 21-16 |

Groveton | W | 29-0 |

Madison | W | 34-0 |

W & L | W | 34-0 |

Wakefield | W | 27-0 |

Bishop Ireton | W | 26-8 |

Annandale | W | 28-0 |

Woodrow Wilson | W | 36-14 |

Andrew Lewis | W | 27-0 |

T.C. Williams plays in a number of close games over the course of the film, but the actual Titans played in only one nail-biter: their fifth game of the season, against Marshall. The Titans trailed for most of the game and were held scoreless until midway through the second quarter. It wasn’t until Frankie Glascoe — who rushed for negative yardage in the first half — broke loose for a 75-yard touchdown run with 5:20 left in the game that the Titans took the lead.

Conversely, the real Herman Boone was a much more complicated character than the one made famous by Washington’s portrayal. The real Boone narrowly survived a player mutiny in 1977 when the football team threatened to quit unless he apologized for a particularly vitriolic tirade after a loss. Boone was removed from his job two years later after he was accused of abusing his players “orally and physically.”

“Remember the Titans” doesn’t completely gloss over the troubling parts of Boone’s character. Some of his hard edge is revealed when he tells tired and dehydrated player Darryl “Blue” Stanton that “Water is for cowards. Water makes you weak” and then proceeds to have the team do up-and-downs “until Blue is no longer tired, and thirsty.” The evidence is clear that this type of coaching is harmful for players and counterproductive to team-building. In fact, it can be deadly. Reports compiled by the University of North Carolina show that 106 football players across all levels of play died from heatstroke from 1971 to 2019. And eight young men died in 1970 — the most fatalities due to heatstroke ever recorded in a single year — just one year prior to the season depicted in the movie. Despite this — mainly because of the strength of Washington’s acting — Boone manages to rate highly on our good and lawful scales.

Matthew McConaughey’s Jack Lengyel is a joyful and big-hearted but ultimately one-dimensional character. Lengyel’s goal in taking over the Marshall University football team a year after nearly the entire coaching staff and all but three players were killed in an airplane crash isn’t to win. It’s merely to suit up and take the field. Lengyel’s expectations fit the moment: The Thundering Herd won two games in 1971. Over his career at Marshall, Lengyel went just 9-33.

Lengyel rates highly on the “good” scale of the alignment chart. He took a job no one else wanted because he thought he could help a town with deep wounds heal. McConaughey’s Lengyel doesn’t pretend to have the answers to the movie’s bigger questions about grief, but he does know when to shut up and listen.

Perhaps because of this, his leadership style drifts toward the chaotic. His accomplishments in the movie are largely due to him pushing back against institutions and bucking convention. He convinces Marshall’s president to petition the NCAA to allow true freshmen to play for the team, overturning a long tradition. And he brazenly asks coach Bobby Bowden at rival West Virginia University to allow him and his staff to plunder their archives to learn the Veer, an offensive system better suited to their players than a more standard offense. In the movie’s best scene, Lengyel and his assistant coach discover that WVU’s team is wearing Marshall stickers on their helmets to honor the team and the dead. As it turns out, the portrayal is accurate — and part of the reason for Bowden’s generosity was that he nearly took the coaching job at Marshall, and very well might have been on that plane.

In the movie based on the book,^{12} Gary Gaines coaches the Permian High School Panthers in a small Texas town obsessed with high school football. The town’s skewed view of the importance of Friday nights seems to rub off on a character that the movie tries to cast as its hero. Gaines — played by Billy Bob Thornton — says he doesn’t do the Oklahoma drill “just to make heads rattle,” but it doesn’t stop him from lining his kids up and doing just that. Perhaps more than any other moment, this knocks Gaines down our “good” and “lawful” scales. The drill’s benefits are not worth the risks involved. The NFL banned the controversial drill last year out of concern that it causes preventable head injury, and college programs have begun to follow suit. And while it’s perhaps unfair to judge a coach too harshly for the practice in 1988, it remains bewildering that sober adults ever thought having young men in high school slam their heads into one another during practice is either smart or acceptable.

On the field, Gaines is wildly successful despite losing his best player, running back James “Boobie” Miles, to a career-ending knee injury in a blowout Week 1 win. Gaines eventually leads the Panthers to the state championship against Dallas Carter High School at the Astrodome in front of 55,000 fans, losing the game on the last play after a courageous come-from-behind effort.

The real-life Gaines ended his first tenure at Permian with a 46-7-1 record,^{13} good for a 86.1 win percentage, and a national championship.

Strothers is the most interesting and complex of the movie coaches we viewed. Perhaps that’s because his character was modeled after the Dallas Cowboys’ Hall of Fame head coach Tom Landry, an undeniably great tactician and one of the first people in the NFL to use an analytical process to manage a team. In one scene in a meeting at team headquarters, Strothers — played by G.D. Spradlin — upbraids his star quarterback for a low completion percentage: “Pass completions were 49 percent. That’s 6.3 percent less than reasonable, and 19 percent less than our outstanding.” The dehumanization of players by the business of football is a major theme of “North Dallas Forty,” and Strothers neatly embodies the cold, calculating executive who stares into a computer screen for insights into his players rather than looking them in the eye.

Set in 1979, the film isn’t clear on who actually calls the plays in the final sequence against Chicago. Cowboys quarterback Don Meredith was known for changing Landry’s plays in the 60s, but if Strothers did have a hand in things, he did a good job. To save the season, the Bulls dial up a clever draw on third-and-long that’s worth 1.25 expected points added, and then North Dallas overcomes a costly holding call with 24 seconds left by calling back-to-back passes so that the team won’t have to huddle and lose time off the clock. Nick Nolte’s character — wide receiver Phil Elliott — runs a devastating “blaze-out route” (one that looks only *slightly* slower than Julio Jones’s more recent incarnations) with no time left on the clock for an improbable last-second touchdown.

Plays run on the last drive by the Bulls in the movie “North Dallas Forty,” with expected points added (EPA) and touchdown probability* for each

Play type | down | To go | Field position | Time left (secs.) | epa | TD odds |
---|---|---|---|---|---|---|

Run | 1 | 10 | 25 | 120 | -0.84 | 38% |

Pass | 2 | 13 | 28 | 80 | -0.49 | 23 |

Draw | 3 | 13 | 28 | 65 | +1.25 | 17 |

Holding, no play | 1 | 10 | 18 | 24 | -1.55 | 35 |

Pass | 1 | 20 | 28 | 16 | +0.18 | 8 |

Pass, touchdown | 2 | 7 | 15 | 1 | +4.09 | 4 |

Strothers is also cast as a hypocrite on drug use. In a league still struggling with how to properly deal with recreational cannabis, Strothers’s attitude is understandable. The movie ends with Elliott forced out of the league ostensibly for smoking grass (but in reality, it was mostly to save payroll costs), while Strothers encourages the use of painkillers to keep his players on the field.

Perhaps unsurprisingly, the NFL was not a fan of the film. There were real-life repercussions for the experts who served as advisers on the movie. Hall of Famer and scout Tom Fears reportedly had contracts with three NFL teams prior to the film’s release, and each of them severed ties with him after “North Dallas Forty” opened in theaters. Raiders receiver Fred Biletnikoff, who coached Nolte and after whom Nolte modeled his character’s playing style, wasn’t asked back to camp. And Tommy Reamon — who played needle-shy receiver Delma Huddle — was cut by the San Francisco 49ers and claimed he had been blackballed by the league.

