If A_{n} Has 6n Dyes in a Box, With Which He Has To Fling [at least] n Sixes, Then A_{n} Has An Easier Task Than A_{n+1}, at Eaven Luck

Doron ZEILBERGER

The probability of A_{n} succeeding, 1 − ∑_{k=0}^{n−1} \binom{6n}{k}(\frac{1}{6})^{k}(\frac{5}{6})^{6n−k}, from which the monotonicity is not obvious, can be rewritten (using zeillim in the package EKHAD accompanying [PWZ]) as:

1 − ∑_{m=0}^{n−1} \frac{2 (94500 m^{4} + 214830 m^{3} + 171573 m^{2} + 56243 m + 6250) (6m)! 5^{5m+2}}{(5m + 5)!m! 6^{6m+5}} ,

from which the monotonicity is obvious. This generalizes, from n = 1, 2, to general n, a statement first proved, in 1693, by Mr. Isaac Newton, in response to a question of Mr. Samuel Pepys.

References

[PN] S. Pepys and I. Newton, correspondence, reproduced in American Statistician, Oct. 1960, 27-30.

[PWZ] M. Petkovsek, H.S. Wilf, and D. Zeilberger, “A=B”, A.K.Peters, 1996. The accompanying Maple package EKHAD can be downloaded from the URLs given below.

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Department of Mathematics, Temple University, Philadelphia, PA 19122, USA. zeilberg@math.temple.edu

http://www.math.temple.edu/~zeilberg   ftp://ftp.math.temple.edu/pub/zeilberg .

Supported in part by the NSF. This version: 11/24/95. First version: Sep. 11, 1995.