Prisoners' hats
7 minute read
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Adapted from this SoME2 video by Going Null.
Problem
$n$ prisoners are given hats, each with a number between $1$ and $n$ (possibly with some repeats). They each get to see the other people’s hats, but not their own, and they cannot communicate. The then each say a number simultaneously, and they win if at least one of them says the number of their own hat. Can they come up with a strategy in advance which allows them to win all the time?
Solution
Going through the case $n=2$ is very helpful: after some trial and error, it’s easy to figure out that the following works:
- prisoner #1 says the same number as prisoner #2’s hat;
- prisoner #2 says the opposite number as prisoner #2’s hat.
Indeed,
- if the hats are the same, then prisoner #1 will guess its hat correctly;
- if the hats are different, then prisoner #2 will guess its hat correctly.
And from there it’s fairly easy to generalize the idea of splitting the “overall events” into $n$ distinct cases, and assign one prisoner to succeed in each case. One way to achieve this is to split by the sum modulo $n$ (this ensures that the prisoner’s choices are well-defined: prisoner $i$ just chooses the value that would make the total sum congruent to $i$ modulo $n$); this is not unique at all.
But you can actually get this idea of “splitting responsibilities up into cases” from scratch as well! Indeed, if the hats were chosen at random, then whatever the strategy is, each player only has a $1/n$ chance to guess right. And we want the union to be $100\%$, so that means that the events “prisoner $i$ gets right” for $i\in [n]$ are all disjoint! So in fact, the key is to avoid any overlap at all costs.