Oliver Stone’s “Any Given Sunday” is a mess of a movie, with camera movements so violent that it’s difficult to sit through the first third without having a seizure. When things reach their nadir and a player loses an eye on the field in an absolutely ridiculous scene, the only thing keeping a viewer interested is Al Pacino’s Tony D’Amato. D’Amato is an old-school coach caught in a league that’s changing in ways he seems unprepared for. Front office meddling and a young offensive coordinator who “knows the probabilities” are conspiring to force him out of the game he loves. But while D’Amato starts out as a selfish, manipulative, run-first control freak (“You talk about a running game. … People want to see passes, touchdowns, high scores. That’s the game today.”), he ultimately morphs into a selfish and manipulative but relatable player’s coach. Even with his characteristic late-career overacting, Pacino’s “Inches” speech is still one of the best of the genre.

On the field, D’Amato evolves as well, ultimately embracing the passing game and a new style of mobile quarterbacks. After his defense gets a stop on fourth-and-1, he calls a long pass play that connects for 80 yards but is called back for holding. D’Amato’s Miami Sharks overcome the setback by passing the rest of the way down the field (save for the final two plays of the drive) against the Dallas Knights to win (though they eventually lose the Pantheon Cup to San Francisco). It would be nice to be able to say more about D’Amato’s tactics, but Stone’s claustrophobic camerawork is so tight on each actor’s face that it’s nearly impossible to determine what’s actually happening on the field.

Burt Reynolds’s portrayal of the woman-beating, drunk-driving former NFL QB Paul Crewe is probably more beloved than it deserves to be. The characters in “The Longest Yard” are largely unlikable (“I never gave a shit about football, or anything else.”), and the comedy — which may once have been shocking — is tired and flat by today’s standards.

As a coach of a semi-pro football team in prison, Crewe integrates the squad and at one point opts to go for a 2-point conversion, both of which are laudable (though to different degrees). Another memorable moment is when a decision to punt by the warden’s team is described by announcers as “kind of a timid or a chicken kind of football.” Still, Crewe’s failures outweigh his successes. In one scene, he wastes two straight downs hitting a guard below the belt with a football, which about sums up the viewing experience for this 1974 “classic.”

Bud Kilmer and Wayne Hisler are lumped together because they’re mostly uninteresting caricatures. Kilmer, played by Jon Voight, represents the worst parts of Strothers, Gaines and Boone all rolled into one seething package. He’s a racist who sends a lineman suffering from a concussion back into the game despite suspecting head trauma, then later cuts him when the injury hampers his performance. An authoritarian who is also, unsurprisingly, a run-first play-caller, Kilmer is ultimately thrown off the team by the players and is not on the sideline for West Canaan High School’s climactic win.

In “Johnny Be Good,” Hisler — portrayed by Paul Gleason — is a craven narcissist trying to parlay his relationship with his star high school QB (played by “The Breakfast Club” nerd-turned-jock Anthony Michael Hall) into a higher-paying job at a university. An abomination of a character, Hisler is unlikable and unwatchable, much like the movie itself.

*Illustration by Madison Ketcham.*

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{14} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

First, an unlucky puzzle that comes a week late:

Depending on the year, there can be one, two or three Friday the 13ths. Last week happened to be the second Friday the 13th of 2020.

What is the *greatest* number of Friday the 13ths that can occur over the course of four consecutive calendar years?

*Extra credit: *What’s the greatest number of Friday the 13ths that can occur over a four-year period (i.e., a period that doesn’t necessarily begin on January 1)?

The solution to this Riddler Express can be found in the following column.

From Patrick Lopatto comes a riddle we can all be thankful for:

To celebrate Thanksgiving, you and 19 of your family members are seated at a circular table (socially distanced, of course). Everyone at the table would like a helping of cranberry sauce, which happens to be in front of you at the moment.

Instead of passing the sauce around in a circle, you pass it randomly to the person seated directly to your left or to your right. They then do the same, passing it randomly either to the person to *their* left or right. This continues until everyone has, at some point, received the cranberry sauce.

Of the 20 people in the circle, who has the greatest chance of being the *last* to receive the cranberry sauce?

The solution to this Riddler Classic can be found in the following column.

Congratulations to 👏 Nathan Ainslie 👏 of Bloomington, Indiana, winner of last week’s Riddler Express.

Last week, you were a contestant on the TV show Jeopardy! You were competing in the (single) Jeopardy! round and your opponents were simply no match for you. You chose first and never relinquished control, working your way horizontally across the board by first selecting all six $200 clues, then all six $400 clues, and so on, until you finally selected all the $1,000 clues. You responded to each clue correctly before either of your opponents could.

One randomly selected clue was a Daily Double. Rather than award you the prize money associated with that clue, it instead allowed you to double your winnings (up to that point) or wager up to $1,000 should you have less than that. Being the aggressive player you are, you always bet the most you could have. (In reality, the Daily Double was more likely to appear in certain locations on the board than others, but for this problem you assumed it had an equal chance of appearing anywhere on the board.)

How much money did you expect to win during the Jeopardy! Round?

There were a total of 30 clues on the board, and any one of those 30 clues could have been the Daily Double. That meant there were 30 cases to consider: when the Daily Double was the first clue you selected, when it was the second clue, the third clue, and so on. And when it was one of the first six clues selected, the Daily Double was worth $1,000; otherwise, it doubled your money. (Technically, if it was the sixth clue, it was worth $1,000 *and* doubled your money, since you had won exactly $1,000 by that point.)

Many solvers listed out all 30 cases and averaged the winnings. Alternatively, you could have added up the total amount of prize money across all the cases and then divided by 30. Across these cases, each clue was the Daily Double once, which meant it *wasn’t* the daily double 29 times. Adding up the values of all the clues — without worrying about the Daily Double — gave you 6·(200 + 400 + 600 + 800 + 1000), or $18,000, which meant that 29 boards of clues were worth $522,000.

Now to add up the Daily Doubles. As we already said, for the first six cases the Daily Double was worth $1,000. When the Daily Double was the seventh clue, it was worth $1,200. From there, its value increased by $400 until it was $3,600 as the 13th clue. Then, it increased by $600 until it was $7,200 as the 19th clue. Next, it increased by $800 until it was $12,000 as the 25th clue. Finally, it increased by $1,000 until it was a whopping $17,000 as the 30th and final clue. That was some James Holzhauer wagering right there.

Even without a spreadsheet, you needed some hefty addition to tally up these Daily Doubles. Their total value across the 30 cases turned out to be $192,000 — roughly 36 percent of what the non-Daily Double clues had been worth.

Combining the regular clues with the Daily Doubles gave a sum of $714,000 for all 30 cases. That meant the average — the amount you’d “expect” to win — was **$23,800**.

For extra credit, instead of working your way horizontally across the board, you selected random clues from anywhere on the board, one at a time. Now how much money did you expect to win during the Jeopardy! Round?

This was a much thornier version of the problem. First, there were five rows in which the Daily Double might have appeared. Then, for each row, you had to consider the order in which you worked your way across the board. So instead of 30 cases, there were in fact 5·30!/((6!)^{4}5!), or about 4.1×10^{19} cases to consider.

From here, most solvers used Monte Carlo methods, simulating the game of Jeopardy! thousands or even millions of times to approximate the answer. As a few brave coders (like Josh Silverman, Lowell Vaughn and Alex Vornsand) found the answer was **approximately $26,150**. (This was quite close to the result you’d get — $26,146.67 — if you assumed every clue had been worth the average amount of $600. The answer was *slightly* larger than this, since the Daily Double more than doubled your winnings whenever you had less than $1,000.)

It made sense that picking clues randomly would net you more winnings on average, because you now had a greater chance of selecting higher-value clues (like the $800 and $1,000 rows) before hitting the Daily Double.

For this puzzle, we’ll let Alex Trebek have the last word. We miss you.

Congratulations to 👏 Alex Zorn 👏 of Brooklyn, New York, winner of last week’s Riddler Classic.

Last week, you modeled blown football leads, something the Atlanta Falcons know a thing or two (or three) about. The Georgia Birds and the Michigan Felines were playing a game in which a fair coin was flipped 101 times. In the end, if heads came up at least 51 times, the Birds won; but if tails came up at least 51 times, the Felines won.

