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This apparent paradox, that probabilities are not transitive, continues with patterns involving patterns of length 3 We showed above the average waiting times for each of the eight patterns that can occur when a fair coin is tossed (Table 76) Now let us play the coin game again where A is the rst player and B is the second player We show the probabilities that B beats A in Table 77 Note that letting > means beats (probably) TTH > HTT > HHT > THH > TTH!
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Waiting Time Problems Table 77 A s Choice HHH HHT HTH HTT THH THT TTH TTT B s Choice THH THH HHT HHT TTH TTH HTT HTT P (B beats A) 7/8 3/4 2/3 2/3 2/3 2/3 3/4 7/8
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Nor is it true that a pattern with a shorter average waiting time will necessarily beat a pattern with a longer waiting time It can be shown that the average waiting time for THTH is 20 tosses and the average waiting time for HTHH is 18 tosses Nonetheless, the probability that THTH occurs before HTHH is 9/14 Probability contains many apparent contradictions
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WHO PAYS FOR LUNCH
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Three friends, whom we will call A, B, and C, go to lunch regularly The payer at each lunch is selected randomly until someone pays for lunch for the second time On average, how many times will the group go to lunch Let X denote the number of dinners the group enjoys Clearly, X = 2, 3, or 4 We calculate the probabilities of each of these values If X = 2, then we have a choice of any of the three to pay for the rst lunch Then the same person must pay for the second lunch as well The probability of this is 1/3 If X = 3, then we have a choice of any of the three to pay for the rst lunch Then we must choose one of the two who did not pay for the rst lunch and nally, we must choose one of the two previous payers to pay for the third lunch There are then 3 2 2 = 12 ways in which this can be done and since each way has probability 1/27, the probability that X = 3 is 12/27 = 4/9 Finally, if X = 4 then any of the three can pay for the rst lunch; either of the other two must pay for the second lunch and the one remaining must pay for the third lunch The fourth lunch can be paid by any of the three so this gives 3 2 1 3 = 18 ways in which this can be done Since each has probability (1/3)4 , the probability that X = 4 is 18/81 = 2/9 These probabilities add up to 1 as they should The expected number of lunches is then E(X) = 2 3/9+3 4/9+4 2/9 = 26/9 This is a bit under three, so they might as well go to lunch three times and forget the random choices except that sometimes someone never pays Now what happens as the size of the group increases Does the randomness affect the number of lunches taken
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Who Pays for Lunch
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Suppose that there are four people in the group, A, B, C, and D Then X = 2, 3, 4, or 5 We calculate the probabilities in much the same way as we did when there were three for lunch 4 1 1 P(X = 2) = = 4 4 4 4 3 2 3 P(X = 3) = = 4 4 4 8 P(X = 4) = And nally, P(X = 5) = 4 3 2 1 4 3 = 4 4 4 4 4 32 4 3 2 3 9 = 4 4 4 4 32
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and these sum to 1 as they should Then E(X) = 2 1/4 + 3 3/8 + 4 9/32 + 5 3/32 = 103/32 = 3 218 75 This is now a bit under 4, so the randomness is having some effect To establish a general formula for P(X = x) for n lunchers, note that the rst payer can be any of the n people, the next must be one of the n 1 people, the third one of the n 2 people, and so on The next to the last payer is one of the n (x 2) people and the last payer must be one of the x 1 people who have paid once This means that n n 1 n 2 n 3 n (x 2) x 1 P(X = x) = n n n n n n This can be rewritten as 1 n P(X = x) = x (x 1)(x 1)!, x = 2, 3, , n + 1 n x 1 If there are 10 diners, the probabilities of having x lunches are shown in Table 78
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Table 78 x 2 3 4 5 6 7 8 9 10 11 P(X = x) 01 018 0216 02016 01512 009072 0042336 00145152 000326592 000036288
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