Werewolves vs Town Jailer – does this work like the Monty Hall problem?

Question

In the game of Werewolf (also known as Mafia), a small group of werewolves is hiding in a larger group of villagers. During the “night” phase of the game, the werewolves choose one villager to kill. During the “day” phase of the game, all players (including the werewolves) debate and vote to choose one player to execute. The villager team wins if they successfully execute all the werewolves before they become outnumbered.

Werewolf can be played with all kinds of different special rules, but in this scenario the rules are:

1. At the start of the game, one of the villagers is secretly assigned the unique role of Town Jailer. They gain the ability to jail one player each night.
2. Each night, when the werewolves decide which villager to kill, they must also choose one werewolf to perform the kill.
3. If the jailed player was the werewolves’ target, the attempt to kill them fails and there are no deaths during the night.
4. If the jailed player was the werewolf performing the kill, their attempt to kill fails and there are no deaths during the night.
5. The werewolves cannot withhold the kill, or choose to kill no one. They must always select a villager to attempt to kill, therefore a lack of death during the night can only mean that the Town Jailer prevented it.
6. There are no other special abilities, and all players are aware of this “open setup”.

Given this scenario, does the probability that the jailed player is a werewolf change after the Jailer successfully prevents the kill?

Some argue that the jailed player now has 50/50 odds of being a werewolf. The blocked kill could have only occured by jailing either the werewolf killer, or the intended villager victim.

Others argue that the jailed player’s probability of being a werewolf remains unchanged. I initially agreed with this position, since it seemed similar to the Monty Hall problem. The Jailer chooses one player, who has some initial probability of being a werewolf. (any example number of players and werewolves can be chosen, but the chance can never be above 50/50 or the game would already be over with a werewolf victory.) After learning the kill was successfully blocked, this “opens all but one other door” by ruling out all the other possibilities. Instead of revealing goats, we reveal non-targeted villagers, and non-killing werewolves. Only two doors remain: our initial choice, which contains either the killer or intended victim, and a door that contains the other half of that pair. The blocked kill gives the illusion of that chance now being 50/50.

But now I’m wondering if the comparison is misleading. Unlike the Monty Hall problem, there is no option to switch. And the Jailer’s choice has a direct effect on the outcome of whether the kill is blocked to begin with.

The way I picture it now is, Monty Hall lines up a bunch of doors and asks you to try and choose one with a werewolf behind it. Most of the time, you choose a villager, since at least 50% of the players to choose from are villagers. But you get no other doors revealed, and no option to choose a different door. Instead, Monty just pauses briefly before opening the door, for dramatic effect. And if you did choose the one werewolf who was performing the kill, or the one villager who was targeted for the kill, a big flashy siren goes off as Monty approaches the door.

If the siren goes off, should you still believe your odds are entirely unchanged? Or does this siren provide information that is worth taking into account - that the odds of this jailed player being a werewolf have now increased to 50/50?

Answer 1

I will base my answer on the simple calculations: the chances to choose the werewolf that makes the kill is $\frac{1}{w+v}$, where w and v are werewolves and villagers. The chances to guess the victim are the same. Since stopping the kill is the same as guessing either the victim (=villager) or the killer (=werewolf) and both events have the same probability, then the the chances that we found the werewolf go up to 50%.

Answer 2

I would say that the difference between your scenario and the Monty Hall problem is the possibility of a whiff. If it was possible for Monty to open a door and go, "Oh dear, that's the car. Show's over, folks," then the fact that he didn't would be new information, Information that would affect the odds of a goat being behind your door. It is only the fact that Monty knows what's behind each door and ensures that he opens a goat that keeps the probability the same.

In your scenario it is possible for the jailer to whiff and fail to prevent a murder. When they do prevent a murder that is new information and should be taken account of.

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Let's plug in some concrete numbers. Let there be $w$ werewolves and $v$ (non-jailer) villagers. Let $J$ be the event that the jailer prevents a murder, $W$ the event that they pick a wolf, and $V$ the event that they pick a villager. Then:

$P(W)= \frac{w}{w+v}$

$P(V)= \frac{v}{w+v}$

$P(J|W) = \frac{1}{w}$ (Given that they picked a wolf, they have that chance of picking the chosen killer)

$P(J|V) = \frac{1}{v}$ (Given that they picked a villager, they have that chance of picking the victim)

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Plugging all of this into Bayes theorem, we get that:

$P(J)= P(J|W) P(W) + P(J|V) P(V) = \frac{1}{w} \frac{w}{w+v} + \frac{1}{v} \frac{v}{w+v} = \frac{2}{w+v}$

(Not surprising: The jailer has two chances in the entire village to pick someone involved in the murder)

$P(W|J) = \frac{P(J|W) P(W)}{P(J)} = \frac{\frac{1}{w} \frac{w}{w+v}}{\frac{2}{w+v}} = \frac{1}{2}$

As expected, the jailer preventing the murder increases the odds of their target being a wolf to $\frac{1}{2}$: either they picked the chosen murderer, or the victim.

Answer 3

Let's say that there are $w$ werewolves and $v$ villagers.

Let's assume: That the werewolves always target a villager and never a werewolf. Although, as I understand the rules, the werewolves could choose to kill one of their own, in line with some larger strategy to defer suspicion.

All villagers are equally likely to be targeted.

All werewolves are equally likely to be the killer.

All participants are equally likely to be targeted by the jailer.

The probability that the jailer picked a werewolf and that was the werewolf assigned to make the kill is $\frac {1}{w+v}$

The probability that the jailer picked the werewolf's target is also $\frac {1}{w+v}$

Indeed, it is equally likely that the jailer's target was a wolf as it is that they were a villager.

We can plug this into Bayes formula -- but it isn't really necessary.