TYI56 asked the following question of Timothy Chow: You have fifteen boxes labelled with the English letters from A to O. Two identical prizes are placed in two (distinct) boxes chosen at random. Andrew’s search order is ABCDEFGHIJKLMNO. Barbara’s search order is AFKBGLCHMDINEJO. (For the full description see the original post.) The first player to hit one of the prizes is declared the winner (ties are possible).
Test your intuition 56: Is Andrew more likely to win this game, or is Barbara? Or are they equally likely to win?
We ran a poll over twitter (589 votes). The most popular answer by a small margin (45.3%) was that Andrew and Barbara are equally likely to win. I suppose that this reflects several intuitions. One intuition is that the ordering cannot matter but there are simple examples where the ordering does matter. Another intuition is that if the ordering does not matter for one prize then it does not matter for two prizes. In our poll 41.1% answered that Andrew is more likely to win and 13.6% that Barbara is more likely to win.
The correct answer is that Andrew is more likely to win. It is indeed correct that if there is just one prize, then Andrew and Barbara are equally likely to win.
Below the fold I will tell you more about the origin of the puzzle and some links and also about related puzzles by Daniel Litt.
The origin of the Chow’s puzzle
The puzzle (or, in fact, a small variation) was proposed by Timothy Chow in 2010. Here is the relevant item from Timothy’s home page:
“T. Chow, Fair permutations and random k-sets, Problem 11523, Amer. Math. Monthly117 (October 2010), 741. A solution by Jim Simons was published in the Monthly119 (November 2012), on pages 801, 802, and 803. My solution is available here. One open question that I raised was solved by Richard Stanley on MathOverflow.”
The puzzle was discussed on the Puzzling StackExchange website a few years ago.
Some related riddles and polls by Daniel Litt
Daniel Litt (famous for this beautiful probabilistic riddle featured in Quanta Magazine and various other riddles on X (Twitter)) posted some variations of Chow’s puzzle. Daniel asked:
Flip 100 coins, labeled 1 through 100. Alice checks the coins in order (1, 2, 3, …) while Bob checks the odd-labeled coins, then the even-labeled ones (so 1, 3, 5, …, 99, 2, 4, 6, …). Who is more likely to see two heads *first*?
Flip 100 coins, marked 1-100. Each second, Alice and Bob simultaneously check one coin. Alice goes in order (1, 2, 3, …); Bob checks the odd coins, then the even (so 1, 3, 5, …, 99, 2, 4, 6, …). Who is more likely to see 26 total heads *first*?