When Jake’s father (Bradley Whitford) comes to town, Jake is excited to see him, but Charles is wary of his intentions; Holt challenges Amy, Terry, Gina and Rosa with a brain teaser in exchange for Beyonce tickets.

- Brooklyn 99, S02 E18,
Captain Peralta

I have no interest in Detective Peralta’s family issues. Captain Holt’s puzzle, on the other hand? Count me in.

The brain teaser is as follows:

There are 12 men on an island. 11 weigh exactly the same amount, but one of them is slightly lighter or heavier. You must figure out which.

The island has no scales, but there is a seesaw. You can only use it three times.

According to IMDB, the puzzle is unsolvable. IMDB is wrong. Can you outsmart Holt, Santiago and the others? Go ahead. My solution is below the line.

This is a classical puzzle of the coin-weighing class, a class that (like people wearing hats) I don’t usually have much time for. But I’ll make an exception here.

My big insight with this was to look at the possible solutions: there are 24 possible solutions, since each of the men could be heavier or lighter than the rest. Each weighing can have three possible outcomes - one side drops, the other side drops, or the sides balance. In principle, if each weighing can distinguish between three situations, three weighings should be able to distinguish 27, so it’s plausible that we can find a solution.

So how do we do it? We start by saying “the first weighing must split the solutions into three equal groups”. A way to do that would be to weigh any four men (call them 1, 2, 3 and 4) against any four others (5, 6, 7, 8).

There are three outcomes:

- A:
**If the pans balance**, one of the remaining men (9-10-11-12) is either heavy or light, we don’t know which. - B1:
**If the 1-2-3-4 pan drops**, either one of these men is heavy, or one of 5-6-7-8 is light. - B2:
**If the 5-6-7-8 pan drops**, either one of these men is heavy, or one of 1-2-3-4 is light.

(The last two situations are the same, just with different labels.)

In situation A, we have four men, of which one is heavier or lighter than the others. We need the next weighing to reduce those eight possibilities to three.

One way to do this is to weigh 9-10-11 against 1-2-3, which we know are all the same weight.

**If the pans balance**, 12 is the odd man out and we can weight him against one of the others to find whether he’s heavy or light.**If 9-10-11 drop**, one of them is heavy. Weigh 9 against 10; if one of them drops, he’s the heavy one; otherwise, 11 is.**If 9-10-11 rise**, one of them is light. Weigh 9 against 10; if one of them rises, he’s the light one; otherwise, 11 is.

This situation is less straightforward, but can still be made to work. The trick is, again, to find weighings that cut the solution space more or less into thirds. (I’m going to suppose we’re in situation B1 - B2 is just the same, but you’ll need to mentally switch the labels.

One way to do this is to put two possibly-heavy men (1-2) and a light man (5) on one end of the see-saw; in the other, put 3 and 4 (possibly heavy) and 6 (possibly light).

**If the pans balance**, either 7 or 8 is light, and weighing them against each other will decide which.**If 1-2-5 drops**, we have three possibilities: 1 is heavy, 2 is heavy or 6 is light. Weigh 1 against 2 to find out which.**If 3-4-6 drops**, either 3 is heavy, 4 is heavy or 5 is light. Weigh 3 against 4 to find out which.

There is much about Brooklyn 99 that’s implausible, but the idea that Captain Holt would be stumped by this - and unable to find an answer - for 20 years is one of the silliest yet. Do they not have mathematicians there? For heaven’s sake.