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coupon collector's problem
In probability theory, the coupon collector's problem describes the "collect all coupons and win" contests. It asks the following question: Suppose that there is an urn of ''n'' different coupons, from which coupons are being collected, equally likely, with replacement. What is the probability that more than ''t'' sample trials are needed to collect all ''n'' coupons? An alternative statement is: Given ''n'' coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once? The mathematical analysis of the problem reveals that the expected number of trials needed grows as .〔Here and throughout this article, "log" refers to the natural logarithm rather than a logarithm to some other base. The use of Θ here invokes big O notation.〕 For example, when ''n'' = 50 it takes about 225〔E(50) = 50(1 + 1/2 + 1/3 + ... + 1/50) = 224.9603, the expected number of trials to collect all 50 coupons. The approximation for this expected number gives in this case .〕 trials to collect all 50 coupons.
==Understanding the problem==
The key to solving the problem is understanding that it takes very little time to collect the first few coupons. On the other hand, it takes a long time to collect the last few coupons. In fact, for 50 coupons, it takes on average 50 trials to collect the very last coupon after the other 49 coupons have been collected. This is why the expected time to collect all coupons is much longer than 50. The idea now is to split the total time into 50 intervals where the expected time can be calculated.
抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』
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