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 next week’s column. Or in this case, two weeks — Happy Thanksgiving! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Dave Moran comes a question about baseball’s unusual 2019 World Series:

In the World Series, one team hosts Games 1, 2, 6 and 7, while the other team hosts Games 3, 4 and 5. When the Nationals beat the Astros last month, it marked the first time in World Series history that the home team lost all seven games. On average, the home team actually wins about 54 percent of the time in baseball. Running the numbers, you’ll quickly see that seven home losses is a once-in-a-lifetime event.

But putting seven aside for a moment, what’s the probability that the home team will lose at least *six* consecutive games?

*Extra credit:* What’s the probability the home team will lose at least five consecutive games? Four consecutive games?

## Riddler Classic

From Charlie Cordova comes a puzzle that brings logic and number theory to the lottery:

Five friends with a lot in common are playing the Riddler Lottery, in which each must choose exactly five numbers from 1 to 70. After they all picked their numbers, the first friend notices that no number was selected by two or more friends. Unimpressed, the second friend observes that all 25 selected numbers are composite (i.e., not prime). Not to be outdone, the third friend points out that each selected number has at least two distinct prime factors. After some more thinking, the fourth friend excitedly remarks that the product of selected numbers on each ticket is exactly the same. At this point, the fifth friend is left speechless. (You can tell why all these people are friends.)

What is the product of the selected numbers on each ticket?

*Extra credit:* How many *different ways* could the friends have selected five numbers each so that all their statements are true?

## Solution to last week’s Riddler Express

Congratulations to 👏 Robert Reid 👏 of North Bend, Washington, winner of last week’s Riddler Express.

Last week’s puzzle was a geometric conundrum inspired by Catriona Shearer:

In the picture above, you were told that the lighter region (inside the larger semicircle but outside the smaller one) had an area of 7. What was the area of the darker region?

Many readers pointed out that the riddle should have been more specific — in particular, the straight edges of the two semicircles are *parallel*. It’s possible to inscribe a non-parallel semicircle, as shown in the animation below, but that’s beyond the scope of this question.

Solver Ria Skies made quick work of this puzzle with some pen and paper:

Using the Pythagorean theorem, you can show that the radius of the larger semicircle is √2 times longer than the radius of the smaller semicircle. That means the *area* of the larger semicircle is twice that of the smaller semicircle — in other words, the area of the smaller semicircle is *half* the area of the larger semicircle.

Since the smaller semicircle takes up half the larger one, the remaining lighter area takes up the other half. And so the two regions — the lighter gray and the darker gray — have the same area! They both take up half the area of the larger semicircle. If the lighter region has area 7, then the darker region also has area **7**.

It’s almost as if the answer was hiding in plain sight all along…

## Solution to last week’s Riddler Classic

Congratulations to 👏 Jesse Cohn 👏 of St. Louis, Missouri, winner of last week’s Riddler Classic.

You were given a fair, unweighted 10-sided die with sides labeled 0 to 9 and a sheet of paper to record your score. To start the game, you rolled the die. Your current “score” was the number shown, divided by 10. For example, if you rolled a 7, then your score would be 0.7. Then, you kept rolling the die over and over again. Each time you rolled, if the digit shown by the die was less than or equal to the last digit of your score, then that roll became the *new* last digit of your score. Otherwise, you just went ahead and rolled again. The game ended when you rolled a zero. (If your first roll was a zero, your score was simply zero.) What was your *average* final score in this game?

First off, thank you to all the readers who shared links and photos of your favorite 10-sided dice and simulators of 10-sided dice. (It’s fair to say that the die in this problem likely came from a Dungeons & Dragons set.)

Getting back to the puzzle, solver Guy D. Moore made a key observation: Whenever you roll a number, from that point forward, it’s as though you were rolling a die with that many sides (since you ignore any higher numbers you might come across). This enabled him to set up what’s known as a “recurrence relation,” an equation that tells you later values in a sequence in terms of earlier values in the sequence.