What was the probability that the Birds had at least a 99 percent chance of winning at *some point* during the game — meaning their probability of victory was 99 percent or greater given the flips that remained — and then proceeded to lose?

This was a challenging riddle, to be sure. For starters, you first had to make sense of what it meant to have “at least a 99 percent chance of winning,” while avoiding the tempting answer of 1 percent. Before any flips were made, the Birds had a 50 percent chance of winning. But suppose, through incredible luck, that the first 50 tosses all came up heads. From there, the Birds could still technically lose if the final 51 tosses all came up tails — an event whose probability was 1/2^{51}.

That was just one (albeit very unlikely) way for the Birds to have a 99 percent chance of winning and then blow the game. But there were many other, more likely scenarios, each of which involved an excess of heads flipped toward the beginning. One way to add up the probabilities of these scenarios was to analyze (preferably via code) what was happening at each combination of wins (*W*) and losses (*L*) for the Birds.

Here’s a graph that shows the pairs (*W*, *L*), shown in red, from which the Birds had at least a 99 percent chance of winning. You could determine these directly using combinatorics, or working backwards recursively, noting that the probability of winning from (*W*, *L*) was the average of the probabilities from (*W*+1, *L*) and (*W*, *L*+1).

Next, you wanted to find the probability that the Birds passed through at least one of the red locations in the graph above. That is, for each (*W*, *L*), what was the probability that at *some* point the Birds had a 99 percent chance of winning?

This graph was very similar to the first. But there were some places — in green and aqua — where the Birds had ventured into the 99 percent region and then back out. These paths through the graph were what made it possible for the Birds to blow their lead.

From here, you had to focus on which of these paths resulted in an overall loss for the Birds, and work backwards. The final graph below shows the Birds’ chances of reaching a 99 percent chance of victory at some point and then blowing the lead, for each (*W*, *L*).

These chances were greatest when the Birds had won 50 games and lost 51, when there was a roughly 2 percent chance that they had blown a 99 percent lead somewhere along the way. Working backwards, their chances of blowing a lead when they had zero wins and zero losses were about 10 times smaller, or **0.21 percent**. This was the solution to the riddle: the Birds’ chances of at some point having a 99 percent chance at victory and then proceeding to lose.

For extra credit, instead of 101 total flips, there were many, many more (i.e., you had to consider the limit as the number of flips went to infinity). Again, the Birds won if heads came up at least half the time. *Now* what was the probability that the Birds had a win probability of at least 99 percent at some point and then proceeded to lose?

To solve this, some coders continued increasing the number of flips beyond 101 and looked for asymptotic behavior. This week’s winner, Alex Zorn, approached the extra credit analytically by first defining three key probabilities:

*h*, the probability that the Birds ever hit 99 percent win probability

*c*, the probability that the Birds choke after attaining their 99 percent win probability

*w*, the probability that the Birds hit 99 percent and go on to win

With these three variables defined, Alex was able to set up a few equations. For example, *w* = 0.99·*h* and *c* = 0.01·*h*, since that’s what was meant by a 99 percent win probability. Moreover, the probability that the Birds were the ultimate winners was 0.5, which *had to* equal *w*. That meant *h* = 50/99, which in turn meant that *c* — the probability of blowing that 99 percent lead, was **1/198**.

Finally, solver Allen Gu noticed some rather peculiar oscillating behaviors as the number of coin flips increased and the threshold for “choking” was closer to 50 percent:

Personally, I’d be very curious to see this plotted on a logarithmic graph. My hunch is that the oscillations are due to the discrete nature of the problem, when the border between the yellow and blue regimes in that first graph shifts.

In any case, I have been informed that the Birds are doing a little better as of late. Those Los Angeles Lightning Bolts, on the other hand … not so much.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{15} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Last Sunday we lost Alex Trebek, a giant in the world of game shows and trivia. The show he hosted over the course of four decades — Jeopardy! — has previously appeared in this column. Today, it makes a return.

You’re playing the (single) Jeopardy! Round, and your opponents are simply no match for you. You choose first and never relinquish control, working your way horizontally across the board by first selecting all six $200 clues, then all six $400 clues, and so on, until you finally select all the $1,000 clues. You respond to each clue correctly before either of your opponents can.

One randomly selected clue is a Daily Double. Rather than award you the prize money associated with that clue, it instead allows you to double your current winnings or wager up to $1,000 should you have less than that. Being the aggressive player you are, you always bet the most you can. (In reality, the Daily Double is more likely to appear in certain locations on the board than others, but for this problem assume it has an equal chance of appearing anywhere on the board.)

How much money do you expect to win during the Jeopardy! round?

*Extra credit:* Suppose you change your strategy. Instead of working your way horizontally across the board, you select random clues from anywhere on the board, one at a time. Now how much money do you expect to win during the Jeopardy! round?

The solution to this Riddler Express can be found in the following week’s column.

From Angela Zhou comes a bad football puzzle. The puzzle’s great, but the football is bad:

Football season is in full swing, and with it have been some incredible blown leads. The Atlanta Falcons know a few things about this, not to mention a certain Super Bowl from a few years back. Inspired by these improbabilities, Angela wondered just how likely one blown lead truly is.

The Georgia Birds and the Michigan Felines play a game where they flip a fair coin 101 times. In the end, if heads comes up at least 51 times, the Birds win; but if tails comes up at least 51 times, the Felines win.

What’s the probability that the Birds have at least a 99 percent chance of winning at *some point* during the game — meaning their probability of victory is 99 percent or greater given the flips that remain — and then proceed to lose?

*Extra credit: *Instead of 101 total flips, suppose there are many, many more (i.e., consider the limit as the number of flips goes to infinity). Again, the Birds win if heads comes up at least half the time. *Now* what’s the probability that the Birds have a win probability of at least 99 percent at some point and then proceed to lose?

The solution to this Riddler Classic can be found in the following week’s column.

Congratulations to 👏 Christoph 👏 of Southampton, United Kingdom, winner of last week’s Riddler Express.

Last week, FiveThirtyEight’s editor Santul Nerkar completed two 20-mile runs on a treadmill while training for the New York City Marathon. For the first run, he set the treadmill to a constant speed so that he ran every mile in 9 minutes.

The second run was a little different. He started at a 10 minute-per-mile pace and accelerated continuously until he was running at an 8-minute-per mile pace at the end. Moreover, Santul’s minutes-per-mile pace (i.e., *not* his speed) changed linearly over time. So a quarter of the way through the duration (in time, not distance) of the run, his pace was 9 minutes, 30 seconds per mile, halfway through it was 9 minutes per mile, etc.

Which training run was faster (i.e., took less time) overall? And what were Santul’s times for the two runs?

The first run wasn’t particularly tricky. If Santul ran 20 miles and each mile took 9 minutes, then he ran for a total of **180 minutes**, or 3 hours.

It was the second run that tripped a few readers up. His initial pace was 10 minutes per mile, meaning he was running 6 miles per hour. His final pace was 8 minutes per mile, or 7.5 miles per hour. At this point, you might have assumed that his speed steadily (i.e., linearly) increased from 6 to 7.5 miles per hour. But that was not the case!

As the problem stated, it was Santul’s minute-per-miles *pace* that steadily decreased (i.e., he sped up) from 10 minutes per mile down to 8 minutes per mile. If it took him *T* minutes to run the 20 miles, then his pace after *t* minutes was 10−2*t*/*T* minutes per mile. (You can check that this equals 10 when *t* = 0, and 8 when *t* = *T*.) If his pace was 10−2*t*/*T* minutes per mile, then what was his speed? It was 60/(10−2*t*/*T*), which gave you the correct units of miles per hour.

