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

From Thomas Sneller comes a puzzle that brings us back to the game show to end all game shows, “Who Wants to Be a Riddler Millionaire?” As you’ll remember, for each question you must pick the correct option from four choices.

You’ve made it to the $1 million question, but it’s a tough one. Out of the four choices, A, B, C and D, you’re 70 percent sure the answer is B, and none of the remaining choices looks more plausible than another. You decide to use your final lifeline, the 50:50, which leaves you with two possible answers, one of them correct. Lo and behold, B remains an option! How confident are you now that B is the correct answer?

## Riddler Classic

From Joel Lewis, this week’s Riddler Classic is a birthday puzzle for the ages:

The classic birthday problem asks about how many people need to be in a room together before you have better-than-even odds that at least two of them have the same birthday. Ignoring leap years, the answer is, paradoxically, only 23 people — fewer than you might intuitively think.

But Joel noticed something interesting about a well-known group of 100 people: In the U.S. Senate, *three* senators happen to share the same birthday of October 20: Kamala Harris, Brian Schatz and Sheldon Whitehouse.

And so Joel has thrown a new wrinkle into the classic birthday problem. How many people do you need to have better-than-even odds that at least *three* of them have the same birthday? (Again, ignore leap years.)

## Solution to last week’s Riddler Express

Congratulations to 👏Fernando Mendez 👏 of San Carlos, California, winner of last week’s Riddler Express.

Last week, you were asked to find the probability of a baseball team winning two of its previous four games *and* four of its previous eight games, if that team had a 50 percent chance of winning each game.

Solver Amy Leblang tackled this riddle by first looking at the probability of the team winning two of its last four games. If we mark a win with a W and a loss with an L, there are 2^{4}, or 16, total ways to write the results of those four games, each with the same probability of occurring. (You get the 2 in the 2^{4} because there are two options: win or lose. And you get the 4 because we’re looking at four games.) But only *six* of these sequences have exactly two wins: WWLL, WLWL, WLLW, LWWL, LWLW and LLWW. That means the chances of the team winning two of its last four games was 6/16, or 3/8.

If a team has already won two of its last four games, then in order to win four of its last eight, it must have won *another *two games in the four games immediately preceding the last four. There are 16 ways to write the results of *those* four games, and six of them will result in exactly two wins, again giving us a result of 3/8.

Putting this all together, the probability of winning two of the last four games *and* winning four of the last eight is the product of these two probabilities: 3/8×3/8 = **9/64**, or a little more than 14 percent of the time.

## Solution to last week’s Riddler Classic

Congratulations to 👏Tom Mahon 👏 of Kingsland, Texas, winner of last week’s Riddler Classic.

Last week, you analyzed three baseball teams: (1) the Mississippi Moonwalkers, whose batters each had a 40 percent chance of walking and a 60 percent chance of striking out; (2) the Delaware Doubloons, whose batters each had a 20 percent chance of doubling — driving in any teammates on base — and an 80 percent chance of striking out; and (3) the Tennessee Taters, whose batters each had a 10 percent chance of hitting a home run and a 90 percent chance of striking out. If these three teams faced each other in a season of baseball, which team was most likely to have the best record?

Riddler Nation was divided over this problem. Among readers who submitted on Friday, more than half said the Moonwalkers would win the most games. But among the slow and steady solvers who submitted solutions over the weekend, more than 70 percent thought the Taters would win the most games. What gives?

Many attacked this riddle by analyzing how many runs each team would score on average, reasoning that more runs would lead to more wins. Solver Michael Campbell found these averages precisely, using techniques from the branch of mathematics known as combinatorics. He found that the Doubloons average only about 2.4 runs per nine innings, the Taters average 3.0 runs per nine innings, and the Moonwalkers average about 3.4 runs per nine innings. From this analysis, it would seem that the Moonwalkers are the best team in Riddler League Baseball.

But not so fast — there was something deeper going on here. Solver Michael Goss took a more direct approach, simulating a million baseball games between each pair of teams. He found that the Taters beat the Doubloons in about 63 percent of games, and that the Taters *also* beat the Moonwalkers in 51.6 percent of games. From these percentages, solver Jason Ash calculated the average number of wins each team would have in a 162-game season: approximately 93 wins for the Taters, 87 wins for the Moonwalkers and 64 wins for the lowly Doubloons. So on average, the winningest team turns out to be the **Taters**.

How can a team that scores fewer runs than its opponent win more games? (Such a feat apparently runs counter to baseball’s Pythagorean expectation.) The answer lies in the probability distribution of runs per game. Imagine Team A scores exactly one run every game, while Team B is scoreless in nine games out of 10 but then piles on 20 runs in the remaining games. Team B scores twice as many runs as Team A on average, but Team A will win 90 percent of their head-to-head matchups!

While the math behind the Taters-Moonwalkers matchup was a little more involved, the reasoning was the same. The graph below shows the run differential in 1 million simulated matchups between the Taters and the Moonwalkers. While the Moonwalkers were more likely to light up the scoreboard, as shown by the graph’s longer left tail, the Taters won a slight majority of the games, often by just one or two runs.

This is definitely a case for the long ball over small ball. (I sure hope Ned Yost isn’t reading this.)

## 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.