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Coupon collector's problem - Wikipedia

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In probability theory, the coupon collector's problem refers to mathematical analysis of "collect all coupons and win" contests. It asks the following question: If each box of a brand of cereals contains a coupon, and there are n different types of coupons, what is the probability that more than t boxes need to be bought 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 Θ ( 𝑛 log ⁡ ( 𝑛 ) ) .[a] For example, when n = 50 it takes about 225[b] trials on average to collect all 50 coupons. By definition of Stirling numbers of the second kind, the probability that exactly T draws are needed is Let time T be the number of draws needed to collect all n coupons, and let ti be the time to collect the i-th coupon after i − 1 coupons have been collected. Then 𝑇 =

Coupon collector's problem - Wikipedia Jump to content From Wikipedia, the free encyclopedia Problem in probability theory n\\in\\mathbb N</math> - number of faces on die"},"support":{"wt":"<math>k\\in\\mathbb N</math> - rolls taken for all faces to appear"},"pdf":{"wt":"<math>\\frac{(n-1)^{\\{k-1\\}}}{n^{k-1}}</math>"},"cdf":{"wt":"<math>\\frac{n^{\\{k\\}}}{n^k}</math>"},"mean":{"wt":"<math>nH_n</math>"},"variance":{"wt":"<math>n^2H^{(2)}_n-nH_n</math>"},"skewness":{"wt":"<math>\\frac{2n^3H^{(3)}_n-3n^2H^{(2)}_n+nH_n}{\\left(n^2H^{(2)}_n-nH_n\\right)^{3/2}}\\ \\underset n\\sim\\ 6^{3/2}2\\fra

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