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Consider the birthday problem with m equally likely dates and k people. Here is a way to approximate the probability that there is no birthday match among the k people. There are C(k ,2) pairs that might match, and any such pair has probability 1/m of actually matching. If k is a small fraction of m , we can approximate the distribution for the number of matches by a Poisson distribution with parameter λ = C(k ,2)/m . Then the probability p that there are no matches is approximated by e–λ . Thus we have the following approximate relationship among m , k and p: exp{–C(k,2)/m} ≈ p. For example, we estimated that p < 1/2 for C(k ,2) > m log(2). Approximating C(k ,2) with k 2 /2, the last inequality gives k > 2m log(2) 29 ≈ 1.2 m . Choosing m = 80,089,128 we found in the text that only 10,740 Easy Pick tickets have to be sold to give a better than even chance that at least one pair would match. Now consider birthday triples. There are C(k ,3) triples that might match, and any such triple has probability 1/m 2 of actually matching. Thus, letting q denote the probability that there is no birthday triple among the m people, we have the Poisson approximation exp{–C(k ,3)/m 2 } ≈ q. Similarly, if r is the probability that there are no birthday quadruples, we obtain exp{–C(k ,4)/m 3 } ≈ r. In the text we considered the case of m = C(45,5) = 1,221,759 sets of 5 white numbers chosen from 45 (the old Powerball format) and k = 17,001 tickets. The approximations gives q ≈ .58 and r ≈ .998. The complementary probabilities were of interest in our earlier discussion.PDF Image | USING LOTTERIES IN TEACHING A CHANCE COURSE
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