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\bf
\centerline
{If 
$A_n$ 
Has 6n Dyes in a Box, With Which He Has To Fling [at least]  n Sixes, }
\centerline
{ Then $A_n$ Has An Easier Task Than $A_{n+1}$, at Eaven Luck}
\rm
\bigskip
\centerline{\it Doron ZEILBERGER}
 
The probability of $A_n$ succeeding, $1
- \sum_{k=0}^{n-1} {{6n} \choose {k}}({{1} \over {6}})^k
({{5} \over {6}})^{6n-k}$, from which the monotonicity is not obvious,
can be rewritten (using {\tt zeillim} in the package EKHAD
accompanying [PWZ]) as:
$$
1-\sum_{m=0}^{n-1}
{{ 2\,(94500\,m^4+214830\,m^3+171573\,m^2+56243\,m+6250)
\,(6m)!\,5^{5m+2}}\over {(5m+5)! m!\,6^{6m+5}}} \quad ,
$$
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.
 
{\bf References}
 
[PN] S. Pepys and I. Newton, {\it correspondence}, reproduced in
American Statistician, Oct. 1960, 27-30.
 
[PWZ] M. Petkovsek, H.S. Wilf, and D. Zeilberger, {\it ``A=B''},
A.K.Peters, 1996. The accompanying Maple package EKHAD can
be downloaded from the URLs given below.
 
--------------
 
{\eightrm  \raggedright
Department of Mathematics, Temple University,
Philadelphia, PA 19122, USA. 
{\eighttt zeilberg@math.temple.edu \break
http://www.math.temple.edu/\Tilde zeilberg \quad
 ftp://ftp.math.temple.edu/pub/zeilberg} \quad .
\break
Supported in part by the NSF. This version: 11/24/95. First
version: Sep. 11, 1995.
}
 
 
\bye
 
 
 
 
 
 
 
 
 
 
