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%\section{preamble}
\title[{\sc\lowercase{the electronic journal of combinatorics 
\textbf{7}} (2000), \#Rxx\hfill}]
{A simple card guessing game revisited}



\date{}


\begin{document}



\maketitle
\begin{center}
{ {\sc Arnold Knopfmacher}}\\

\medskip
The John Knopfmacher Centre for Applicable Analysis and Number Theory\\
 Department of Applied Mathematics\\
University of the Witwatersrand, P.~O. Wits\\
2050 Johannesburg, South Africa\\
 email:
{\tt arnoldk@gauss.cam.wits.ac.za}\\
homepage:
{\tt http://www.wits.ac.za/science/number{\underline{\ }}theory/arnold.htm}

\vskip 1.5cm

{ {\sc Helmut Prodinger}}\\

\medskip
The John Knopfmacher Centre for Applicable Analysis and Number Theory\\
 Department of Mathematics\\
University of the Witwatersrand, P.~O. Wits\\
2050 Johannesburg, South Africa\\
 email:
{\tt helmut@gauss.cam.wits.ac.za}\\
homepage:
{\tt http://www.wits.ac.za/helmut/index.htm}


\bigskip

{Submitted: January 4, 2000; Accepted: March 3, 2000.} 

\end{center}


\bigskip
\begin{abstract}
A deck of cards consisting of $m$ red and $n$ black cards is given. A guess is
made as to the colour of the top card, after which it is revealed and discarded.
To maximise the number of correct guesses one chooses the colour corresponding
to the majority of cards remaining in the deck. We determine the expected number
of correct guesses with this strategy, as well as the probability of beating an
opponent 
who uses the naive stategy of random guessing. \end{abstract}


\medskip
\noindent
{\bf AMS Subject Classification.} 05A15 (primary) 60C05 (secondary) 

\bigskip

\section{Introduction}
\label{sec:intro}
\pagestyle{myheadings}
\thispagestyle{empty}


A number of authors have considered the following simple
combinatorial game;
\cite{Read62, Levasseur88, Zagier90, Sulanke95}. 
One starts with a deck of cards consisting of $m$ red and $n$ black cards. A
guess is made as to the colour of the top card, after which it is revealed and
discarded. To maximise the number of correct guesses one chooses the colour
corresponding to the majority of cards remaining in the deck. We determine the
expected number of correct guesses with this strategy, as well as the
probability of beating an opponent\footnote{Previous authors \cite{Levasseur88,  
Zagier90} used phrases like ``How often should you beat your kids?'' etc.
In accordance with the editor's wishes
 it was decided to avoid such phrases.} 
who uses the naive stategy of random
guessing. The first of these problems was analysed by 
Levasseur \cite{Levasseur88}  but only
for the case $m=n$.  The second question was investigated by 
Zagier \cite{Zagier90},
again for the case $m=n$. He obtained an asymptotic answer only, whereas we can
derive exact formulas for this probability for general $m$ and $n$. We also
re--derive the probability distribution for the number of correct guesses for a
pack of $m+n$ cards, found originally by 
Sulanke \cite{Sulanke95}.

Whereas the previous authors made use of lattice paths, our approach will be
based on recurrence relations and probability generating functions. We also
consider a variation of the game in which you and your opponent alternate in
guessing the colour of the card. We find that there is a slight advantage to
playing second in this variation.

In what follows, $m$ always refers to number of cards of the
colour that is predominant, $n$ to the other colour. 


\section{The number of correctly guessed cards}

We begin by re--proving some results of \cite{Sulanke95}.


