EXAMPLE 35

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Let s Make a Deal

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In this TV game show, a contestant is presented with three doors, one of which contains a valuable prize while the other two are empty The contestant is allowed to choose one door Regardless of the choice made, at least one, and perhaps two, of the remaining doors is empty The show host, say Monty Hall, opens one door and shows that it is empty He now offers the contestant the opportunity to change the choice of doors; should the contestant switch, or doesn t it matter It matters The contestant who switches has probability 2/3 of winning the prize If the contestant does not switch, the probability is 1/3 that the prize is won In thinking about the problem, note that when the empty door is revealed, the game does not suddenly become choosing one of the two doors that contains the prize The problem here is that sometimes Monty Hall has one choice of door to show empty and sometimes he has two choices of doors that are empty This must be accounted for in analyzing the problem

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Bayes Theorem

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An effective classroom strategy at this point is to try the experiment several times, perhaps using large cards that must be shuf ed thoroughly before each trial; some students can use the never switch strategy whereas others can use the always switch strategy and the results compared This experiment alone is enough to convince many people that the switching strategy is the superior one; we analyze the problem using geometry To be speci c, let us call the door that the contestant chooses as door 1 and the door that the host opens as door 2 The symmetry of the problem tells us that this is a proper analysis of the general situation Now we need some notation Let Pi , i = 1, 2, 3, denote the event prize is behind door i and let D be the event door 2 is opened We assume that P1 = P2 = P3 = 1/3 Then, P(D|P1 ) = 1/2, since in that case the host then has a choice of two doors to open; P(D|P2 ) = 0, since the host will not open the door showing the prize; and P(D|P3 ) = 1, since in this case door 2 is the only one that can be opened to show no prize behind it

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Our unit square is shown in Figure 37 It is clear that the shaded area in Figure 37 represents the probability that door 2 is opened The probability that the contestant wins if he switches is then the proportion of this area corresponding to door 3 This is 1 1 3 1 1 1 + 0+ 2 3 3 2 3

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P(P3 |D) =

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Another, perhaps more intuitive, way to view the problem is this: when the rst choice is made, the contestant has probability 1/3 of winning the prize The probability that the prize is behind one of the other doors is 2/3 Revealing one of the doors to be empty does not alter these probabilities; hence, the contestant should switch

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BAYES THEOREM

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Bayes theorem is stated here although, as we have seen, problems involving it can be done geometrically

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3

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Conditional Probability

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Bayes theorem: If S = A1 A2 An , where Ai and Aj have no sample points in common if i = j, then if B is an event, / P(Ai |B) = P(Ai |B) = or P(Ai |B) = P(Ai ) P(B|Ai )

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n j=1

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P(Ai B) P(B) P(Ai ) P(B|Ai ) P(A1 ) P(B|A1 ) + P(A2 ) P(B|A2 ) + + P(An ) P(B|An )

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P(Aj ) P(B|Aj )

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In Example 35 (Let s Make a Deal), Ai is the event Prize is behind door i for i = 1, 2, 3 (We used Pi in Example 35) B was the event door 2 is opened

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