Applying Bayesian probability to a generalized Monty Hall problem

I posted this question about the Monty Hall problem and Monty's knowledge of the probability distribution several months ago. I got some good answers and this one in particular helped me gain some intuition by using the strategy of relating the problem to a similar one with 100 doors. With the analogous problem, I can more clearly see that switching is a good idea in the first scenario, whereas it doesn't matter if you switch in the second scenario.

However, while it gives me some more intuition, I don't feel like I really understand explicitly why there's a difference in the probabilities between the two scenarios. Certainly not well enough that I could explain it to someone else, which is my usual test for how well I understand something.

More specifically, I think the thing that's repeatedly getting me stuck is that I keep thinking back to the original problem and how it's explicitly stated that the first door Monty opens has a goat behind it. It seems like, regardless of how that came to happen, (whether Monty did it on purpose or he just happened to choose the goat door by chance and he could just as easily have chosen the door with the car), we have the same information about the system. And, from a Bayesian perspective, the information we have is what's important, right? That's what determines the prior probability.

In other words, why does it matter what information Monty has about the system, when he's not the one doing the probability calculation? Or, more generally (since I'm pretty sure my confusion here isn't inherently about the Monty Hall problem itself, but rather a deeper misunderstanding about Bayesian probability) shouldn't it only matter what information the person doing the calculation, using Bayes' theorem, has?

To put it another way, what if we assume a more abstract version of the problem: Agent 1 chooses randomly 1 door out n doors, knowing that there's target behind one door and non-targets behind all the others, but doesn't look behind the door.

Then, a second agent comes in, opens another door (with agent 1 having no idea about how they chose that particular door) and agent 1 sees that the opened door reveals a non-target. Does agent 1's lack of knowledge the method through which the open door was chosen affect how they should calculate the probabilities they their initially chosen door has the target, vs the probability that one of the other other unopened doors has it? For example, instead of asking if agent 2 knows where the target is, what if we ran various simulations of this generalized problem, which a different probability function used to choose the door to open each time? Say, in one simulation, a uniform probability distribution is used to randomly choose a door number to open (excluding the one already chosen by agent 1), and in another simulation, the number is again chosen randomly, but from (a discrete version of) a normal distribution. And in other case, we use some other weighted probability distribution. Would those different distributions used to choose which door agent 2 opens effect the probability of finding the target behind agent 1's door vs find it behind one of the other unopened doors? If so, how specifically? Like, how would exactly they be involved in the calculations using Bayes' theorem? And if not, why is this not analogous to the standard Monthly Hall problem vs a version where Monty has no more knowledge of which door the car is behind than the contestant?

• I think your understanding issue is that it's not "regardless of how that came to happen". The key point is that Monty knows where the goat is, so he opens that door. So, there's no new information when he opens that door. If the goat was behind the other door, he would have opened that one instead. Stlll, no new information in that case either. Commented Jun 28 at 16:02
• @mlofton, so you're saying the key is that the contestant ALSO knows, ahead of time, that Monty is always going to open a door with a goat behind it? So, it is the same, mathematically, as if Monty didn't know (or the contestant didn't know Money knows), but then they get the additional information that the open door has a goat behind it? Commented Jun 28 at 16:21
• Thought I should mention the frequentist solution to the Monty Hall problem as this clarifies what the structure of the problem implies (in contrast to what the contestant knows). Prior to choosing a door, a contestant who intends to randomly select a door has 1/3 probability of choosing the right door and 2/3 chance that the correct door is one of the other 2 doors. Of these 2 doors, the presenter will always open 1 door with a goat behind it. Commented Jun 28 at 20:36
• So for this contestant the remaining unselected door will have a 2/3 probability of being the correct door, (independently of whether or not the contestant knows that the presenter always will open a door with a goat behind it). Commented Jun 28 at 20:37
• Once the contestant selects a door, and the presenter opens one of the other doors, then at that point the contestant is entitled to have 33% confidence in her initial choice and 67% confidence that the car is behind the remaining door (if she understands the structure of the game). Commented Jun 28 at 20:59

I agree with you, but there's a big "however" following this opening section: it doesn't matter why he picks the door, as long as the door has the goat behind it. The presence of the goat throws out all the "universes" in which he randomly opens the door and the car is behind it.

A little simulation will support this conclusion:

wins <- 0
for (j in 1:10000) {
car <- sample(c(1,2,3), size=1)
me <- sample(c(1,2,3), size=1)

# MH random pick
monty <- sample(c(1,2,3), size=1)
while (monty == me | monty == car) {
monty <- sample(c(1,2,3), size=1)
}

# Switch
if (monty == 1) {
me <- ifelse (me == 2, 3, 2)
} else if (monty == 2) {
me <- ifelse (me == 1, 3, 1)
} else {
me <- ifelse (me == 1, 2, 1)
}

if (me == car) wins <- wins + 1
}

wins / 10000
[1] 0.6567


... obviously indicating that switching gives us a 2/3 chance of getting the car.

The key to the divergent opinions, IMO, lies in the timing of when we evaluate our strategy. If our strategy is evaluated before Monty chooses, then, because Monty at that point has a 1/3 chance of picking the car in those situations where we haven't already picked the car, our chance of winning with a "switch" strategy is 1/3 - because we won't get to exercise that strategy when Monty opens the door with the car, and when he doesn't, we have a 50/50 chance of having already picked the car. If our strategy is evaluated after Monty chooses, and we are in the "goat" world, our chance of winning with a switch strategy is 2/3 - because we've thrown out the 1/3 of the cases when we lose as soon as Monty opens the door. When Monty is constrained to open only a goat door, those 1/3 of the cases aren't possibilities at any stage of the game, so it doesn't matter when we evaluate our strategy - our chance of winning is 2/3.

Here's the simulation where we lose when Monty opens the door with the car:

wins <- 0
for (j in 1:10000) {
car <- sample(c(1,2,3), size=1)
me <- sample(c(1,2,3), size=1)

# MH random pick
monty <- sample(c(1,2,3), size=1)
while (monty == me) {
monty <- sample(c(1,2,3), size=1)
}

# Switch
if (monty != car) {
if (monty == 1) {
me <- ifelse (me == 2, 3, 2)
} else if (monty == 2) {
me <- ifelse (me == 1, 3, 1)
} else {
me <- ifelse (me == 1, 2, 1)
}

if (me == car) wins <- wins + 1
}
}

wins / 10000
[1] 0.3331


Since the "standard" version of the problem evaluates our strategy before Monty opens the door, which is fair because it's an evaluation of our probability of winning if we choose to play the game (at all) with an optimal strategy, the correct answer is that it doesn't matter what we do if Monty chooses at random, but it does if his choice is limited to a goat door.

• "The key to the divergent opinions, IMO, lies in the timing of when we evaluate our strategy" Ah! THIS is what was throwing me off. I think I get it now. Thank you! Commented Jun 28 at 21:18
• Graham and Jbowman: Thank you for explaining this in all the great detail. I just came back to this after initially commenting and see all the great stuff. I'm old so monty is old for me but it's so interesting that it still never goes away. Also, Mikayla, I'm glad that you understand now. I always think of it as: "Monte showed me NOTHING and door prob sums to 1.0." Commented Jun 29 at 6:54