Questions are frequently asked about the odds of winning a lottery, and it usually seems to be the same type of lottery, whether it's in the UK or in various US states. The player selects k distinct integers from the range 1 to N. Then the "house" randomly chooses k distinct numbers from the same range, and the player's winnings depend on how many of his numbers match the "house" numbers. Someone usually explains that the probability of matching j numbers is C(N-k,k-j) C(k,j) ----------------- C(N,k) where C(m,n)=m!/(n!(m-n)!). Although this gives a pedagogically useful motivation for (and proof of) the elementary identity k ______ \ / \ / \ / \ > | N-k | | k | | N | / | k-j | | j | = | k | ------ \ / \ / \ / j = 0 it isn't a particularly challenging structure. Suppose we drop the requirement for DISTINCT integers. Let's say when you purchase a "chance" you are given five randomly selected integers from the range 1 to 45, DUPLICATIONS ALLOWED. Then the house randomly selects five integers from the same range (again, duplications allowed). What is the probability of matching the house's multi-set of five numbers exactly? (Note that the order of selection is unimportant.) This seems like a more interesting process because the result occurs in two stages. First, your personal numbers are selected and, although this doesn't determine whether you win or lose, the partition you draw does determine your odds of ultimately winning. (Ideally you would hope to get five distinct numbers, because that gives you the best odds, whereas five of a kind gives the worst odds.) Then the house draws its numbers to actually determine the winner(s). Whether this scheme actually has more popular appeal than the conventional method, it certainly is a bit more challenging to figure the odds. The seven possible partitions are aaaaa 45 = (45) 1 aaaab 1980 = (45)(44) 5 aaabb 1980 = (45)(44) 10 aaabc 42570 = (45)(44)(43)/2 20 aabbc 42570 = (45)(44)(43)/2 30 aabcd 595980 = (45)(44)(43)(42)/6 60 abcde 1221759 = (45)(44)(43)(42)(41)/120 120 The product in the center column is the number of possible occurrences of the respective partition, without regard to order of selection. The right hand column gives the number of distinct sequences that can give each of the occurrences of the respective partition. Suppose you draw a multi-set of numbers with the partition "aabcd". Out of the 45^5 possible selection sequences, there are (60)(595980) that give this partition, so your probability of drawing one of these is (60)(595980)/(45^5). Then to actually win you need the house to get one of the 60 selection sequences that give your specific values of a, b, c, and d. So your odds of actually winning, given a multi-set of the form aabcd, are 60/(45^5). The odds for the other partitions can be figured out in the same way, and the combined result is that your overall probability of winning is { 45(1)^2 + 1980(5)^2 + 1980(10)^2 + 42570(20)^2 + 42570(30)^2 + 595980(60)^2 + 1221759(120)^2 } / (45^10) = 19794608145/45^10 = 5.81E-7 One of the nice things about answers to probability questions is that they immediately translate into combinatorial identities. The conventional lottery gives a summation for C(N,k), whereas the less common scheme just described gives a summation for N^k.

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