Earlier today I present to you the following puzzle, which has been making the rounds in the academic community because of its counterintuitive result. Here it is again with the solution.
The 15 boxes
Andrew and Barbara are playing a game in which fifteen boxes are arranged in a grid as shown below.
Prizes are placed in two randomly chosen boxes. Andrew will search the boxes row by row, so his search order is ABCDEFGHIJKLMNO. Barbara will search column by column, so her order is AFKBGLCHMDINEJO.
If Andrew and Barbara open their boxes together each turn, that is, on the first turn they both open A, on the second Andrew opens B and Barbara opens F, on the third Andrew opens C, and Barbara opens K open, and so on, who is more likely to find a prize first?
a) Andrew.
b) Barbara.
c) Both equally likely.
Solution a) Andrew.
Intuitively, it feels like both players should get to the box at the same time. If the prices are in randomly selected boxes, why should one search method have an advantage over another?
Indeed, if there were only a single prize in a single box, that would be the answer: they would be equally likely to win. What changes everything is that there are two prizes, and the game is over when the first one is found.
Let’s imagine that there is a single price. The illustration below shows who would win the game if the prize is in that box, next to the number of the turn. So, if the prize is in box A, H or O, they have an equal chance, marked ‘=’, because they both open that box on the same turn. In six boxes (in turns 2,3,4,5,9,10) Andrew wins because he is the first to open those boxes, and in six boxes (in turns 2,3,5,6,9, 12) Barbara wins.
Now let’s consider the situation when there are two prizes, in two randomly chosen boxes. If both prizes are in boxes Andrew gets first then he wins. If both prizes are in boxes which Barbara reaches first, then she wins. Both of these occurrences are equally likely, so no one has an advantage. Likewise, we don’t have to worry about prizes in boxes that they both reach together, because if it’s A they both win, if it’s O it reduces to the one-price version, and if it’s H, Andrew wins if the other prize is in B,C,D or E, and Barbara does if it is in F,G,K and L – giving both equal chances.
The crucial case is when one prize is in a box that Andrew gets first, and one is in a box that Barbara does. In this case, Andrew holds a slight edge because he wins on average on earlier turns. (Andrew wins on 2,3,4,5,9,10, Barbara on 2,3,5,6,9,12.) For example, if the prize is in box N, which Barbara gets on turn 12, she will always lose because Andrew will have won by turn 10.
The puzzle was first posed (in a slightly different form) by Timothy Chow in 2010. A few years ago it was picked up by StackExchangeand more recently by Gil Kalai, a former plenary speaker at the International Congress of Mathematicians, on his blog.
The puzzle is interesting to professional mathematicians because even they struggle to find an intuitive explanation for why Andrew gets a prize first. They were also surprised by the solution.
If you can come up with a simple, intuitive explanation – please share it below!
I hope you enjoyed the puzzle. I’ll be back in two weeks.
Had I known about Chow’s puzzle, I might have included it in my latest book, Think Twice, a compilation of many counterintuitive puzzles. (In the US it is called Puzzle Me Twice.) The idea behind the book is for it to be read on its own, or in a group, as these puzzles are great fun to reason about.
Think Twice: Solve the Simple Riddles (Almost) Everyone Gets Wrong (Square Peg, ££12.99). To support the Guardian and Observer, order your copy from guardianbookshop.com. Delivery charges may apply.
I’ve been doing a puzzle here on alternate Mondays since 2015. I’m always on the lookout for great puzzles. If you want to suggest one, email me.
