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.

## Riddler Express

Football season is almost here, which means *fantasy* football season is already here.

You and your two best friends are in a three-person league, drafting just three positions each for your teams: quarterback, running back and wide receiver. Yes, this is a simplified version of fantasy football.

The following table shows the top three athletes in each position, as well as the number of fantasy points they are expected to earn over the course of the season. You and your friends must each select exactly one player from each position.

##### Can you build a Riddler fantasy football dream team?

Top players in Riddler fantasy football for each position, by expected number of fantasy points over the course of the season

Quarterback | Running Back | Wide Receiver | |||
---|---|---|---|---|---|

Matrick Pahomes | 400 | Caffrey McChristian | 300 | Avante Dadams | 250 |

Osh Jallen | 350 | Calvin Dook | 225 | Hyreek Till | 225 |

Myler Kurray | 300 | Herrick Denry | 200 | Defon Stiggs | 175 |

The draft is a “snake draft.” If person A drafts first, B drafts second and C drafts third, then the order of the picks is as follows: A-B-C-C-B-A-A-B-C.

Your friends — being the kind people that they are — agree that you can choose your pick number. Which draft position should you choose to maximize your expected fantasy score?

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

## Riddler Classic

From Alon Brodie comes a puzzle that’s worth pursuing:

Hames Jarrison has just intercepted a pass at one end zone of a football field, and begins running — at a constant speed of 15 miles per hour — to the other end zone, 100 yards away.

At the moment he catches the ball, you are on the very same goal line, but on the other end of the field, 50 yards away from Jarrison. Caught up in the moment, you decide you will always run directly toward Jarrison’s current position, rather than plan ahead to meet him downfield along a more strategic course.

Assuming you run at a constant speed (i.e., don’t worry about any transient acceleration), how fast must you be in order to catch Jarrison before he scores a touchdown?

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

## Solution to last week’s Riddler Express

Congratulations to 👏 Austin Calico 👏 of Ashland, Kentucky, winner of last week’s Riddler Express.

Last week, there was a cricket on my floor. I wanted to trap it with a cup so that I could safely move it outside. But every time I got close, it hopped exactly 1 foot in a random direction.

I took note of its starting position and came closer. Boom — it hopped in a random direction. I got close again. Boom — it took another hop in a random direction, independent of the direction of the first hop.

What was the *most probable* distance between the cricket’s position after two random jumps and its starting position? (Note: This puzzle was not asking for the *expected* distance, but rather the *most probable *distance. In other words, if you considered the probability distribution over all possible distances, where was the peak of this distribution?)

First off, thanks to the circular symmetry of the puzzle, you could assume the cricket hopped in any direction at first, “without loss of generality” as mathematicians say, and restricted your attention to the second hop. So let’s suppose the cricket’s first hop was to the right. After its second hop, it could be anywhere along the circle centered 1 foot to the right of its origin.

After the second hop, the farthest the cricket could be from its starting point was 2 feet — if the cricket again hopped to the right. Meanwhile, the closest it could be was right where it started — 0 feet away — if the cricket hopped left. The answer was therefore somewhere in this range.

A few readers were tempted to say that all distances between 0 and 2 feet were equally likely, since there were two ways to achieve each distance (i.e., in the upper semicircle and, symmetrically, in the lower semicircle). However, this line of thinking neglected the *continuous* nature of the problem. With infinitely many locations between 0 and 2, the probability of any specific distance was technically zero. But here, you were asked to find the maximum of the continuous probability distribution between 0 and 2.

One way to find the most likely distance was to simulate the two hops, as solver Lowell Vaughn did. Lowell ran a total of 4 million simulations, dividing each of the two hops into 2,000 uniformly distributed angles. Here was the resulting distribution Lowell found:

From this graph, it appeared that the distance of **2 feet** was the most probable. In other words, after two hops, the cricket was more likely to be farther away (i.e., more than 1 foot) from the origin and less likely to be closer (i.e., less than 1 foot) to it. But how could that be?

Jackson Curtis explained this unexpected result by looking at the circle representing the second hop, examining which parts of it were approximately the same distance from the starting point. The following animation highlights which points along the second circle were close to a particular distance from the origin, indicated in red:

As you can see, there was “more red” for distances close to 2 feet.^{2} This was because the two circles — the circle representing the second hop and the fully expanded probing circle — were perfectly tangent, an alignment that meant they had many points close together.