At this point, you knew Santul’s speed as a function of time. To figure out how long he was running for, you had to integrate this speed over time — yes, you needed calculus — and you had to set that equal to 20, his distance in miles. This is precisely what solver Eli Wolfhagen did, and solving this integral for *T *gave the correct answer: 2/(3·ln(5/4)) hours, or about **179.26 minutes**. Accelerating this way meant that Santul ran about 45 seconds faster, which can feel like an eternity when running a marathon.

No matter how he trained, I think it’s fair to say that Santul ran one hell of a race!

Congratulations to 👏 Nick Meyer 👏 of Oakland, California, winner of last week’s Riddler Classic.

Last week, inspired by John von Neumann, you tried to simulate a fair coin using a biased coin that had a probability *p* of landing heads.

Suppose you wanted to simulate a fair coin in at most three flips. For which values of *p* was this possible?

Von Neumann’s approach worked for any value of *p*, but didn’t guarantee a simulation that lasted a finite number of flips. However, for certain values of *p*, it was indeed possible to simulate a fair join in just a few flips.

Of course, with one flip there was only one way to simulate a fair coin — when *p* was 1/2.

With two flips, things got a little more complicated. When *p* was 1/2, you could still simulate a fair coin (e.g., the same outcome on both flips could simulate heads, while different outcomes simulated tails). But there were two other values of *p* that were also possible. When *p* was 1/√2, the probability of getting two heads (HH) was 1/2, while the probability of getting anything else (HT, TH or TT) was also 1/2. You could similarly simulate a fair coin when *p* was 1−1/√2, in which case the probability of getting TT was 1/2.

So with two coin flips, there were three values of *p* that could be used to simulate a fair coin: 1/2, 1/√2 and 1−1/√2. Just one problem — the riddle asked about *three* flips, not two. But at this point, your strategy was set. You were looking for different ways to cobble together different outcomes so that their combined probabilities equaled exactly 1/2.

With three coin flips, there were eight total outcomes:

- HHH, which had probability
*p*^{3} - HHT, HTH and THH, which each had probability
*p*^{2}(1−*p*) - HTT, THT and TTH, which each had probability
*p*(1−*p*)^{2} - TTT, which had probability (1−
*p*)^{3}

From here, it was a matter of adding these probabilities together and seeing when these sums equaled 1/2. In all, there were 64 ways to combine them: *p*^{3}, *p*^{3} + *p*^{2}(1−*p*), *p*^{3} + 2*p*^{2}(1−*p*), etc. Here are all 64 polynomials plotted when *p* was between 0 and 1:

It’s a little hard to see from the graph (solver David Lewis zooms in on a similar graph), but there were 19 locations where these curves cross the 0.5 mark, meaning there were **19** distinct values of *p* that could simulate a fair coin in at most three tosses. Solver Josh Silverman used a combinatorial approach to count up the solutions without needing to find them all, while Emma Knight was able to list them all out.

For extra credit, you were asked to simulate a fair coin in at most *N* flips, rather than just three flips. For how many values of *p* is this possible?

We already saw that for one flip, there was one value of *p*. For two flips, there were three values. And for three flips, there were 19 values. Brave souls who looked at four flips found there were 271 possible values of *p*, while there were 8,635 values for five flips and 623,533 values for six flips. As it often turns out in this column, there was an OEIS sequence for that… but wait a minute — solver Tracy Hall just posted this sequence last weekend, meaning it was a brand new discovery. Amazing!

Solver Angela Zhou decided to plot a histogram showing all 623,533 values of *p* that could yield a fair coin in at most six flips:

There’s a lot of mathematical richness to unpack there, like the chasm surrounding 0.5 (although 0.5 is itself a solution), not to mention all those other peaks and valleys.

It’s hard to compete with von Neumann, but Riddler Nation has definitely given him a run for his money.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>For decades, I’ve tuned into the trivia game show “Jeopardy!” for the facts. On Sunday, the show lost its judicious leader, Alex Trebek, who died at age 80 after a battle with cancer.

Trebek had hosted “Jeopardy!” for my entire life. He began in 1984 and hosted every episode since — save for April Fool’s Day in 1997 when he and Pat Sajak of “Wheel of Fortune” swapped places — more than 8,000 half-hour shows in all. For many, watching Trebek was ritualistic — the way a day ends and a night begins. It was also a way to get real facts, a brief respite from the “alternative facts” so prevalent over the past few years.

Trebek and the show — the two difficult now to divorce — were a beacon of democratic ideals, flattening the world for all to consume. High and low, popular and obscure, new and old, holy and profane, Trebek put all of them on equal terms. Look no further than the episode of “Jeopardy!” that aired this past Friday, for example. It featured clues about Rihanna, Madonna and Katy Perry, and clues about the city of Vaduz, the Russian navy and the German chancellor. All were on the same game board — though Gene Wilder and the Cook Strait were worth more money than any of them.

“Jeopardy!” and Trebek have been a haven for facts as monuments of truth have crumbled in the public sphere. The holder of the country’s highest office now baselessly disputes legitimate election results, espousers of baseless and dangerous conspiracy theories are elected to Congress, and fact-checking is a booming industry.

Trebek’s performance hasn’t changed to combat the untruths of the last few years, in part because it already was a kind of totem to the value of fact. Since 1984, he has stood each weekday in a suit behind a podium with a sheet of notes and delivered more than 400,000 clues, each a minor daily inoculation against the creep of lies — or whatever you want to call them.

“Alex was so much more than a host,” tweeted James Holzhauer, who set a series of unreal records on “Jeopardy!” last spring. “He was an impartial arbiter of truth and facts in a world that needs exactly that.”

The show’s eclectic and unpredictable subject matter is the result of an unassuming, wide-angle lens cast upon a large and complicated world. We call the material Trebek delivered “trivia,” but few things are less trivial than turning a generous eye toward the world — in all its strange and diverse splendor — and calling things by their right name.

The headline on a great 2014 profile of Trebek in the New Republic declared him the “Last King of the American Middlebrow.” This is true not in a pejorative sense but in a statistical one, as in the average between high and low. Trebek’s essential demeanor — and therefore “Jeopardy!” itself — is stripped of pretension and pretext.

The show’s archive showcases its diverse interests, each of which was presented on the show’s giant board simply and equally in the iconic all-caps white text on a blue square. Knowledge of the former British Prime Minister Benjamin Disraeli is worth the same as that of “The Office” actor Steve Carrell, as that of the toy Mr. Potato Head. The playwright Shakespeare is equal to the running back Walter Payton. The element helium is as valuable as the airport Heathrow.

Trebek’s performance as host emphasized this egalitarianism. His affect was steady and his pronunciation — of French, most famously — was impeccable, and he took pre-show notes with diacritical marks to ensure that he’d get it right. And at the same time he was, among other things, an underrated rapper, as Holzhauer joked, enthusiastically engaging with the repertoires of Lil Wayne, Drake and Kendrick Lamar.

“I’m not too good at it, but I was getting into it,” Trebek said after delivering the verses, and I believed him.

Moreover, Trebek’s performance didn’t just lack pretension, it was *anti*-pretentious. He once famously lightly chastised three contestants, who had correctly answered clues about Molière and Thor Heyerdahl, for not knowing enough about football. “I have to talk to them,” Trebek said disapprovingly before going to commercial. The democratic nature of trivia had been thrown off-kilter, and Trebek had to correct it.

By all accounts, Trebek embodied these trivia ideals while also being a genuine and gracious human.

“Alex wasn’t just the best ever at what he did,” tweeted Ken Jennings, the show’s most famous contestant, who won a record 74 consecutive times. “He was also a lovely and deeply decent man, and I’m grateful for every minute I got to spend with him.”

