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The “Monty Hall Problem”

The “Monty Hall Problem”. The Problem: Monty Hall is (was?) the host of a TV game show called “Let’s Make a Deal”. On the show there are 3 doors, behind one of which is a prize. Monty Hall, the host, asks you to pick a door. Let’s say you pick door A.

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The “Monty Hall Problem”

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  1. The “Monty Hall Problem” The Problem: Monty Hall is (was?) the host of a TV game show called “Let’s Make a Deal”. On the show there are 3 doors, behind one of which is a prize. Monty Hall, the host, asks you to pick a door. Let’s say you pick door A. MH opens (e.g.) door B and shows there is nothing behind door B. You are now given the choice of either sticking with your original choice of door A, or switching to door C. Should you switch? To answer this question we will use Bayes’s Theorem (Included in Lecture 1 but not discussed in class) P416 Monty Hall

  2. Bayes’s Theorem relates conditional probabilities. It is widely used in many areas of the physical and social sciences. Bayes’s Theorem Let A1, A2,..Ai be a collection of mutually exclusive and exhaustive events with P(Ai)>0 for all i. Then for any other event B with P(B)>0 we have: Lecture 1 page 16 We call: P(Aj) the aprori probability of Aj occurring P(Aj|B) the posterior probability that Aj will occur given that B has occurred P(B|Aj) the likelihood We want to calculate: P(A|MH opens B) = probability the prize is behind door A and MH opens door B P(C|MH opens B) = probability the prize is behind door C and MH opens door B P416 Monty Hall

  3. P(A|MH opens B) = probability the prize is behind door A and MH opens door B P(A|MH opens B) = P(MH opens B|A)*P(A)/P(MH opens B) P(C|MH opens B) = probability the prize is behind door C and MH opens door B P(C|MH opens B) = P(MH opens B|C)*P(C)/P(MH opens B) Apriori, the prize can be behind doors A, B, C with equal probability: P(A)=P(B)=P(C)=1/3 We need to calculate the probability that MH opens door B: P(MH opens B)=P(MH opens B|A)*P(A)+P(MH opens B|B)*P(B) +P(MH opens B|C)*P(C) We can calculate the likelihoods assuming you’ve chosen door A: The probability that MH opens door B if the prize is behind door B: P(MH opens B|B) = 0 (game would be over, no fun!) The probability that MH opens door B if the prize is behind door A: P(MH opens B|A) = 1/2 (could have opened B or C) The probability that MH opens door B if the prize is behind door C: P(MH opens B|C) = 1 (no other choice door A is already picked, C gives game away) P(MH opens B)=1/3*1/2 +1/3*0+1/3*1=1/6+1/3=1/2 P(A|MH opens B) = (1/3*1/2)/(1/2)=1/3 P(C|MH opens B) = (1/3*1)/(1/2)=2/3 You Should SWITCH !!!! P416 Monty Hall

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