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Let us take 3 doors A, B, C. Now, let us say car is in A.

Now if player selects A, host opens B, on switching player loses.

If player selects A, host opens C, on switching player loses.

If player selects B, host opens C, on switching player wins.

If player selects C, host opens B, on switching player wins.

So 50% winning prob instead of 66%. Although when I use total prob theorem, I do end up getting 66%. Now, can somebody explain to me the reason why the above mentioned cases are not equally likely?

RobPratt
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aditya gupta
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  • @RobArthan: Probably not, since this poster is asking a rather different question. – Brian M. Scott Aug 12 '20 at 21:43
  • How does the player know she should select A with twice as much probability as B? Because that’s what you’re asking her to do if you want those four cases to be equally likely. – David K Aug 13 '20 at 15:41

3 Answers3

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On the assumption that the player chooses $A,B$, and $C$ with equal probability $\frac13$, and that if the player chooses $A$, the host opens $B$ and $C$ with equal probability $\frac12$, the actual probability of your first event is $\frac13\cdot\frac12=\frac16$. The same goes for your second event. Each of your other two events, however, has probability $\frac13$.

As long as the host has a positive probability of opening each of doors $B$ and $C$ when the player chooses $A$, the probabilities of your first two events will be less than $\frac13$.

Brian M. Scott
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The case "Player select $A$ and host opens $B$" happens in one time on six and so for the case "Player select $A$ and host opens $C$".

user
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The four outcomes you have described do not have equal probabilities.

They have probabilities $1/6, 1/6, 1/3, 1/3$ respectively (the player selects a door randomly, and if they choose A the host randomly chooses between B and C, which is why the first two outcomes are $\frac{1}{3} \cdot \frac{1}{2}$), so the probability of winning is $1/3 + 1/3 = 2/3$.

angryavian
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    @amWhy: If you actually read and understand the answers at the link that Rob Arthan posted, you’ll find that this isn’t a duplicate. The OP is asking a rather different question. – Brian M. Scott Aug 12 '20 at 21:43
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    @amWhy The OP has formulated a specific attempt at solving the problem and seeks guidance on why it is wrong. The "duplicate" question is a broader open-ended question about the problem, and reading through all the answers may not illuminate the problem with OP's attempt. – angryavian Aug 12 '20 at 21:44
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    @amWhy The OP has listed 4 outcomes, where the player wins in 2 of them. They mistook the events as being equiprobable, leading to a winning probability of $1/2$. – angryavian Aug 12 '20 at 21:47
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    @amWhy The OP also mentioned that using a different method, they got the correct answer of $2/3$. In the last sentence of the post, they clearly ask that they are trying to find the error in their first attempt, and specifically "why the four cases are not equally likely." – angryavian Aug 12 '20 at 21:54