It was with this same ethos of decency and calm that Trebek announced, in March 2019, that he had been diagnosed with stage 4 pancreatic cancer. He said he “wanted to prevent you from reading or hearing some overblown or inaccurate reports regarding my health.” The rest of the short statement was filled with facts, and he vowed to fight the cancer and continue working. “Truth told, I *have* to,” he said, with a characteristic pinch of humor. “Because under the terms of my contract, I have to host ‘Jeopardy!’ for three more years.”

I wish he were able to do so. The show may go on, but it’s hard to imagine anyone personifying its ideals as well as Alex Trebek did. Of course, the contestants on “Jeopardy!” just ask the questions. It was Trebek who had all the answers.

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{16} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Last weekend’s New York City Marathon was canceled. But runners from Des Linden — one of the top American marathoners — to FiveThirtyEight’s own Santul Nerkar — my number one editor — still went out there and braved the course. Santul finished in a time of 3:41:43 (3 hours, 41 minutes, 43 seconds), which averaged to 8 minutes, 27 seconds per mile.

Suppose, while training, Santul completed two 20-mile runs on a treadmill. For the first run, he set the treadmill to a constant speed so that he ran every mile in 9 minutes.

The second run was a little different. He started at a 10 minute-per-mile pace and accelerated continuously until he was running at an 8-minute-per mile pace at the end. Moreover, Santul’s minutes-per-mile pace (i.e., *not* his speed) changed linearly over time. So a quarter of the way through the duration (in time, not distance) of the run, his pace was 9 minutes, 30 seconds per mile, halfway through it was 9 minutes per mile, etc.

Which training run was faster (i.e., took less time) overall? And what were Santul’s times for the two runs?

Mathematician John von Neumann is credited with figuring out how to take a biased coin (whose probability of coming up heads is *p*, not necessarily equal to 0.5) and “simulate” a fair coin. Simply flip the coin twice. If it comes up heads both times or tails both times, then flip it twice again. Eventually, you’ll get two different flips — either a heads and then a tails, or a tails and then a heads, with each of these two cases equally likely. Once you get two different flips, you can call the second of those flips the outcome of your “simulation.”

For any value of *p* between zero and one, this procedure will always return heads half the time and tails half the time. This is pretty remarkable! But there’s a downside to von Neumann’s approach — you don’t know how long (i.e., how many flips) the simulation will last.

Suppose I want to simulate a fair coin in at most* three* flips. For which values of *p* is this possible?

Extra credit: Suppose I want to simulate a fair coin in at most* N *flips. For how many values of *p* is this possible?

Congratulations to 👏 Nilesh Shah 👏 of Seattle, Washington, winner of last week’s Riddler Express.

Last week, while waiting in line to vote early, I overheard a discussion between election officials. Apparently, there was a political sign within 100 feet of the polling place, against New York State law.

This got me thinking. Suppose a polling site was a square building whose sides were 100 feet in length. An election official took a string that was also 100 feet long and tied one end to a door located in the middle of one of the building’s sides. She then held the other end of the string in her hand.

What was the area of the region outside of the building she could reach while holding the string?

Had there been no building, but if she were still constrained to be within 100 feet of a fixed point, then the region she could have reached was anywhere within a 100-foot circle, whose area was 𝜋(100)^{2}, or about 31,416 square feet. Due to the building, that area served as an upper bound, meaning the answer had to be less than that.

At this point, there were two approaches. A few readers subtracted the overlapping area between the building and the 100-foot circle. The geometry here was a little involved, but it yielded an answer of (5𝜋/6−√3/4)·100^{2}, or about 21,849 square feet. However, this wasn’t quite right either.

Why was that? Because this assumed that the string was able to pass through the building. While it *was* Halloween last week, that sort of “ghostly” string behavior would simply have been too paranormal. So whenever the election official lost line of sight with the door where the string was tied, 50 feet of string were taut against the wall of the building, leaving her with a 50-foot radius with which to move around.

Solver Tamera Lanham sketched out the accessible area:

The red region on the right was a semicircle with a radius of 100 feet, meaning its area was 𝜋(100)^{2}/2, or about 15,708 square feet. Meanwhile, the two blue regions were both quarter-circles with a radius of 50 feet. Together, these quarter-circles formed a semicircle whose area was 𝜋(50)^{2}/2, or about 3,927 square feet.

All together, the area of the region the election official could reach was **6,250𝜋**, or about 19,635 square feet.

This riddle was a version of what are known as goat problems, in which a goat is tethered to part of a fence and you are tasked with calculating its grazing area. If you thought the square shape of the building wasn’t challenging enough, try re-running the problem for a circular building. It’s tricky!

Congratulations to 👏 Brian Corrigan 👏 of Los Angeles, California, winner of last week’s Riddler Classic.

Last week, you and 60 of your closest friends decided to play a socially distanced game of hot pumpkin.

Before the game started, you all sat in a circle and agreed on a positive integer *N*. Once the number had been chosen, you (the leader of the group) started the game by counting “1” and passing the pumpkin to the person sitting directly to your left. She then declared “2” and passed the pumpkin one space to *her* left. This continued with each player saying the next number in the sequence, wrapping around the circle as many times as necessary, until the group had collectively counted up to *N*. At that point, the player who counted “*N*” was eliminated, and the player directly to his or her left started the next round, again proceeding to the same value of *N*. The game continued until just one player remained, who was declared the victor.

In the game’s first round, the player 18 spaces to your left was the first to be eliminated. Ricky, the next player in the sequence, began the next round. The second round saw the elimination of the player 31 spaces to Ricky’s left. Zach began the third round, only to find himself eliminated in a cruel twist of fate. (Woe is Zach.)

What was the smallest value of *N* the group could have used for this game?

A good first step was to nail down the important information from the problem. In the first round, when there were 61 players, the pumpkin went around the circle some number of times (it could have been zero times, once, twice, etc.) and then 18 more spaces. In other words, *N* had to be some multiple of 61 (one multiple for each time the pumpkin went around the circle) *plus* 19. In modular arithmetic notation, this meant that *N* ≡ 19 (mod 61).

Why did you have to add 19, and not 18? Because after the pumpkin had finished going around the circle, *you* and the 18 people to the left of you all counted off. And you plus the 18 others made 19 people in total.

In the second round, when there were 60 players, the pumpkin again went around some number of times and then 31 more spaces. That told you that *N *≡ 32 (mod 60). And in the third round, when poor Zach took himself out of the competition, you learned that *N *was 1 more than a multiple of 59, i.e., *N* ≡ 1 (mod 59).

Because the numbers 59, 60 and 61 were all relatively prime, the Chinese remainder theorem told you that there was a single value of *N* between 1 and 59·60·61, or 215,940, that satisfied all three of the aforementioned modular relations. But while it was one thing to prove there was a solution, finding it was another matter entirely.

One way to do this was through brute force coding — searching for the one number less than 215,940 that had a remainder of 19 when divided by 61, 32 when divided by 60 and 1 when divided by 59. That smallest number was **136,232**. Thanks to the wonders of modular arithmetic, any multiple of 215,940 plus 136,232 would also have resulted in the same sequence of eliminations.

For extra credit, you had to play this game to its conclusion, where you were identified as Player No. 1, the player to your left as Player No. 2 and so on. Once again, your computer was your friend. If you simulated the rest of the game, it was **Player No. 58** who emerged as the winner.

And for extra extra credit, you had to find the smallest value of *N* for which you, Player No. 1, won the game. This question was ambiguous as to whether you simply wanted the smallest value of *N* or the smallest value of *N* that also resulted in the eliminations listed in the original problem. The answer to the former was **140**, while the answer to the latter was a whopping **42,892,352**. Only five solvers found this latter result: Angela Zhou, Peter from Hanover, Germany, Emma Knight, Jim Waters and Asha O’Shaughnessy. They will each be winning an imaginary hot pumpkin!

