The Monty Hall Problem is Really About Policy Assumptions

10 Feb 2020

9 minute read

TL;DR: the Monty Hall problem doesn’t have a well-defined solution.

The Monty Hall problem is a classic “probability paradox”1, where the answer is counter-intuitive to most people’s intuitions. To paraphrase Wikipedia’s introduction to the problem2:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a prize; behind the others, nothing. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has nothing. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

This can be reframed as “what is the probability that the prize is behind door 2?” The “intuitive answer” is 1/2 (meaning there is no advantage to switching), since initially the prize could be behind any of the 3 doors, and you find out that it isn’t behind door #3, leaving just 2 doors left with equal probability of having the prize.

In contrast, the “correct answer” is 2/3. This discrepancy is usually explained away as the host giving away information with their choice of which door to open. Again, paraphrasing Wikipedia:

An intuitive explanation is that, if the contestant initially picks an empty door (2 of 3 doors), the contestant will win the prize by switching because the other empty door can no longer be picked, whereas if the contestant initially picks the prize (1 of 3 doors), the contestant will not win the prize by switching.

However, in this post I am arguing that neither of these answers are “correct”, because the answer ultimately depends on your assumptions on what actions the game show host would take in counterfactual situations. Since this is not specified in most versions of the problem, the problem can’t really be said to have a “correct answer”. After introducing the math necessary to solve the problem, I will work through some example assumptions and their resulting implications, and conclude with the surprising result that under the right assumptions, the answer to the problem could be any probability.

Bayesian Formulation of the Problem

Let:

• $$c$$ be the contestant’s coice of door
• $$o$$ be the door the host opens
• $$a$$ be the answer (the door the prize is actually behind)

In a Bayesian decision framework, the way to handle this problem is to:

1. Use your knowledge of the situation to construct a probability distribution over the relevant variables ($$p(o,a,c)$$).
2. Use the values of the observed variables $$c$$ and $$o$$ to calculate the posterior probability distribution over the answers: $$p(a\mid o,c)$$.
3. Use this probability distribution to make an informed choice. In this situation, reasonable choice would be to choose the most likely answer, $$a^* = \arg\max_a p(a\mid o,c)$$.

Therefore solving the problem is the same as finding the posterior distribution. The posterior distribution can be found using Bayes theorem:

$p(a\mid o,c) = \frac{p(o,a,c)}{p(o,c)} = \frac{p(o,a,c)}{\sum_{a'} p(o,a',c)}$

This is far as we can go without making any assumptions. The most basic assumptions (which aren’t that interesting) is that the game is fair ($$a$$ is chosen randomly in advance) and real (the player doesn’t know anything about $$a$$, so their choice is effectively random). This means that the only interesting behaviour is that of the host, who can choose which door to open, knowing both where the prize is and the player’s choice. Mathematically, this means that the distribution $$p(o,a,c)$$ can be factored as:

$p(o,a,c) = p(a)p(c)p(o\mid a,c)$

Substituting this into the posterior equation above, and nothing that $$p(c)$$ is a constant and cancels, we find:

$p(a\mid o,c) = \frac{p(o,a,c)}{\sum_{a'} p(o,a',c)} = \frac{p(o\mid a,c)p(a)p(c)}{\sum_{a'}p(o\mid a',c)p(a')p(c)} = \frac{p(o\mid a,c)p(a)}{\sum_{a'}p(o\mid a',c)p(a')}$

Assuming further that all answers are equally probable, this further simplifies to:

$p(a\mid o,c) = \frac{p(o\mid a,c)}{\sum_{a'}p(o\mid a',c)}$

This means that for this particular problem, the posterior can be computed with only 3 values: $$p(o=3\mid a_i, c=1)$$, $$i\in\{1,2,3\}$$.

Examples

Standard assumptions: the argument for 2/3

When most people hear this problem, beyond the assumptions outlined in the previous section, they usually assume the following things:

1. the host will not open the door with the prize behind it
2. the host will not open the player’s door
3. the host will always open a door
4. if the host can open multiple doors, they will choose one at random

Assumptions 1-2 make sense, because otherwise the game show might not be any fun to watch. Assumption 3 isn’t necessary but is usually assumed because when people imagine this game show they usually imagine that this switching gimmick is just part of the show. Assumption 4 makes sense, since there is no reason a priori to assume that the host will be biased towards any particular door.

With this, we can calculate the 3 values of $$p(o\mid a,c)$$:

1. $$p(o=3\mid a=1,c=1)=1/2$$: the host could open doors 2 or 3 here, and would therefore pick at random. This would have them pick door 3 50% of the time.
2. $$p(o=3\mid a=2,c=1)=1$$: here the host cannot choose door 1 (since the player chose it) or door #2 (since it has the prize), so opening door #3 is the only option
3. $$p(o=3\mid a=3,c=1)=0$$: if door #3 had the prize, the host wouldn’t have opened it, so this event isn’t possible.

So, $$p(a=2\mid o=3,c=1) = \frac{1}{1/2+1} = 2/3$$, suggesting that the best choice is to pick door 2 (switch doors). This is the classic “correct answer” of 2/3.

Biased random choice of host

Suppose that assumption 4 above isn’t true: when the host has multiple doors that they could choose, they don’t make a truly random decision3. One example of this is if the host tends to pick higher numbers over lower numbers: that is, given a choice of opening doors 2 or 3, they will pick door 3 with probability $$z$$, which might not be $$1/2$$.

