Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and 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. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Rich Holmes, some probabilistic supervillainy:

Thanos, the all-powerful supervillain, can snap his fingers and destroy half of all the beings in the universe.

But what if there were 63 Thanoses, each snapping his fingers one after the other? Out of 7.5 billion people on Earth, how many can we expect would survive?

If there were N Thanoses, what would the survival fraction be?

## Riddler Classic

From Kevin Costello, you’re back on Riddler Nation’s favorite game show, “Guess Your Hat.” Only this time, things are a little more colorful:

You and three friends are contestants on the newest edition of “Guess Your Hat.” As in previous editions of the show, you will each have a hat placed on your head. But now the hats can be one of four colors: red, yellow, blue or green. Some (or even all) of you may have the same color hat.

You can see everyone else’s hat but not your own, and your goal is to guess the color of your own hat. Everyone who guesses right wins, and everyone who guesses wrong loses. The guesses are made in private.

You can’t communicate with your friends in any way during the game, but you *can* communicate with them beforehand to decide on a strategy. However, the showrunners will be listening in, so they will know whatever strategy you decide on. They will then choose your hats in the worst possible way (for you), trying to make as many people as possible lose.

Given the behavior of the nefarious showrunners, you probably can’t pick a strategy that will let everybody win. But can you design a strategy that guarantees *at least one* of you will win? How about guaranteeing at least two winners?

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Daniella Scruggs ÑÑâÐ of Philadelphia, winner of last week’s Riddler Express!

Last week brought a sequence of numbers, and your job was to figure out what the sequence was and what numbers came next. (The numbers in parentheses provided a hint.)

1 (1)

2 (2)

4 (3)

6 (4)

12 (6)

24 (8)

36 (9)

48 (10)

60 (12)

120 (16)

180 (18)

240 (20)

360 (24)

720 (30)

840 (32)

The sequence is the numbers that have **more factors** than any smaller number, and the numbers in parentheses are the number of factors that can go into each entry in the sequence. For example, 12 has six factors (1, 2, 3, 4, 6 and 12), which is more than any number from 1 through 11.

So to figure out what comes after 840, we need to look for the next number that has more than 32 factors. That turns out to be **1,260**, which has 36 factors — 1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 14, 15, 18, 20, 21, 28, 30, 35, 36, 42, 45, 60, 63, 70, 84, 90, 105, 126, 140, 180, 210, 252, 315, 420, 630 and 1260.

After that comes 1,680 (40), 2,520 (48), 5,040 (60), 7,560 (64), 10,080 (72), …

You can find this full sequence (and the one in parentheses) in the On-Line Encyclopedia of Integer Sequences — but prepare to spend your whole day on the most engrossing website on the internet.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Sanandan Swaminathan ÑÑâÐ of San Jose, California, winner of last week’s Riddler Classic!

Last week also brought a circular conundrum: If \(N\) points are generated at random places on the perimeter of a circle, what is the probability that you can pick a diameter such that all of those points are on only one side of the newly halved circle?

The probability is \(N/\left(2^{N-1}\right)\).

An elegant expression, but how do we get there? Here is the approach taken by this week’s winner, Sanandan:

Imagine \(N\) points that are all within a semicircle. One of those points is at one end and the other \(N-1\) points are to its right, in a clockwise direction. Now pick any one point. The probability that another point falls within the clockwise semicircle starting from the point you just picked is \(1/2\) — because it could fall in the counterclockwise semicircle with equal probability. Therefore, the probability that all \(N-1\) points fall within the clockwise semicircle from the starting point is \((1/2)^{N-1}\). But *any* of the \(N\) random points could be the leftmost point (and exactly one of them will be). Therefore the overall probability that all \(N\) points fall within a semicircle is \(N / \left(2^{N-1}\right)\).

Solver Jordan plotted what this probability looks like as the number of points gets larger and larger, along with some examples for various numbers of points. Of course, the chances that you’ll be able to fit all these random points in some semicircle become vanishingly small as the number of points increases.

Another variation of this problem: What’s the probability that a circular table with \(N\) legs placed randomly along its perimeter will stand up?

## 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 me at oliver.roeder@fivethirtyeight.com.