Finally, hats off to those who attempted this not with code, but by hand or with spreadsheets. One such solver, Shane Tilton, said he was content with simply having been invited to play a game of hot pumpkin in the first place.

Who needs to play a game that requires 42 million turns to win when, more importantly, you have 60 good friends to play it with?

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{17} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

While waiting in line to vote early last week, I overheard a discussion between election officials. Apparently, there may have been a political sign that was within 100 feet of the polling place, against New York State law.

This got me thinking. Suppose a polling site is a square building whose sides are 100 feet in length. An election official takes a string that is also 100 feet long and ties one end to a door located in the middle of one of the building’s sides. She then holds the other end of the string in her hand.

What’s the area of the region outside of the building she can reach while holding the string?

The solution to this Riddler Express can be found in the following week’s column.

From Ricky Jacobson comes a party game that’s just in time for Halloween:

Instead of playing hot potato, you and 60 of your closest friends decide to play a socially distanced game of hot pumpkin.

Before the game starts, you all sit in a circle and agree on a positive integer *N*. Once the number has been chosen, you (the leader of the group) start the game by counting “1” and passing the pumpkin to the person sitting directly to your left. She then declares “2” and passes the pumpkin one space to *her* left. This continues with each player saying the next number in the sequence, wrapping around the circle as many times as necessary, until the group has collectively counted up to *N*. At that point, the player who counted “*N*” is eliminated, and the player directly to his or her left starts the next round, again proceeding to the same value of *N*. The game continues until just one player remains, who is declared the victor.

In the game’s first round, the player 18 spaces to your left is the first to be eliminated. Ricky, the next player in the sequence, begins the next round. The second round sees the elimination of the player 31 spaces to Ricky’s left. Zach begins the third round, only to find himself eliminated in a cruel twist of fate. (Woe is Zach.)

What was the smallest value of *N* the group could have used for this game?

*Extra credit:* Suppose the players were numbered from 1 to 61, with you as Player No. 1, the player to your left as Player No. 2 and so on. Which player won the game?

*Extra extra credit:* What’s the smallest *N* that would have made *you* the winner?

The solution to this Riddler Classic can be found in the following week’s column.

Congratulations to 👏 Eric Mentzell 👏 of Takoma Park, Maryland, winner of last week’s Riddler Express.

Last week, four-time WNBA champion Sue Bird and Seattle Storm teammate Breanna Stewart were interested in testing whether Bird had a “hot hand” — that is, if her chances of making a basket depended on whether or not her previous shot went in. Bird happened to know that her chances of making any given shot was *always* 50 percent, independent of her shooting history, but she agreed to perform an experiment.

In each trial of the experiment, Bird took three shots, while Stewart recorded which shots Bird made or missed. Stewart then looked at all the trials with at least one shot that was immediately preceded by a made shot. She randomly picked one of these trials, and then randomly picked a shot that was immediately preceded by a made shot. (If there was only one such shot to pick from, she chose that shot.)

What was the probability that Bird made the shot that Stewart picked?

At first glance, you might have thought that the answer was 50 percent. After all, Bird acknowledged that she had a 50 percent chance of making any given shot. Right?

Wrong. And Riddler Nation was not fooled. To see why the answer wasn’t 50 percent, many solvers, like Deborah Abel, listed out all eight possible shot sequences. If we let “X” represent a made shot and “O” a missed shot, then the eight possible shot sequences were: OOO, OOX, OXO, XOO, OXX, XOX, XXO, and XXX. Because Bird had an equal chance of making or missing any shot, all eight of these shot sequences were equally likely.

During her analysis, Stewart first looked at shots that were “immediately preceded by a made shot.” This didn’t occur for sequences OOO and OOX, so we know that Stewart was limiting her analysis exclusively to the remaining six sequences, and that each had a one-in-six chance of being selected.

Next, Stewart “randomly picked a shot that was immediately preceded by a made shot.” Here’s what happened if you looked at the six possible sequences one at a time:

- OXO: Stewart looked at the last shot. For this sequence, Bird made zero percent of the shots Stewart could have picked.
- XOO: Stewart looked at the second shot. For this sequence, Bird made zero percent of the shots Stewart could have picked.
- OXX: Stewart looked at the last shot. For this sequence, Bird made 100 percent of the shots Stewart could have picked, which gives a 1/6 chance of selecting a made shot in this sequence.
- XOX: Stewart looked at the second shot. For this sequence, Bird made zero percent of the shots Stewart could have picked.
- XXO: Stewart could have looked at the second or third shots. For this sequence, Bird made 50 percent of the shots Stewart could have picked, which gives a 1/12 chance of selecting a made shot in this sequence.
- XXX: Stewart could have looked at the second or third shots. For this sequence, Bird made 100 percent of the shots Stewart could have picked, which gives a 1/6 chance of selecting a made shot in this sequence.

Putting this all together, the probability Bird had made the shot that Stewart picked was 1/6 + 1/12 + 1/6, or **5/12** — the solution!

So while Bird had a 50 percent chance of making any given shot, Stewart’s methodology for selecting a shot was somehow biased toward shots that Bird missed. What was going on here?

Solver Madeline Argent of Launceston, Australia offered an explanation. Among the six shot sequences listed above, there were eight cases in which Bird made one of her first two shots. Of these eight, Bird made four of the next shots and missed four of them. So if Stewart had selected among all *shots* that were immediately preceded by a made shot, Bird would have made half the selected shots. But because Stewart first randomly selected a *trial*, the last two shot sequences — XXO and XXX, in which Bird made most of her shots — were unfairly weighted equally alongside the other four sequences, even though they had twice as many made shots for Stewart to choose from.

That was what happened when Bird took three shots per trial. Meanwhile, solver Dean Ballard found similar results when Bird took more shots. The probability she had made a selected shot approached 50 percent as the number of shots increased, but it never quite reached 50 percent.

Clearly, this methodology for determining whether a basketball player had a “hot hand” was flawed. It may surprise some readers that this was precisely the methodology used in an attempt to debunk the “hot hand” back in 1985 — a debunking that itself was later debunked. If you’d like to read more on this, check out this 2019 article that connects all of this to the Monty Hall problem. (Kudos to submitter Drew Mathieson for the link!)

So if Stewart performed the experiment as outlined, and found that Bird had made *half* the selected shots (rather than 5/12 of them), she would have rightfully concluded that Bird *did* have a “hot hand.”

Congratulations to 👏 Lucas Robinson 👏 of Oakwood, Ohio, winner of last week’s Riddler Classic.

Last week, four-time NBA champion LeBron James was playing a game of sudden-death, one-on-one basketball with Los Angeles Lakers teammate Anthony Davis. They flipped a coin to see which of them had first possession, and whoever made the first basket won the game.

Both players had a 50 percent chance of making any shot they took. However, Davis was the superior rebounder and would always rebound any shot that either of them missed. Every time Davis rebounded the ball, he dribbled back to the three-point line before attempting another shot.

Before each of Davis’s shot attempts, James had a probability *p* of stealing the ball and regaining possession before Davis could get the shot off. What value of *p* made this an evenly matched game of one-on-one, in which both players had an equal chance of winning *before* the coin was flipped?

Suppose James’s probability of winning when he had possession was *J*, while James’s probability of winning when *Davis* had possession was *D*. We can write an equation for *J*: For James to win when he had possession, he either had to score (with probability 1/2) or, upon missing (also with probability 1/2), he’d have to somehow win after Davis got the rebound and gained possession. In other words, *J* = 1/2 + 1/2·*D*. We can similarly write an equation for *D*: For James to win when Davis had possession, he either had to steal the ball (with probability *p*) and regain possession, or, upon not getting the steal (with probability 1−*p*), he needed Davis to miss (with probability 1/2) so James could have another chance at victory. In other words, *D* = *pJ* + (1−*p*)*D*/2.