Let $p(m,n;k)$  denote the probability that, assuming that one has $m$
cards of one colour and
  $n$
cards of a second colour, 
with $m\ge n\ge0$, that one guesses $k$ cards correctly. Sulanke
\cite{Sulanke95}, using lattice path arguments, found the following explicit
expression: 
\begin{equation} 
\label{pr_distr}
p(m,n;k)=\frac{1}{\binom{m+n}{m}
}\bigg[\binom{m+n}{k}-\binom{m+n}{k+1} \bigg]\qquad \text{for $m\le k\le m+n $}.
\end{equation}

One can prove this also using the following recursions. Introducing
probability generating functions
\begin{equation*}
\varphi_{m,n}(u)=\sum_{m\le k\le m+n  }p(m,n;k)u^k,
\end{equation*}one sees the recursions
\begin{align*}
\varphi_{m,n}(u)&=u\frac{m}{m+n}\varphi_{m-1,n}(u)
+\frac{n}{m+n}\varphi_{m,n-1}(u) \qquad \text{for $m>n\ge0 $},\\
\varphi_{m,m}(u)&=(1+u)\frac{1}{2}\varphi_{m,m-1}(u)
\qquad \text{for $m\ge1 $},\  \varphi_{0,0}(u)=1
\end{align*}
almost immediately. If one defines $\Phi_{m,n}(u)=\binom{m+n}{m}
\varphi_{m,n}(u)$ instead, the recursions are nicer, viz.

\begin{align*}
\Phi_{m,n}(u)&=u\Phi_{m-1,n}(u)
+\Phi_{m,n-1}(u) \qquad \text{for $m>n\ge0 $},\\
\Phi_{m,m}(u)&=(1+u)\Phi_{m,m-1}(u)
\qquad \text{for $m\ge1 $},\  \Phi_{0,0}(u)=1.
\end{align*}
  From these,  Sulanke's result 
(\ref{pr_distr})  is readily proved by induction. 


Instead of either guessing or using lattice path arguments, we 
can use 
a generating function method 
that deserves perhaps to be known more widely
to deal with such recursions with
boundary conditions. 
It is due to Knuth \cite[page~537]{Knuth68}; a tutorial version of it 
appears in \cite{PaPr98}.


If we set
\begin{align*}
F(z,x)&=\sum_{m\ge n\ge 0} \Phi_{m,n}(u)z^mx^{m-n},\\
F_0(z)&=\sum_{m\ge 0} \Phi_{m,m}(u)z^m,\\
F_1(z)&=\sum_{m\ge 1} \Phi_{m,m-1}(u)z^m,\\
\end{align*}


then the first and second recursion for $\Phi_{m,n}(u)$ give
\begin{equation*}
F(z,x)-F_0(z)=uzx\,F(z,x)+\frac1x\big[F(z,x)-F_0(z)-F_1(z)x\big]
\end{equation*}
and $F_0(z)=1+(1+u)F_1(z)$, respectively. 
Hence
\begin{equation*}
(1-x+uzx^2)F(z,x)=1-x+(1+u-ux)F_1(z).
\end{equation*}

Now the fact that the power series $F(z,x)$ remains finite at
$x=\lambda:=2/(1+\sqrt{1-4uz}\,)$ gives
(this is Knuth's trick)
$F_1(z)=(\lambda-1)/(1+u-\lambda u)$ and finally
\begin{equation*}
F(z,x)=\bigg(
\frac{1+\sqrt{1-4uz}}{2}-
\frac{1-\sqrt{1-4uz}}{2}u
\bigg)^{-1}
\bigg(1-\frac{1-\sqrt{1-4uz}}{2}x\bigg)^{-1}.
\end{equation*}

  From this, using the expansion of the geometric series and
the identity
 \cite[(5.61)]{GrKnPa94}
\begin{equation*}
\sum_{n\ge0}\binom{2n+k}{n}X^n=\frac{2^k}{\sqrt{1-4X}(1+\sqrt{1-4X}\,)^k }
\end{equation*}
the quantities in (\ref{pr_distr}) can be obtained.

