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

As of today, The Riddler Social Network is being rebranded as μετα — that’s mu epsilon tau alpha. Those Greek letters really *augment* the brand, don’t you think?

A group of 101 people join μετα, and each person has a random, 50 percent chance of being friends with each of the other 100 people. Friendship is a symmetric relationship on μετα, so if you’re friends with me, then I am also friends with you.

I pick a random person among the 101 — let’s suppose her name is Marcia. On average, how many friends would you expect each of Marcia’s friends to have?

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

## Riddler Classic

From Emma Knight comes a unit conversion conundrum:

The sum of the factors of 36 — including 36 itself — is 91. Coincidentally, 36 inches rounded to the nearest centimeter is … 91 centimeters!

Can you find another whole number like 36, where you can “compute” the sum of its factors by converting from inches to centimeters?

*Extra credit:* Can you find a third whole number with this property? How many more whole numbers can you find?

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

## Solution to last week’s Riddler Express

Congratulations to ÐÐ¯Ð¡Ð Ryan Morgan ÐÐ¯Ð¡Ð of Toronto, Canada, winner of last week’s Riddler Express.

Last week, I had a spherical pumpkin. I carefully calculated its volume in cubic inches, as well as its surface area in square inches. Next, I got up to have a piece of Halloween candy (which, naturally, was a Reese’s Peanut Butter Cup).

But when I came back to my calculations, I saw that my units — the square inches and the cubic inches — had mysteriously disappeared from my calculations. But it didn’t matter, because both numerical values were the same!

What was the radius of my spherical pumpkin?

If the pumpkin had radius *R* inches, then its volume was 4/3ÐÐÐ¬Ð*R*^{3} cubic inches, while its surface area was 4ÐÐÐ¬Ð*R*^{2} square inches. Solver Hope Weeda of Grandville, Michigan, recognized that these two expressions had the same numerical value (i.e., ignoring their units) when the equation 4/3ÐÐÐ¬Ð*R*^{3} = 4ÐÐÐ¬Ð*R*^{2} was true. Multiplying both sides by 3 and dividing by 4ÐÐÐ¬Ð*R*^{2} meant that *R* was equal to **3 inches**. And that was it!

While that was a fairly “express” Riddler, the fun was lurking in the extra credit. Instead of a 3D pumpkin, I had an *n*-hyperspherical pumpkin. Once again, I calculated its volume (with units in* ^{n}*) and surface area (with units in

^{n}^{−1}). Miraculously, the numerical values were once again the same! What was the radius of my

*n*-hyperspherical pumpkin?

Solver Steve Curry of Albuquerque, New Mexico, looked at the volumes and surface areas for hyperspheres in small numbers of dimensions and noticed a pattern. In 2D, the “volume” (i.e., area) and “surface area” (i.e., perimeter) were numerically the same when ÐÐÐ¬Ð*R*^{2} was equal to 2ÐÐÐ¬Ð*R*. And that was true when *R* was 2. So in 2D, *R* was 2. In 3D, *R* was 3. In 4D, would *R* be 4?

Without venturing through a proof, I’ll say that the volume of a 4-hypersphere was 1/2ÐÐÐ¬Ð^{2}*R*^{4}, while its surface area was 2ÐÐÐ¬Ð^{2}*R*^{3}. Sure enough, setting these expressions equal to each other meant that *R* was 4. It appeared that my *n*-hyperspherical pumpkin had a radius of *n*** inches**.

To *prove* this was the case for any dimension *n*, I immediately knew that my *n*-hypersphere had a surface area of the form *aR*^{n}^{−1}, where *a* was some constant — undoubtedly with a power of ÐÐÐ¬Ð in there somewhere. To find the volume, I could integrate shells of surface area, from radius *r* = 0 up to *r* = *R*. The result of this definite integral was *aR** ^{n}*/

*n*. Finally, setting the surface area equal to the volume gave the equation

*aR*

^{n}^{−1}=

*aR*

*/*

^{n}*n*. The solution was indeed

*R*=

*n*.

Working backwards, as solver Benjamin Dickman did, you might notice that derivatives of expressions for volume give you expressions for surface area. There are no coincidences!