That gave you two equations with three unknowns. What was missing? Based on the coin flip, James started with possession half the time, and Davis started with possession the other half the time. James’s probability of winning was therefore *J*/2 + *D*/2, which the problem stated was to equal 1/2, so the third equation was *J*/2 + *D*/2 = 1/2, or *J* + *D* = 1.

Solving this system of three equations gave you the result that *J* = 2/3 (James had a two-thirds chance of winning when he had possession), *D* = 1/3 (James had a one-third chance of winning when Davis had possession) and *p* = 1/3. In other words, for the game to be fair, James had to have a **one-third** chance of stealing the ball.

Solver Quoc Tran extended the puzzle, simulating how James’s fortunes changed in the more realistic scenario where Davis didn’t nab *every* rebound. For the game to be fair, we already saw that James needed to steal one-third of the time when Davis rebounded every shot. Meanwhile, if James and Davis had had equal chances of getting a rebound, then to maintain parity, James couldn’t steal at all. But when Davis had a rebounding edge, interesting mathematics was happening, as shown below:

How fitting that this game was fair where purple met gold.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{18} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Drew Mathieson comes an exploration of basketball’s historied hot hand:

This season, on the way to winning her fourth WNBA championship in her 17-year career, Sue Bird made approximately 50 percent of her field goal attempts. Suppose she and Seattle Storm teammate Breanna Stewart are interested in testing whether Bird has a “hot hand” — that is, if her chances of making a basket depend on whether or not her previous shot went in. Bird happens to know that her chances of making any given shot is *always* 50 percent, independent of her shooting history, but she agrees to perform an experiment.

In each trial of the experiment, Bird will take three shots, while Stewart will record which shots Bird made or missed. Stewart will then look at all the trials that had at least one shot that was immediately preceded by a made shot. She will randomly pick one of these trials and then randomly pick a shot that was preceded by a made shot. (If there was only one such shot to pick from, she will choose that shot.)

What is the probability that Bird made the shot that Stewart picked?

The solution to this Riddler Express can be found in the following week’s column.

Now that LeBron James and Anthony Davis have restored the Los Angeles Lakers to glory with their recent victory in the NBA Finals, suppose they decide to play a game of sudden-death, one-on-one basketball. They’ll flip a coin to see which of them has first possession, and whoever makes the first basket wins the game.

Both players have a 50 percent chance of making any shot they take. However, Davis is the superior rebounder and will always rebound any shot that either of them misses. Every time Davis rebounds the ball, he dribbles back to the three-point line before attempting another shot.

Before each of Davis’s shot attempts, James has a probability *p* of stealing the ball and regaining possession before Davis can get the shot off. What value of *p* makes this an evenly matched game of one-on-one, so that both players have an equal chance of winning *before* the coin is flipped?

The solution to this Riddler Classic can be found in the following week’s column.

Congratulations to 👏 Frank Probst 👏 of Houston, Texas, winner of last week’s Riddler Express.

Last week, the residents of Riddler City were electing a mayor from among three candidates. The winner was the candidate who received an outright majority (i.e., more than 50 percent of the vote). But if no one achieved this outright majority, there would be a runoff election among the top two candidates.

If the voting shares of each candidate were uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what was the probability of a runoff?

The “uniformly distributed” wording in the problem was ambiguous and was interpreted several ways by readers. How can you randomly choose three numbers between 0 and 100 that add up to 100? Here, I will write about three popular interpretations.

First, imagine randomly picking three values between 0 and 100, and call them *x*, *y* and *z*. Each choice of (*x*, *y*, *z*) corresponded to a point in a cube that measured 100 by 100 by 100. But only *some* points in this cube had coordinates that summed to 100 — those that also lay on the plane *x*+*y*+*z* = 100. This intersection between a cube and a plane might have been hard to visualize — it was an equilateral triangle (shown below). If you divided this triangle into quarters, then three of those quarters had one value (*x*, *y* or *z*) that exceeded 50. That meant the probability of a runoff, with *no* voting shares that exceeded 50, was **1/4**.

That was one interpretation. Another way to “uniformly” pick three values was to draw a number line between 0 and 100 and break it into three segments by randomly picking two points, *a* and *b*. Assuming *b* was greater than *a*, the three lengths that summed to 100 were *a*, *b*−*a* and 100−*b*. The challenge was to find when each of these three values exceeded 50 inside the triangle defined by 0 ≤ *a *≤ *b* ≤ 100. Each of the three inequalities (a ≥ 50, *b*−*a* ≥ 50 and 100−*b* ≥ 50) carved out a quarter of the larger triangle. And so, once again, that meant the probability of a runoff was **1/4**.

Yet another way to “uniformly” pick three values was to go ahead and pick three numbers between 0 and 100 (again, let’s call them *x*, *y* and *z*) and then “normalize” them — that is, divide each number by *x*+*y*+*z* and multiply by 100, so that they were guaranteed to add up to 100. Like before, each choice of (*x*, *y*, *z*) corresponded to a point in a cube. But this time, to avoid a runoff, you needed one of the values to exceed the sum of the other two, meaning it would exceed 50 percent of the sum of all three numbers. There were three regions in the cube where a runoff would *not* occur: *x* > *y*+*z*, *y* > *x*+*z* and *z* > *x*+*y*. Each region made up one-sixth of the cube, so all together they represented half the cube, as shown below. In the other half, a runoff was necessary. So according to this interpretation of the problem, the answer was **1/2**. (Some solvers, like Benjamin Dickman, noted that this approach was identical to finding the probability that three lengths from a random, uniform distribution satisfy the triangle inequality.)

For extra credit, you wanted the probability of a runoff when there were *N* candidates instead of three. Once again, the answer depended on your interpretation of the problem. Based on the first interpretation, this general solution was 1−*N*/2^{N−1} (nicely explained by Josh Silverman), since each of the N candidates had a 1/2^{N−1} chance of winning an outright majority. Similarly, based on the second interpretation, the answer was again 1−*N*/2^{N−1}. But based on the third interpretation, the answer was 1−1/(*N*−1)!.

That’ll do it for Riddler City’s mayoral election. Don’t forget to vote in *any other elections* that may be happening!

Congratulations to 👏 Asher S. 👏 of Chicago, Illinois, winner of last week’s Riddler Classic.

Last week, you were playing a modified version of “The Price is Right.” In this version’s bidding round, you and two (not three) other contestants had to guess the price of an item, one at a time.

The true price of this item was a randomly selected real number between 0 and 100. Among the three contestants, the winner was whoever guessed the closest price *without going over*. For example, if the true price was 29 and you guessed 30, while another contestant guessed 20, then they would be the winner even though your guess was technically closer.

In the event all three guesses exceeded the actual price, the contestant who had made the lowest (and therefore closest) guess was declared the winner.

If you were the first to guess, and all contestants played optimally (taking full advantage of the guesses of those who had gone before them), what were your chances of winning?

At first, this three-player game might have seemed unsolvable. As the first to guess, you would want to know what the second and third players’ strategies would be. But their strategies depended on yours and on each other’s. Was there any way out of this mess?

Indeed there was. One approach was to work backwards. Suppose you (the first player) guessed a price *A* and the second player guessed a price *B*. What should the third player do? For now, let’s assume *A* was less than *B*. The third player would then choose from among three options:

- Guess a value of zero, in which case they’d win if the true price was between 0 and
*A*— a range of*A*. - Guess a value infinitesimally greater than A, in which case they’d win if the true price was between
*A*and*B*— a range of*B*−*A*. - Guess a value infinitesimally greater than
*B*, in which case they’d win if the true price was between*B*and 100 — a range of 100−*B*.

But which of the three values should the third player guess? Whichever corresponded to the greatest range. (If *A* had been greater than B, there would again be three options, but with *A* and *B* reversed.)