It helps if we define *S*_{N} as the average score you’ll get when playing with a die numbered from 0 to *N*. So if you’re playing with die numbered from 0 to 9, your average score will be *S*_{9}. Another way to compute this average is to break it down into 10 cases, depending on what your first roll is. If your first roll is a 9, then your average score will be 0.9 *plus* the average of all the scoring you’ll do thereafter. Because all the numbers are still in play, the only thing that’s different now is that your scoring has moved to the right by one decimal place, meaning your rolls now count for one-tenth of what they used to. In other words, when your first roll is a 9, your average score will by 0.9 + 0.1*S*_{9}.

And what happens if your first roll is an 8? Your average score will be 0.8 plus the average of your scoring thereafter, which turns out to be 0.1*S*_{8}. Continuing all the way down to *S*_{0}, the resulting equation for *S*_{9} is — take a deep breath — *S*_{9} = 0.1(0.9 + 0.1*S*_{9}) + 0.1(0.8 + 0.1*S*_{8}) + 0.1(0.7 + 0.1*S*_{7}) + 0.1(0.6 + 0.1*S*_{6}) + 0.1(0.5 + 0.1*S*_{5}) + 0.1(0.4 + 0.1*S*_{4}) + 0.1(0.3 + 0.1*S*_{3}) + 0.1(0.2 + 0.1*S*_{2}) + 0.1(0.1 + 0.1*S*_{1}) + 0.1(0 + 0.1*S*_{0}). The coefficient of 0.1 in front of each term comes from the fact that each case — rolling a 9 first, rolling an 8 first, etc. — has an equal probability of one-tenth.

What a recursive mess. What if we were to look at smaller values of *N* instead? Well, we know *S*_{0} = 0, since it represents your average when all you can roll is a 0. The equation for *S*_{1}, the average score when you can only roll a 0 or a 1, is slightly more complicated: *S*_{1} = 0.5(0.1 + 0.1*S*_{1}) + 0.5(0 + 0.1*S*_{0}). Using the fact that *S*_{0} = 0, we can solve this equation for *S*_{1}, finding that it equals 1/19. Next, we can use these values for *S*_{0} and *S*_{1} to find *S*_{2}, which turns out to be 2/19. Lo and behold, a beautiful pattern emerges: *S*_{N} equals *N*/19. That means *S*_{9} equals **9/19**, our answer. Honestly, it’s pretty satisfying to see that in this game, where all the scores are decimals between 0 and 1, the average score winds up being fairly close to 0.5.

Meanwhile, solver Ignas from London took a different approach, noting that your first roll can be anywhere from 0 to 9 with equal probability, so the *average* value of the digit in the tenths place will be 4.5 (halfway between 0 and 9). Whatever your first roll was, your second roll is constrained to be between 0 and that first roll, so the average value of your second roll will again be halfway between 0 and your roll. This also means that the average digit in the hundredths place will be 2.25 (halfway between 0 and 4.5), the average digit in the thousandths place will be 1.125 (halfway between 0 and 2.25), and so on. Summing the resulting geometric series again gives an average final score of 9/19.

The puzzle’s author, Ricky Jacobson, also found the average final score in the general case of an *k*-sided die: (*k*−1)/(2*k*−1).

Finally, as I had hoped, several members of Riddler Nation turned to their computers for help, simulating loads and loads of games. Robert DiMartino kindly provided a link to his code that will run 10,000 games in under a second. Solver Angelos Tzelepis went so far as to play 3 million games, finding an average score of 0.47356, within spitting distance of 9/19. He even achieved a top score of 0.99999995. Impressive!

Solver Hernando Cortina plotted a histogram of his scores, revealing which are more likely, less likely and downright impossible:

Think that’s cool? It turns out that graph is actually fractal! As you zoom in on scores closer to 1, the pattern repeats itself. Crazy, I know.

## Want more riddles?

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!

## Want to submit a riddle?

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