This scenario is similar to the standard one, except $$p(o=3\mid a=1,c=1)=z$$, so $$p(a=2\mid o=3,c=1)=\frac{1}{1+z}$$, which can vary between 1/2 and 1 for different values of $$z$$. Already, this formulation (which isn’t completely unrealistic) gives a huge range of possible answers, all contingent on your belief about how the host picks random numbers.

Inattentive host

Suppose the host didn’t see what door the player chose. The host’s decision then can’t depend on $$c$$ (breaking assumption 2). Instead, the host will randomly open one of the doors that doesn’t have the prize, regardless of whether the player chose this door. This means $$p(o=3\mid a=i,c=1)=1/2$$ for $$i=1,2$$ ($$i=3$$ remains unchanged since the host won’t open the door with the answer behind it). Therefore, $$p(a=2|o=3,c=1)=1/2$$. This scenario is certainly plausible in a real game show (although fairly unlikely).

No door as an option

If the host can choose not to open a door, this opens up a wide range of possibilities:

Dramatic host 1

To make the show more entertaining to watch, the host wants to bait the player into switching doors, not getting the prize, and then watch them agonize over their decision to switch. One way to do this would be to only open a door if the player’s initial guess was correct.

This means $$p(o=3\mid a=1,c=1)=1/2$$, and the other 2 probabilities are 0, so the posterior probability $$p(a=2\mid o=3,c=1)=0$$.

Dramatic host 2

Dramatic host 1 is too predictable (since being offered to switch doors means you should never switch). Dramatic host 2 becomes slightly less predictable by choosing with probability $$q$$ to open a wrong door if the player initially chose correctly. This means that $$p(o=3\mid a=1,c=1)=1/2$$ (like last time), but $$p(o=3\mid a=2,c=1)=q$$, making the posterior probability $$p(a=2\mid o=3,c=1)=\frac{q}{1/2+q}$$, which can range from 0 to 2/3.

Merciful host

Instead, a host could open a door only if the player’s initial choice was wrong, yielding a posterior probability $$p(a=2\mid o=3,c=1)=1$$. In this case, you would always want to switch doors if offered.

Only door 3 can be opened

Imagine that the host will act normally (i.e. standard assumptions), but if the final decision is to open doors 1 or 2 they will simply not open a door at all (they are only willing to open door 3)4. This means that $$p(o=3\mid a=1,c=1)=1$$, so the posterior $$p(a=2\mid o=3,c=1)=1/2$$.

Rigged game: the answer could be anything

Suppose that our initial model of the causal process is wrong, and the game is unfair. Specifically, the player chooses a door, and then the correct door is chosen afterwards with this information5. Specifically, suppose that the producers select the player’s door as correct with probability $$q$$, and one of the other 2 doors with probability $$1-q$$. Note that if $$q=1/3$$, this corresponds to the standard assumptions of a fair game.

To calculate the posterior in this scenario, we need $$p(a\mid c)$$ and $$p(o\mid a,c)$$, which are the following:

• $$p(a=1\mid c=1) = q$$ (by assumption above)
• $$p(a=2\mid c=1) = p(a=3\mid c=1) = (1-q)/2$$ (since the probabilities must add to 1)
• Since the hosts actions don’t change, $$p(o\mid a,c)$$ is the same as before ($$p(o=3\mid a=1,c=1)=1/2$$, $$p(o=3\mid a=2,c=1)=1$$, $$p(o=3\mid a=3,c=1)=0$$)

Reapplying Bayes rule, we get:

$p(a\mid o,c) = \frac{p(o\mid a,c)p(a\mid c)}{\sum_{a'}p(o\mid a',c)p(a'\mid c)}$ $p(a=2\mid o=3,c=1) = \frac{1\times(1-q)/2}{q/2 + 1\times(1-q)/2 + 0\times(1-q)/2} = 1-q$

Since $$q$$ can take on any value between 0 and 1, the posterior value can be any value between 0 and 1, meaning that the right decision could be absolutely anything. Of course, this is far from what most people assume when they encounter the Monty Hall problem, but if you were actually on such a game show in real life and offered this deal, would it not be prudent to at least consider this possibility?

Conclusion

In this post, I’ve shown how Bayes rule can be used to solve the Monty Hall problem, which under standard assumptions gives the “correct answer” of 2/3. However, under different assumptions, the posterior probability of the prize being behind door 2 can take on many different values, which between a biased host and a host that wants to make you lose spans the whole range of probabilities between 0 and 1. In fact, just assuming that the game isn’t fair yields a model where it could be better to switch with probabilities ranging from 0 to 1 inclusive!.

Overall, it’s very misleading to suggest that this is just a simple problem that tests if you can do some algebra; the entire solution is completely determined by one’s beliefs about the host’s counterfactual actions, so it’s frankly surprising that this aspect of the problem isn’t more widely discussed. As always, while an explicit Bayesian formulation isn’t required to arrive at the desired answer, the Bayesian framework does aid greatly in making your assumptions explicit.

1. Specifically it would be a veridical paradox, because the result appears absurd but is nonetheless true.

2. the original version used cars and goats, which I changed to just a prize/nothing for clarity.

3. since people are generally quite bad at choosing truly random numbers

4. maybe they are superstitious and like the number 3?

5. The producers of the game show might do this to increase the amount of dramatic tension on the show. Who says that “reality” TV needs to be real?