So for any combination of *A* and *B*, the third player’s strategy was known. Next, it was time to look more closely at the second player.

For each value of *A*, the second player could figure out their chances of winning for any *B *they picked, since they now knew what the third player would do given *A* and *B*. For each *A*, the second player would then pick a *B* that maximized their own chances of winning.

At last, we’re back to you, the first player. By now, you knew exactly what the second and third players would do in response to any guess *A*. As with the second player, that meant you had to pick the value that maximized your own chances of winning.

Amidst all this strategizing, I neglected to mention just what these optimal guesses were. In the end, your best guess was two-thirds of 100 (~66.7). Then the second player’s best move was to guess one-third of 100 (~33.3), and the third player’s best move was to guess anything less than that (e.g., zero). All players had a **one-third** chance of winning. If you deviated from these optimal values, the second and third players would both have the advantage over you, each with a greater than one-third chance of winning.

Some solvers said they would have guessed a price that was *one*-third of 100. But as Emma Knight observed, Player 2 could then have guessed anything less than that and Player 3 slightly more. That would have meant Player 2 *still* had a one-third chance of winning, while Player 3’s chances went up to two-thirds, leaving you with nothing. Player 2 might not have chosen to sabotage your hopes of winning, but why leave it to chance?

Finally, Keith Wynroe of Edinburgh, Scotland offered a neat extension to this problem, asking how the three players’ strategies might change if the goal was not simply maximizing one’s chances of winning, but rather maximizing the *expected value* of the prize won. According to Keith, while this shift would incentivize all three players to bet higher values, no one’s chances of winning actually changed.

After all was said and done, it turned out to be a fair game. How sweet.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{19} and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

As you may have seen in FiveThirtyEight’s reporting, there’s an election coming up. Inspired, Vikrant Kulkarni has an electoral enigma for you:

On Nov. 3, the residents of Riddler City will elect a mayor from among three candidates. The winner will be the candidate who receives an outright majority (i.e., more than 50 percent of the vote). But if no one achieves this outright majority, there will be a runoff election among the top two candidates.

If the voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what is the probability of a runoff?

*Extra credit:* Suppose there are *N* candidates instead of three. What is the probability of a runoff?

The solution to this Riddler Express can be found in the following week’s column.

This week, we return to the brilliant and ageless game show, “The Price is Right.” In a modified version of the bidding round, you and two (not three) other contestants must guess the price of an item, one at a time.

Assume the true price of this item is a randomly selected value between 0 and 100. (Note: The value is a real number and does not have to be an integer.) Among the three contestants, the winner is whoever guesses the closest price *without going over*. For example, if the true price is 29 and I guess 30, while another contestant guesses 20, then they would be the winner even though my guess was technically closer.

In the event all three guesses exceed the actual price, the contestant who made the lowest (and therefore closest) guess is declared the winner. I mean, *someone* has to win, right?

If you are the first to guess, and all contestants play optimally (taking full advantage of the guesses of those who went before them), what are your chances of winning?

The solution to this Riddler Classic can be found in the following week’s column.

Congratulations to 👏 Nick Russell 👏 of Vancouver, Canada, winner of last week’s Riddler Express.

Last week, you were helping me park across the street from a restaurant for some contactless curbside pickup. There were six parking spots, all lined up in a row.

While I *could* parallel park, it definitely wasn’t my preference. No parallel parking was required when the rearmost of the six spots was available or when there were two consecutive open spots. If there was a random arrangement of cars occupying four of the six spots, what was the probability that I had to parallel park?

With six spots and four cars, there were 6 choose 4, or 15 cases to consider. Some solvers, like Lisa Fondren of Montrose, Michigan, listed them all out and counted how many required parallel parking.

But there were other ways to find the solution that didn’t require working through every case. Rather than considering where the four cars were, solver Libby Aiello equivalently looked at where the two empty spots were. Among the total 15 combinations, there were five in which the two empty spots were adjacent: the first and second spots, the second and third, the third and fourth, the fourth and fifth, and the fifth and sixth.

There were also five combinations in which the last spot was open, since there were five spots from which to choose the *other* open spot. Combining these two cases (having two consecutive open spots and having the sixth spot open), there appeared to be 10 combinations that didn’t require parallel parking.

But that wasn’t quite right. As Libby noted, one combination — when the fifth and sixth spots were open — was counted twice, since two consecutive spots were open *and* the sixth spot was open. Subtracting one to account for this double counting meant there were nine combinations that didn’t require parallel parking, and six combinations that did. Therefore, the probability I had to parallel park was 6/15, or **40 percent**.

I was pleased to see how many readers solved this combinatorics challenge. Now if only that many drivers could successfully parallel park…

Congratulations to 👏 Douglas Thackrey 👏 of Loulé, Portugal, winner of last week’s Riddler Classic.

Parking cars was one thing — parking trucks was another thing entirely. Last week, you looked at a *very long* truck (with length *L*) with two front wheels and two rear wheels. (The truck was so long compared to its width that you could consider the two front wheels as being a single wheel, and the two rear wheels as being a single wheel.)

You were asked to determine the truck’s turning radius, given the angles by which you could turn the front or rear wheels. First, you considered what would happen if the front wheels could be turned up to 30 degrees in either direction (right or left), but the rear wheels did not turn.

There was no doubt among readers that this was a geometry puzzle, but the challenge lay in translating the constraints on the wheels (i.e., only turning a certain amount or not at all) into the resulting motion of the truck.

As suggested by the term “turning radius,” the key was to think about the circular motion of the truck. When the driver rotated the front wheels a full 30 degrees in one direction and drove forward, both the front and the rear of the truck would move in circles. The front of the truck always made a 30 degree angle with the tangent line to the circle it was moving around.

Meanwhile, the rear wheels of the truck couldn’t turn. That meant the truck’s rear was always tangent to the circle it was moving around.

If that wasn’t clear, here’s an animation to illustrate how the truck was moving:

The green line segment represents the truck, and the circles represent the paths of the truck’s front and rear. Sure enough, the angle between the truck and the tangent line (represented by the white segment) is always 30 degrees. This means that the front of the truck is moving around a *wider* circle than the rear — and, consequently, that the front of the truck moves *faster* than the rear!

At this point, calculating the turning radius was a matter of geometry and trigonometry. If the green segment was doubled in length so that it formed a chord within the larger circle, the 30 degree angles meant that this chord was one side of an inscribed regular hexagon, whose sides all equal the circle’s radius. And so if the truck had length *L*, the turning radius — that is, the radius of the circle around which the front of the truck moved — was **2**** L**. (Solvers who gave the turning radius for the truck’s midpoint or rear and explained their reasoning were also given full credit.)

That was the case when you could only turn the front wheels. You were also asked for the turning radius when both the front and rear wheels could be independently turned up 30 degrees in either direction. To make the tightest possible circle, the front and rear wheels were both rotated the full 30 degrees, but in opposite directions, allowing the front and rear to move along the same circle. Again, here’s an animation:

This time, the truck made up a complete chord that was a side of an inscribed regular hexagon. That meant the turning radius was equal to the truck’s length, ** L**. (Again, solutions for different locations on the truck were also accepted.)

A few solvers, including Laurent Lessard and Josh Silverman, tackled the general version of this problem, in which the front wheels could be turned an angle *θ*_{1} and the rear wheels could be turned an angle *θ*_{2}. The turning radius at the front of the truck was *L*·cos(*θ*_{2})/sin(*θ*_{1}+*θ*_{2}), while the turning radius at the rear was *L*·cos(*θ*_{1})/sin(*θ*_{1}+*θ*_{2}).

These formulas checked out for both questions in the riddle. And when neither wheel could turn (i.e., *θ*_{1} and *θ*_{2} were both zero), the turning radius went to infinity, which also made sense. In that case, you’d just have to keep on truckin’.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>