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In the Monty Hall problem, if we pick door 1, we say that probability that door 1 has the car is $\frac{1}{3}$. Even if Monty reveals door 2 with the goats (say), this probability doesn't change. This is equivalent to saying that probability of door 1 being bad ($=\frac{2}{3}$) remains same even after Monty reveals door 2 with the goats.

I want to prove this exact statement. This is my approach:

Let $C_i$ be the event that door $i$ is bad.

Let $M_i$ be the event that Monty reveals door $i$.

Hence $P(C_1) = P(C_1|M_2) = \frac{2}{3}$ is what we want to prove.

By Bayes Theorem, $$P(C_1|M_2)$$ $$=$$ $$\frac{P(M_2|C_1)P(C_1)}{P(M_2|C_1)P(C_1) + P(M_2|C_2)P(C_2) + P(M_2|C_3)P(C_3)}$$

The denominator is just $P(M_2)$ by law of total probability. I am not sure how to solve after this. I think solving it this way is tricky as we run into duplicate situations such as while calculating $P(M_2|C_1)$ and $P(M_2|C_2)$.

If we know door 1 is bad, there is a case where Monty reveals door 2 which is bad. Also, if we know door 2 is bad, there is also a case that Monty reveals door 2 , because he could not reveal door A which was also bad as we had chosen it earlier. Both cases had the prize behind door 3, which we overcount in separate cases. For 3 doors, maybe we can consider manually, but if we have the same problem with $n$ doors how to do it systematically?

I have little idea of handling overcounting with sophistication. Is there a neat way to do this? (except like solving for the complement of this problem which will give you $\frac{1}{3}$ and subtracting it from 1).

M.B.
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  • The Monty Hall problem has been written up extensively, with many Bayes' Theorem computations out there. here for instance. It's not clear what you are asking that's different from any of those other presentations. – lulu Jan 23 '25 at 10:50
  • And there are discussions of Bayes' theorem as applied to the Mothy Hall problem on this site such as the answers to this question – Henry Jan 23 '25 at 12:24
  • @lulu The conventional approaches to this problem with Bayes rule , involves defining $C_i$ as the event of the car being behind door $i$. I have approached it by assuming the opposite, i.e. $C_i$ is the event that door $i$ does not have the car. Now I want to show that $P(C_1)$ is the same ($=\frac{2}{3}$)after Monty reveals some door with goats,i.e. $P(C_1|M_2)=\frac{2}{3}$ – M.B. Jan 23 '25 at 15:10
  • @Henry The solutions you referenced to this problem with Bayes rule , involves defining $C_i$ as the event of the car being behind door $i$. I have approached it by assuming the opposite, i.e. $C_i$ is the event that door $i$ does not have the car. Now I want to show that $P(C_1)$ is the same ($=\frac{2}{3}$)after Monty reveals some door with goats,i.e. $P(C_1|M_2)=\frac{2}{3}$ – M.B. Jan 23 '25 at 15:14

1 Answers1

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Let $C_i$ be the event that door $i$ is bad.

Let $M_i$ be the event that Monty reveals door $i$.

By Bayes Theorem, $$\Bbb P(C_1|M_2) = \dfrac{\Bbb P(M_2\mid C_1)\,\Bbb P(C_1)}{\Bbb P(M_2\mid C_1)\,\Bbb P(C_1)+\Bbb P(M_2\mid C_2)\,\Bbb P(C_2)+\Bbb P(M_2\mid C_3)\,\Bbb P(C_3)}$$

The denominator is just $\Bbb P(M_2)$ by law of total probability.

No; that is not correct.   The issue is that these $C_1, C_2, C_3$ are not disjoint events.   So, they do not partition the outcome space.

Rather, exactly two among those events always occur simultaneously, so the correct application of L.o.T.P. gives:

$$\begin{align}\Bbb P(M_2) &= \Bbb P(M_2,C_1,C_2)+\Bbb P(M_2,C_1,C_3)+\Bbb P(M_2,C_2,C_3)\\[1ex] &= 1\cdot \tfrac 13+ 0\cdot \tfrac 13+\tfrac 12\cdot \tfrac 13\\[1ex]&=\tfrac 12\\[2ex] \Bbb P(M_2, C_1)&=\Bbb P(M_2,C_1,C_2)+\Bbb P(M_2, C_1, C_3)\\[1ex] &=1\cdot \tfrac 13+ 0\cdot \tfrac 13\\[1ex] &= \tfrac 13\\[2ex]\therefore\quad\Bbb P(C_1\mid M_2) &=\left. \tfrac 13\middle/\tfrac 12\right.\\[1ex]&=\tfrac 23 \end{align}$$

Graham Kemp
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    The reason the usual approach uses "the car hides behind door [number]" as the events is that there is only one car to hide. – Graham Kemp Mar 03 '25 at 23:19
  • Thats an elegant way to do it ! I did the version with $n$ doors myself with your help i.e. if $M_2,M_3,...,M_{i-1},M_{i+1},...,M_n$ be an event $M$ then $P(C_1)=P(C_1|M)=\frac{n-1}{n}$. When I started thinking about this problem, I thought about using Principle of Inclusion and Exclusion, as it goes hand in hand with overcounting problems but I couldn't do it myself. Any chances to do this with $PIE$ ? – M.B. Mar 26 '25 at 17:39