I am very confused by the Monty Hall Problem. See Wikipedia for the set up.

Suppose we have the three doors $A,B,C$. Suppose $t$ indicates rounds of choosing.

Round 1

Prior to round one, no doors have been opened. Thus, $$P_{t=1} (A)= P_{t=1} (B)=P_{t=1} (C)=\frac{1}{3}$$

The contestant chooses a door.

Round 2

Suppose the contestant choose door $A$. We now learn the prize is not behind door $C$. Therefore, we know $$P_{t=2} (C) = 0$$

But what about A and B.

My Question

Would this problem be less paradoxical if we defined "equally probable" as merely $P(A^c) = P(B^c) = P(C^c) = \frac{2}{3}$?

My Reasoning

In other words, let's suppose we do not know $P(A),P(B),P(C)$ but do know the probabilities for their complements. Then if the contestant chose door A and is shown door C is empty, there would be no surprise since all this has told us is now we have $P(C)$. So we can find $P(B)$ since $P(B) = P(A^c) - P(C)$. This logic makes sense and avoids confusion and leads to the $\frac 2 3$ answer cleanly because we never knew $P(A)$, just $P(A^c)$. So $P(C) = 0$ implies $P(B) = \frac 2 3$ since $P(A^c) = \frac 2 3$.

To me, the Monty Hall solution doesn't make sense if you define the probabilities to the doors explicitly. If you define the probabilities, the logic is "I have $n$ doors, so each should have a $\frac{1}{n}$ chance." But this procedure shouldn't change in the second round because there is no difference in probability between $P(\text{revealed empty door})$ and $P(\emptyset)$, meaning we could have just not included $C$ to begin with. In which case, there are just $2$ doors. So the probabilities for each should be $\frac{1}{2}$.


3 Answers 3


This is a clever way to think about it and it does work. One can clean up the reasoning you give with probability. In particular, essentially, the important statement is the following:

The event that the prize isn't behind door $A$ is independent of the event that the host reveals door $C$.

This is because the last step of your argument where you write $$P(B)=P(A^C)-P(C)$$ needs all those probabilities to be calculated given that the host revealed door $C$ and if $P(A^C)$ weren't independent of that event, your proof fails. Formally, if $R_C$ is the event that the host reveals door $C$, we should be writing: $$P(B|R_C)=P(A^C|R_C)-P(C|R_C).$$

You can see that the host revealing door $C$ is independent from the prize being behind door $A$ intuitively by noting that, if the prize isn't behind door $A$, then the host has to reveal the door the prize isn't behind - so either $B$ or $C$ with $50\%$ probability. Similarly, if the prize is behind door $A$, then the host chooses between doors $B$ and $C$ arbitrarily with $50\%$ probability*. This essentially says $P(R_C|A)=P(R_C|A^C)$ which suffices to show independence.

With that technicality out of the way, we trivially have $P(A^C|R_C)=P(A^C)=\frac{2}3$ and easily achieve the desired result that $P(B)=\frac{2}3$. It's worthy of note that this also means that $A$ is independent of $R_C$, meaning that we, in a way, "lock in" the probability of $A$ being right to $\frac{1}3$ when we choose it, and the host's actions do not affect it since, from our point of view, they act the same regardless of whether the prize is behind our door or not.

If the host doesn't follow this randomized procedure - e.g. if they always chose $C$ when the prize is behind $A$, then the argument fails and $P(B|R_C)$ is in fact less than $\frac{2}3$. This is a subtlety not captured in the argument you give. Moreover, this reveals an important feature: the event that the host reveals door $C$ is different from the event that the prize isn't behind door $C$. That is to say, the naive argument of having a probability behind each door essentially says correctly that $P(B|C^C)=\frac{1}2$, but then incorrectly assumes this is the same as $P(B|R_C)$. This is what defeats the idea of treating door $C$ like it was never even there.

  • $\begingroup$ Wow, that was amazing. Thanks for the details. I learned a lot! $\endgroup$ Jan 6, 2016 at 5:44

You choose a door, hopefully without bias.   Whatever door this is, we'll now call this door A.   The probability that you chose well is: $\mathsf P(A)=1/3$.

A door without a prize behind it is then opened.   This is not entirely a random process.   Whichever door is opened, we'll now call this door C.

One other door remains.   We now call this door B.

In other words, let event $A$ be that you chose well in the first place, and event $B$ be that changing your mind after the elimination pays off.

Either you chose the door with the prize or you didn't.   If you change your mind after one of the doors without a prize is eliminated, what is the probability that the prize is behind the remaining door?

Case 1: If you chose well in the first place, the prize certainly isn't behind the remaining door. $\mathsf P(B\mid A)=0$

Case 2: If you chose poorly in the first place, the prize certainly is behind the remaining door. $\mathsf P(B\mid A^\complement)=1$

So by the rule of total probability: $$\mathsf P(B)=\mathsf P(A)\mathsf P(B\mid A)+\mathsf P(A^\complement)\mathsf P(B\mid A^\complement) \\ = \tfrac 1 3 \cdot 0 + \tfrac 2 3 \cdot 1 \\ = \dfrac 23$$

What if there are $n$ doors rather than $3$?

$$\mathsf P(B_n)=\mathsf P(A_n)\mathsf P(B_n\mid A_n)+\mathsf P(A_n^\complement)\mathsf P(B_n\mid A_n^\complement) \\ = \tfrac 1 n \cdot 0 + \tfrac {n-1} n \cdot \tfrac 1{n-2} \\ = \dfrac {n-1}{n(n-2)}$$

So if there are more than three doors (and only one is eliminated), changing your mind is not a good bet.

  • 1
    $\begingroup$ Why do you say that "if there are more than three doors (and only one is eliminated), changing your mind is not a good bet"? Certainly $\frac{n-1}{n(n-2)}>\frac{1}n$ so it still increases the probability that you get it. $\endgroup$ Jan 6, 2016 at 5:17

I'd say that with a problem like this, it's more important to simply model the problem itself right - using complement notation doesn't seem to make much of a difference to me, and it doesn't help me understand the problem more than I already understood it. Focusing on symbols, and trying to make conclusions based on symbols/symbolic manipulation, and what you think they imply, in a problem like this... it seems to make it harder than it is to me - just model the problem right, and it will all work out fine.

In the last paragraph you say "But this procedure shouldn't change in the second round..." - I don't really understand your argument, but I disagree, you don't have $n$ (in this case $2$) doors in a general way anymore, you have a door that likely doesn't have a car behind it, one other door, and you know that there's car behind one of them, so it seems quite clear to me what the probabilities are.


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