Math and Life

# Josephine’s problem (of unfaithful husbands)

In Josephine's Kingdom every woman has to pass a logic exam before being allowed to marry.1

Every married woman knows about the fidelity of every man in the Kingdom except for her own husband, and etiquette demands that no woman should be told if her husband is cheating.

A gunshot fired in any house in the Kingdom will be heard in any other house.

Queen Josephine announces that at least one unfaithful man had been discovered in the Kingdom, and that any woman knowing her husband to be unfaithful is required to shoot him at midnight following the day she discovered his infidelity.

How did the wives manage this?

If you want to take a whack at it, you certainly can. It’s tricky. The solution is below this line.

Case 1: There is only one unfaithful husband in the kingdom.

The wife of this husband knows that every other man in the kingdom is faithful. She knows that there is at least one unfaithful man in the kingdom (Queen Josephine announced this), so the wife deduces that her husband must be unfaithful, and shoots him in the dead of night.

Case 2: There are two unfaithful husbands in the kingdom.

The wife of one of the unfaithful husbands knows that there are two possibilities: either a. her husband is faithful, and there is only one unfaithful man in the kingdom or b. Her husband is unfaithful. (remember, wives know the fidelity of every husband in the kingdom except their own)

If a. were the case, and there is only one unfaithful husband, Case 1 will apply, and the wife will hear a gunshot on the first night.

Because neither wife knows the fidelity of the husband on the first night, no gunshot will be heard on the first night.

After not hearing a gunshot on the first night, the wives can rule out a.,conclude that her husband must be unfaithful, and shoot him on the second night.

Case 3: There are three unfaithful husbands in the kingdom

The wife of one of the unfaithful husbands again knows that there are two possibilities: a. her husband is faithful, and there are two unfaithful men in the kingdom or b. Her husband is unfaithful.

If a. Were the case, and there are only 2 unfaithful husbands, Case 2 will apply, and the wife will hear a gunshot on the second night.

Because no wife knows the fidelity of their husband on the second night, no gunshot will be heard.

After not hearing a gunshot the second night, each of the three wives can rule out option a., and shoot their husbands on the third night.

You can continue this argument for any natural number of unfaithful husbands with the general case:

Case n: There are n unfaithful husbands in the kingdom.

The wife of one of the unfaithful husbands knows that there are two possibilities: a. her husband is faithful, and there are n-1 unfaithful men in the kingdom or b. Her husband is unfaithful.

If a. Were the case, and there are only n-1 unfaithful husbands, Case n-1 will apply, and the wife will hear a gunshot on the n-1st night.

Because no wife knows the fidelity of their husband on the n-1st night, no gunshot will be heard.

After not hearing a gunshot the n-1st night, each of the wives can rule out option a., and shoot their husbands on the nth night.

Hopefully, I’ve convinced you. If this is your first time seeing this puzzle, you might have to read the solution a couple times to fully swallow it.

Not only is this an interesting logic puzzle, it’s also a great example of mathematical induction, which I’ll be discussing in the next post.

1We are supposed to take from this that every wife is a perfect logician, not that Josephine is a eugenicist

Noah CaplingerComment