I know that this question has been asked frequently but I don't know where my argument went wrong. This is Monty Hall Variant, where the initial setting is exactly the same, except Monty forgot which door contained the prize, so he had to randomly choose one door among the remaining two other doors, and it turns out the door he chose contains a goat. (If he happens to choose the door that contains the prize, then you lose)

For example, WLOG you have initially picked the door A. Then Monty randomly chooses one door either from B or C, but let's say C for the sake of argument. Then the prize is either behind A or B. Then let's define some notations:

Let A,B, and C be events that door A contains the prize, door B contains the prize, and door C contains the prize respectively. Also let Mc be the event that Monty chooses the door C.
Then I want to compare P(A|Mc) and P(B|Mc).

Then all there is left with is using bayes' rule to solve both probabilities, and in order to solve it, I need to find P(Mc|A), P(Mc|B), P(Mc|C), but I argued that they are all 1/2 because no matter where the prize is hidden, he has to choose the door B or C with equal probability, namely 1/2, so conditioning on where the prize is hidden won't affect Monty's decision to choose the door among B or C, except he can't choose A since it has been already chosen by you. (I think this is where I screwed up, but I can't find what went wrong).

Then I find P(Mc) which turns out to be 1/2, and then I compute P(A|Mc) and P(B|Mc), which is 1/3 for both of them. The only part I got it right is that there is no advantage gained by switching doors, but from what I have googled, the probability that the prize is behind in either of these doors must be 1/2. What went wrong? Any input appreciated.

  • $\begingroup$ This has been asked several times with answers here and there. Is it this one ? math.stackexchange.com/questions/41807/… $\endgroup$ – mathreadler Feb 28 '17 at 10:39
  • $\begingroup$ Yea, I already know the answer but I want to know why above approach doesn't work. $\endgroup$ – Ted Feb 28 '17 at 10:44
  • $\begingroup$ @mathreadler: It appears the question is not "what is the right argument?", but "where does this proposed argument go wrong?", so probably not a duplicate. $\endgroup$ – hmakholm left over Monica Feb 28 '17 at 10:44
  • $\begingroup$ Ah ok. In my experience for most questions related to probability the largest chance of error is misinterpretation / miscommunication of the event space. $\endgroup$ – mathreadler Feb 28 '17 at 10:47
  • $\begingroup$ There are 3 cases, which are all equally likely - you choose the prize and he shows you and empty box, you choose and empty box and he shows you an empty box, you choose an empty box and he shows you the prize - the two cases where you can carry on are equally likely, in one you lose if you swap and in the other you win if you swap - therefore your chances of winning are a a third (taking into account the cases where monty picks the prize and you lose) $\endgroup$ – Cato Feb 28 '17 at 11:04

You're forgetting $P(C\mid M_C)$ which is also $1/3$.

If you want to model the situation that Monty randomly opens door $C$ and that is not where the prize is, you need to state that directly in the condition. The probabilities then jump to $$ P(A\mid M_C\land\neg C) = \tfrac12 \qquad\qquad P(B\mid M_C\land\neg C) = \tfrac12 $$

As you note, this doesn't affect $P(M_C)$, but since $P(M_C\land \neg C)$ is smaller than $P(M_C)$ (as it is perfectly possible that Monty opens door C and discovers a door), the corresponding conditional probabilities can still differ.

Computing $P(B\mid M_C)=P(C\mid M_B)=1/3$ correctly tells you that your chances of winning this entire game is $1/3$ if your strategy is "choose door A first, and then if you don't immediately lose, switch doors".

| cite | improve this answer | |
  • $\begingroup$ Yes that is the difference where some assume the condition and some don't. $\endgroup$ – mathreadler Feb 28 '17 at 10:48
  • $\begingroup$ That's weird. When my prof first introduced Monty Hall problem, he never explicitly specified the condition that that the door that Monty opened contained a goat. For example, you initially pick A, and Monty open one of the remaining doors (B or C), say B, containing no prize. Then P(A|M_B) vs P(C|M_B). Then P(M_B|A)=1/2, P(M_B|C)=1, P(M_B|B)=0, So P(M_B)=1/2 using total probability. So P(A|M_B)=1/3, P(C|M_B)=2/3. $\endgroup$ – Ted Feb 28 '17 at 11:05
  • $\begingroup$ @Ted: That's because in the standard variant it is part of the rules that the opened door must give a goat -- so in that probability space the events $M_C$ and $M_C\land \neg C$ are the same event. Since you have changed the rules, this can no longer be assumed. $\endgroup$ – hmakholm left over Monica Feb 28 '17 at 11:08
  • $\begingroup$ The difference is in the idea that immediate loss occurs if Monty shows the car. In the version I have seen the game ends in neither a win nor a loss if he accidentally picks it.. $\endgroup$ – mathreadler Feb 28 '17 at 11:08
  • $\begingroup$ Ahh...okay. Thanks a lot. $\endgroup$ – Ted Feb 28 '17 at 11:13

In the version I have seen you are supposed to forget that $P(C|M_C)$ probability because that never happens : there is no choice left to be made if it happens so there is no game to be played. If you remove that 1/3 perceived as a loss then we stay at 2/3 probability as in the original Monty hall.

So there are at least 3 different interpretations of the event space, all depending on how the event that Monty shows car is treated.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.