The Double Dixie Cup Problem

The familiar childhood occupation of obtaining a complete set of pictures of baseball players, movie stars, etc., which appear on the covers of dixie cups raises some interesting questions. One, which has already been answered, is the "single dixie cup problem," that of determining the expected number, E(n), of dixie cups which must be purchased before a complete set of n pictures is obtained: E(n) = n(I + 1/2 + * * * + 1 /n) ( [1 ] p. 213). Some time ago W. Weissblum asked how long, on the average, it would take to obtain two complete sets of n pictures. This corresponds to the situation observed when two tots collect cooperatively, i.e., "trading" takes place. This "double dixie cup" problem cannot be handled by the same device used for the problem of the single set and in this paper we find a new method which allows us to write down the solution, Em(ln), (as an easily evaluated definite integral) for the problem of collecting m sets. For m fixed and n large the expected number of dixie cups turns out to be n(log n+(m1) loglog n+o(l)). Thus, although the first set "costs" n log n, all further sets cost n loglog n. Suppose m sets are desired. Let pi be the probability of failure of obtaining m sets up to and including the purchase of the ith dixie cup. Then the expected number of dixie cups Em(ln) = Z%=0 pi, by a well-known argument ([1] p. 211). Now pi= Ni/ni where Ni is the number of ways that the purchase of i dixie cups can fail to yield m copies of each of the n pictures in the set. If we represent the pictures by xi, * , xn, then Ni is simply (xi + * * * +xn)i expanded and evaluated at (1, . . ., 1) after all the terms have been removed which have each exponent for each variable larger than m -1. Now consider m fixed and introduce the following notation. If P(x1, . . . , x,,) is a polynomial or power series we define { P(xi, . . . , x,C) } to be the polynomial, or series, resulting when all terms having all exponents _ m have been removed. In terms of this notation pi is { (xi + * +x.) } /ni evaluated at x = *= -1. If we now make the definition