Monty Hall problem: On the game show, Let’s make a Deal, you are shown four door

Question

Monty Hall problem: On the game show, Let’s make a Deal, you are shown four doors: A, B, C, and D, and behind exactly one of them is a big prize. Michael, the contestant, selects one of them, say door C, because he know from having watched countless number of past shows that door C has twice the probability of being the right door than door A or door B or door D. To make things more interesting, Monty Hall, game show host, opens one of the other doors, say door B, revealing that the big prize is not behind door B. He then offers Michael the opportunity to change the selection to one of the remaining doors (door A or door D). Should Michael change his selection? Justify your answer by calculating the probability that the prize is behind door A, the probability that the prize is behind door C, and the probability that the prize is behind door D, given that Monty Hall opened door B (to show that prize is not there), using Bayes Theorem.

Explanation / Answer

Answer to the question)

Given door C has double the chance of winning as compared to the rest of the doors

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So Michael selects door C:

If Michael selects door C and doesnot Switch , his P(w) = probability of winning = 0.4

P(L) = probability of losing is = 0.6

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Now since Monty Hall , opens door B and shows that there is no prize behind it , Michael has the option to switch to door A or D

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Now suppose Michael switches, and we need to find the probabiltiy of winning

This means door C does not have the prize, that is why when he switches he wins

Thus he selects wrong door first , P(C) = 0.4

and then he selects the right door, say suppose A has the prize , so if we need to find the probability of winning , probability of selecting A is 0.2

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Likewise we have a second case where D has the prize , and in order to win Michael selects door D , P(D) = 0.2

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Thus P(W) = 0.4* (0.2 +0.2) = 0.16

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P(L) = He will lose if he on swithcing also selects the worng door

Say the prize is behind gate D , and he chose A , P(A) = 0.2 , or vice versa

Thus P(L) = 0.16 ...[it would be same as P(W) when swtiched]

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Thus now we can see that if Michael does not switch he has 40% chance of winning , but if he swtiches he has only 16% of chances to win. Thus we would suggest Michael not to switch.

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