## Solution to last week’s Riddler Classic

Congratulations to ÐÐ¯Ð¡Ð Amy Leblang ÐÐ¯Ð¡Ð of Wayland, Massachusetts, winner of last week’s Riddler Classic.

Last week, you had made it to the fifth round of The Squiddler — a competition that took place on a remote island. In this round, you were one of the 16 remaining competitors who had to cross a bridge made up of 18 pairs of separated glass squares. Here was what the bridge looked like from above:

To cross the bridge, you had to jump from one pair of squares to the next. However, you had to choose *one* of the two squares in a pair to land on. Within each pair, one square was made of tempered glass, while the other was made of normal glass. If you jumped onto tempered glass, all was well, and you could continue on to the next pair of squares. But if you jumped onto normal glass, it broke, and you were eliminated from the competition.

You and your competitors had no knowledge of which square within each pair was made of tempered glass. The only way to figure it out was to take a leap of faith and jump onto a square. Once a pair was revealed — either when someone landed on a tempered square or a normal square — all remaining competitors took notice and chose the tempered glass when they arrived at that pair.

On average, how many of the 16 competitors would survive and make it to the next round of the competition?

Before working this out precisely, Laurie Blackman approximated the solution by thinking about a typical contestant who was currently at the front, deciding which square to jump onto next. Half the time, they’d jump to normal glass and be eliminated. A quarter of the time, they’d land safely on tempered glass and *then* jump to normal glass and be eliminated. An eighth of the time, they’d land safely twice and then be eliminated on their third jump.

This seemed analogous to asking how many times you’d have to flip a fair coin until it came up heads. If you were willing to flip as many times as necessary, then it would take you two flips on average. So, if each contestant is eliminated after two jumps on average, then you’d expect nine competitors to be eliminated over the course of the 18 jumps. That left a total of seven survivors. Was seven the answer?

No, it was not. As noted by solver Josh Silverman, one way to see this was through symmetry — or a lack thereof. The probabilities of having six survivors (and 10 people eliminated) was the same as having eight survivors (and 8 people eliminated). This was because competitors were just as likely to break 10 out of the 18 normal squares (and leave eight of them intact) as they were to break eight of them (and leave 10 intact).

Similarly, the probabilities of five vs. nine survivors were equal, as were four vs. 10, three vs. 11, two vs. 12 and one vs. 13. But that’s where the symmetries ended. Technically, there were three different ways to have zero survivors — they could have landed on two, one or zero tempered squares. No matter what, this still counted as zero survivors. Meanwhile, at the other probabilistic extreme, you could have had 14, 15 or 16 survivors, stretching the expected value higher. That meant the answer was not seven expected survivors, but rather *slightly more *than seven.

Solvers Laurent Lessard and Rohan Lewis found the exact answer by thinking recursively. One way to do this was to let *p*(*T*, *N*) represent the probability that *T* tempered squares and *N* normal squares were landed on among the first *T*+*N* jumps. The last square was equally likely to be tempered or normal, so *p*(*T*, *N*) = *p*(*T*−1, *N*)/2 + *p*(*T*, *N*−1)/2. If we defined *p*(0, 0) = 1, p(*T*, 0) = 0 for nonzero *T* and p(0, *N*) = 0 for nonzero *N*, then we were off and running.

But one final note: P(1, 16) was not equal to *p*(0, 16)/2 + *p*(1, 15)/2. That first term amounted to 16 normal squares being landed on, and once that happened there was no one left to land on a tempered square. So in addition to the aforementioned recursive rules, we had the edge case *p*(*T*, 16) = p(*T*, 15)/2.

With rules like these in place, you could compute the probabilities of each number of survivors, as shown below. As predicted, the average was slightly more than seven — it was 7+5/2^{16}, or about **7.000076294**.

Solver Emily Boyajian further found that, more than 90 percent of the time, between four and 10 Squiddler contestants survived. This compared rather favorably to the *three* survivors in the comparable Squid Game competition. (Where was Sang-woo’s impressive education when you needed it? Oh.)

## 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.