So it was indeed possible to solve this puzzle without an ounce of algebra or computer code. Nevertheless, several solvers, like Emily Boyajian of Minneapolis, Minnesota, explicitly found the probability distribution that Lowell approximated computationally. According to Emily, the probability *p*(r) that the cricket was a distance *r* from the origin was *p*(r) = 2/(𝜋√(4−*r*^{2})).

In case you were curious, I never did catch that cricket. Why? Because it always hopped a bit farther than my intuition said it would.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Rich Erikson 👏 of Warrenton, Virginia, winner of last week’s Riddler Classic.

When you roll a pair of fair dice, the most likely outcome is 7 (which occurs 1/6 of the time) and the least likely outcomes are 2 and 12 (which each occur 1/36 of the time).

But last week, annoyed by the variance of these probabilities, I set out to create a pair of “uniform dice.” These dice still had sides that were uniquely numbered from 1 to 6, and they were identical to each other. However, they were weighted so that their sum was more uniformly distributed between 2 and 12 than that of fair dice.

Unfortunately, it was impossible to create a pair of such dice so that the probabilities of all 11 sums from 2 to 12 were *identical* (i.e., they are all 1/11). But I wagered that we could get pretty close.

The variance of the 11 probabilities was the average value of the squared difference between each probability and the average probability (which was, again, 1/11). One way to make my dice as uniform as possible was to minimize this variance.

So how should I have made my dice as uniform as possible? In other words, which specific weighting of the dice minimized the variance among the 11 probabilities? That is, what should the probabilities have been for rolling 1, 2, 3, 4, 5 or 6 with one die?

Before getting to the solution, it was helpful to first map out the possible sums for a standard pair of dice. The following diagram shows all 36 outcomes, with sums colored the same. Just looking at the diagram, you can see quite a bit of variance — rolling a 6, 7 or 8 is far more likely (i.e., these regions take up much more area in the diagram) than 2 or 12. The precise probabilities of rolling sums of 2 through 12 are, correspondingly, 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36 and 1/36. The variance among these 11 probabilities was approximately 0.001977.

The challenge, then, was to adjust the probabilities of rolling a 1 through 6 with each die — that is, adjusting the horizontal and vertical borders in the diagram above — so that areas of the colored regions were more uniform.

At first glance, this seemed over-the-top hard. After all, there were six different probabilities to consider. You knew they had to sum to 1, which meant there were potentially five degrees of freedom at play.

A key simplification was to assume each die was symmetric — that the probability of rolling a 1 had to be the same as rolling a 6, and that the same went for the pair 2 and 5, and for 3 and 4. In this case, you only had two variables: the probability of rolling a 1 (which we’ll call *a*) and the probability of rolling a 2 (which we’ll call *b*). For all six probabilities to add up to 1, the probability of rolling a 3 had to be 0.5−(*a*+*b*).

From there, you could find a formula for the variance in terms of *a* and *b* and explore this two-dimensional function, which was only defined when the sum of *a* and *b* was at most 0.5. The variance was greatest when *a* and *b* were at their extreme values, as indicated by the graph below:

But this riddle was asking where the function was *minimized*, and that happened at the lowest point of the valley nestled between the three peaks at (0, 0), (0.5, 0) and (0, 0.5). There were multiple algorithms for finding this minimum, from gradient descent to setting the partial derivatives of the variance expression to zero. However you did it, you got the same result: the probability of rolling a 1 (or 6) was **approximately 0.244**,** **a 2 (or 5) **approximately 0.137 **and a 3 (or 4) **approximately 0.119**. The variance dropped to about 0.00122, a 38 percent reduction from the variance of a standard pair of dice. I’ll take it!

With these updated probabilities, here was the resulting area distribution for the 36 cases:

Rolling a 7 was now more likely than it had been with a fair die, but this peculiarity was more than offset by the remaining 10 sums, which were now all much closer together in likelihood.

Finally, while this puzzle asked you to assume that the two “uniform dice” were identical, a few solvers like Dean Ballard and Laurent Lessard went ahead and explored the case when they were allowed to differ, finding that the resulting probabilities were rational. Specifically, Laurent found the probabilities for 1 through 6 for one die should be {1/2, 0, 0, 0, 0, 1/2}, and for the other die they should be {1/8, 3/16, 3/16, 3/16, 3/16, 1/8}. With these dice, the variance dropped all the way to 0.000258, an 87 percent reduction from the original variance.

Dean even worked out how to construct such a pair of dice, which means he won’t be welcome in Las Vegas anytime soon.

## 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.