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. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

This week’s Riddler Express was related to me by Sunil Singh, who heard it from Sam Vandervelde, who heard it from … well, let’s just say this puzzle has been circulating for a while and is ready for prime time!

An auditorium with 200 seats, numbered from 1 to 200, is filled to capacity. A speaker, who happens to be a mathematician, steps up to the podium overlooking the audience and pauses for a moment. “You know,” she says, “I’m thinking of a rather large whole number. Every seat number in this auditorium evenly divides my number, except for two of them — and those two seats happen to be next to each other.”

As you’d expect, adjacent seats in the auditorium have consecutive numbers. Which two numbers was the speaker referring to?

## Riddler Classic

Last year, Riddler Nation tackled a puzzle from Mike Donner about wartime navigation. An enemy submarine determined to sink your ship was exactly halfway between you and your home port. Your ship had no further information about the submarine’s subsequent movement. The sub had to be directly underneath your ship to sink it, but the sub could track your moves with precision and respond efficiently. How much faster than the sub did your ship have to be to guarantee you could avoid the sub and get home?

It turned out your ship needed to be about 2.33 times faster than the sub. This is demonstrated by the animation below: The expanding pink circle is the region where the sub could be located after a given amount of time, and the blue curve represents your ship’s course around it.

This week, aware of your mathematical prowess, the enemy has deployed *two* submarines so that your ship, the submarines and the harbor you seek are evenly spaced along a straight line.

If both subs travel at the same speed, how much faster than the subs does your ship have to be to guarantee that you can avoid them and reach the harbor?

**Extra Credit:** Suppose there are *N* submarines evenly spaced between your ship and the harbor. How much faster than the subs must your ship be?

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐRyan Potts ÑÑâÐ of Newnan, Georgia, winner of last week’s Riddler Express.

Last week, you had made it to the final question in “Who Wants to Be a Riddler Millionaire?” Out of the four choices, A, B, C and D, you were 70 percent sure the answer was B, and the remaining choices looked equally implausible. You decided to use your final lifeline, the 50:50, which left you with two possible answers, one of them correct. Lo and behold, B remained an option! How confident were you then that B was the correct answer?

Once again, there was a fair bit of disagreement across Riddler Nation. Just over half of the more than 800 respondents were 87.5 percent sure that B was the correct answer. Meanwhile, a fifth of respondents were *still* 70 percent sure that B was the correct answer. The next most popular answers were 90 percent, 80 percent and 84 percent. Despite this disagreement over the exact confidence, I am proud to say that almost all of Riddler Nation recognized that the answer was *at least* 70 percent. If B remains an option, it can’t be *less* likely to be the correct answer. But how confident should you be — still 70 percent, 87.5 percent or something else?

Many readers who said your confidence should remain at 70 percent cited the famous Monty Hall problem as part of their reasoning. In that problem’s game show, there are three doors and you pick the one you think the prize is behind. Next, the host reveals a door that does *not* hold a prize. (Importantly, this will never be the door that you initially picked.) You are then asked whether you would like to stick with your door or switch to the remaining unopened door. Your initial probability of picking the right door was 1/3, and no matter which door you picked, the host can always reveal another door that lacks the prize. The probability that your original pick is correct has not changed. If you switch to the remaining door, you’ll win the prize two-thirds of the time, so switching is your best bet.

But there’s a key difference between the Monty Hall problem and last week’s Riddler Express: The answer you were leaning toward, choice B, could very well have been eliminated by the 50:50. In other words, the game show’s decision of which two options would be removed was *independent* of your preference for B. Meanwhile, in the Monty Hall problem, the door you chose was *never* eliminated from contention. It’s a subtle difference, but it has a profound effect on the answer.

Note that this puzzle may also differ slightly from the original “Who Wants to Be a Millionaire,” where, according to solver Rob Moran (who was a contestant on the show), “…the [two] answers remaining in this situation are not the correct answer and a randomly selected wrong answer. They are the correct answer and the most likely incorrect answer.”

Solvers Liam Lloyd and Jason Ash both used Bayes’ Theorem to determine the answer. Thomas McGlynn, meanwhile, invoked a related approach, thinking about 100 parallel universes, where choice B was the correct answer in 70 of them. In these 70 universes, choice B would always remain after the 50:50 because it was the right answer. But what about the other 30 universes? Well, the correct choice would have to remain, while the three incorrect responses (including B) would each survive the 50:50 a third of the time, or in 10 universes each. So, of the 80 universes in which choice B remains after the 50:50, it’s the correct answer in 70 of them. That means the probability that B is correct is 70/80, or **87.5 percent**.

The Monty Hall problem may be a classic, but sometimes it pays to stick with your gut.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐMichael Branicky ÑÑâÐ of Lawrence, Kansas, winner of last week’s Riddler Classic.

Last week’s puzzle asked how many people you needed in a room so that there were better-than-even odds that at least *three* of them had the same birthday.

This was a variation on the classic birthday problem, which asks how many people are needed so that at least *two* of them have the same birthday. One way to solve this is to first calculate the probability that they all have different birthdays. For example, suppose you had four people — let’s call them A, B, C and D — in a room, lined up against a wall. The probability that B’s birthday is different from A’s is 364/365, since B’s birthday could be any of the 364 days that are not A’s birthday. By the same logic, the probability that C’s birthday is different from both A’s and B’s is 363/365, and the probability that D’s birthday is different from the other three birthdays is 362/365. To find the probability that all four people have different birthdays, we can multiply these probabilities together, giving us (365×364×363×362)/365^4, or about 98.4 percent. That means the other 1.6 percent of the time, at least two of them will have the same birthday. Extending this reasoning to larger groups of people, it turns out that among 23 people, there’s a 50.7 percent chance that at least two of them will have the same birthday.

Unfortunately, in going from at least two people with the same birthday to at least three, the math gets a lot thornier. But Riddler Nation was not so easily intimidated. Approaches ranged from generating functions, which have indeed been used to tackle problems like this, to detailed combinatorics. As puzzle submitter Joel Lewis observed and as Hector Pefo has written, one must carefully count all possible cases where there are pairs — but not triples — of people with the same birthday. In fact, there’s a 51.1 percent chance that with **88 people** in the room, at least three of them will have the same birthday. It’s certainly more than the 23 people from the two-birthday problem … but again, paradoxically, it’s a smaller number than you might expect.

Another way to solve this problem is with brute force and computer simulation. When 87 people were in the room, the chances of at least three people having the same birthday were awfully close to 50 percent, so many solvers had to run *millions* of simulations to figure out if that probability was just under or just over the mark. Meanwhile, Dogan Kazakli’s computer came close to burning a hole through his desk as it cranked out the probability curves for different groups of people and how many had the same birthday:

And there’s good news for those of you wondering how many people you’d need in a room so you have better-than-even odds that at least four, five, six or more of them have the same birthday: There’s a sequence for that!

## 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.