%coupon.tex: ``How Many Singles, Doubles, Triples, Etc.,
%Should The Coupon Collector Expect?"
%for the Coupon Collector's Sisters' Expectations
%a Plain TeX file by Doron Zeilberger (1 page)
%begin macros
\def\L{{\cal L}}
\baselineskip=14pt
\parskip=10pt
\def\Tilde{\char126\relax}
\def\halmos{\hbox{\vrule height0.15cm width0.01cm\vbox{\hrule height
0.01cm width0.2cm \vskip0.15cm \hrule height 0.01cm width0.2cm}\vrule
height0.15cm width 0.01cm}}
\font\eightrm=cmr8 \font\sixrm=cmr6
\font\eighttt=cmtt8
\magnification=\magstephalf
\def\lheading#1{\bigbreak \noindent {\bf#1}
}
\parindent=0pt
\overfullrule=0in
%end macros
\bf
\centerline
{
How Many Singles, Doubles, Triples, Etc.,
Should The Coupon Collector Expect?
}
\rm
\bigskip
\centerline{ {\it
Doron ZEILBERGER
}\footnote{$^1$}
{\eightrm \raggedright
Department of Mathematics, Temple University,
Philadelphia, PA 19122, USA.
%\break
{\eighttt zeilberg@math.temple.edu} \hfill \break
{\eighttt http://www.math.temple.edu/\Tilde zeilberg/ .}
July 5, 2001 Supported in part by the NSF.
}
}
There are $m$ equi-probable baseball cards
placed at random in chewing gums. It is well known and
easy to see that a collector should expect to buy
$m(1+1/2+1/3 + \dots +1/m)$ gums before acquiring all the
kinds of cards. At the end, he would have some singles,
some doubles, some triples, etc.
Let $A(m,i)$ be the expected number of kinds of cards
of which he has exactly $i$ copies of. Here I give
a short proof of:
{\bf Formula} (Foata-Han-Lass[1]):
$\sum_{i=1}^{\infty} A(m,i)t^i=t-1+m!/\prod_{j=2}^{m} (j-t)$.
{\bf Proof:} Let's number the (kinds of) cards, in order of first
arrival by $1, 2, \dots m$. The purchased cards define
a word given by the regular expression
$11^{*}2\{1,2\}^{*}3\{1,2,3\}^{*}4 \dots \{1,2,\dots, m-1\}^{*}m$,
whose (probability) generating function is
$$
f(x_1, \dots, x_m)=
{{x_m} \over {m} }\prod_{j=1}^{m-1}
{{x_j}\over {m-(x_1+ \dots+x_{j})}}
\quad.
$$
Let a {\it marked word} be a pair $[w,i]$ where
$w$ is a word that is an instance
of the above regular expression, and $1 \leq i \leq m$.
By the familiar trick of computing expectations by changing
the order of summation, it follows that the left side of
the Foata-Han-Lass
formula is the sum of the weights of all eligible marked words,
where $weight([w,i]):=(1/m)^{\vert w \vert}$ times $t$ raised to the
power [the
number of times the letter $i$ occurs in $w$].
For example, if $m=3$ and
w=$1111211213$, then $weight([w,1])=(1/3)^{10}t^7$,
$weight([w,2])=(1/3)^{10}t^2$,
$weight([w,3])=(1/3)^{10}t$. Hence
$\sum_{i=1}^{\infty} A(m,i)t^i=m! \sum_{i=1}^{m} f_i$
(the factor of $m!$ is to account for all possible orderings),
where $f_i$ is $f$ with all the $x$'s replaced by $1$,
except for $x_i$ that is replaced by $t$.
But
$$
f_{m-i}=
{{1} \over {m}} \prod_{j=1}^{m-i-1}
{{1} \over {m-j}}
\cdot
{{t}\over {i+1-t}} \cdot
\prod_{j=m-i+1}^{m-1}
{{1}\over {m-j+1-t}}=
{{i! t } \over {m! (2-t)(3-t) \cdots (i+1-t)}} \quad,
$$
when $i>0$ and $f_m=t/m!$.
Hence
$$
\sum_{i=1}^{\infty} A(m,i)t^i
=t+t\sum_{i=1}^{m-1} {{i!} \over {(2-t)(3-t) \cdots (i+1-t)}}
=
t+
{{m!} \over {(2-t)(3-t) \dots (m-t) }}-1 \quad \halmos.
$$
{\bf Reference}
1. D. Foata, G.-N. Han, et B. Lass, {\it
Les nombres hyperharmoniques et la fratrie du collectionneur
de vignettes}, preprint, available from Foata's website.
\bye