\medskip

Considering the expected values 
$T(m,n)=\frac  d{du}\varphi_{m,n}(u)\big|_{u=1}$, simple telescoping leads
to
\begin{equation*}
\binom{m+n}{m}T(m,n)=\sum_{k\ge m}
k\bigg[\binom{m+n}{k}-\binom{m+n}{k+1}
\bigg]

=m\binom{m+n}{m}+\sum_{k>m}\binom{m+n}{k}.
\end{equation*}




\begin{theorem}The expected number of correct guesses is given by
\begin{equation}\label{form1}
T(m,n)=m+\frac{1}
{\binom{m+n}{m}}\sum_{k=0}^{n-1}\binom{m+n}{k}.
\end{equation}


\end{theorem}

{\bf Remark.} Although this 
seems to be the nicest expression for the mean,  we are
able to derive other equivalent ones:

\begin{align}\label{form2}
T(m,n)&=m-1+\frac{2^n}{
\binom{m+n}{m}}\sum_{k=0}^{n}\binom{m+k-1}{k}\frac{1}{2^k},\\
\label{form2a}
T(m,n)&=m+\frac{2^{n-1}}{
\binom{m+n}{m}}\sum_{k=0}^{n-1}\binom{m+k}{k}\frac{1}{2^k},\\
\label{form3}
T(m,n)&=m+n\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{2^{n-1-k}(-1)^{k}}
{m+1+k} .
\end{align}



Since the equivalence of these binomial expressions is not obvious, we
prove these equalities.
To prove that (\ref{form1}) is equivalent to (\ref{form2}), we must show that
\begin{equation*} \sum_{k=0}^n\binom{m+n}{k}=
2^n\sum_{k=0}^n\binom{m+k-1}{k}2^{-k}. \end{equation*}
In terms of generating functions, we must prove that
\begin{equation*}
[x^n]\frac{1}{1-x}(1+x)^{m+n}=[x^n]\frac{1}{1-2x}\frac1{(1-x)^m}.
\end{equation*}
This can be seen by Cauchy's integral formula (the substitution
$x=t/(1-t)$ is used):
\begin{align*}
[x^n]\frac{1}{1-x}(1+x)^{m+n}
&=\frac{1}{2\pi i}\oint\frac{dx}{x^{n+1}}\frac1{1-x}(1+x)^{m+n}\\
&=\frac{1}{2\pi i}\oint\frac{dt}{t^{n+1}}\frac1{1-2t}\frac1{(1-t)^m}\\
&=[t^n]\frac{1}{1-2t}\frac1{(1-t)^m}.
\end{align*}

The equivalence of (\ref{form2}) and (\ref{form2a}) is straightforward
by writing
\begin{equation*}   
\binom{m+k}{k}=\binom{m+k-1}{k}+\binom{m+k-1}{k-1}
\end{equation*}
and performing some elementary operations. 

 For the equivalence of (\ref{form3}) and (\ref{form2a})
we use hypergeometric functions:

\begin{align*}
n\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{2^{n-1-k}(-1)^{k}}
{m+1+k}
&=\frac{n2^{n-1}}{(m+1)}\ {_2F_1}[-n+1,m+1;m+2|\tfrac{1}{2}] \\
&=\frac{n2^{n-1}(n-1)!(m+1)!}{(m+1)(m+n)!}
\ {_2F_1}[-n+1,m+1;-n+1|\tfrac{1}{2}] \\
&=\frac{2^{n-1}}{\binom{m+n}{m}}
\sum_{k=0}^{n-1}\binom{m+k}{k}\frac{1}{2^k}.
\end{align*}
Here, we have used the transformation (5.105) \cite{GrKnPa94}.
\smallskip

{\bf Remark}. As one referee has pointed out, there are shorter
proofs of these equivalences.
 The ones  presented above have the advantage of offering some
insight as opposed to providing merely mechanical verifications.

\medskip

It is interesting to note that the formula
\begin{equation*}   
\sum_{k=0}^m\binom{m+k}{k}\frac{1}{2^k}=2^m
\end{equation*}
follows from the discussion of Banach's match box 
problem, see \cite{GrKnPa94}.


It is also not too hard to compute the variance; we start with the
second moment:
\begin{align*}
&\frac{1}{\binom{m+n}{m}}\sum_{k\ge m}k^2
\bigg[\binom{m+n}{k}-\binom{m+n}{k+1} \bigg]\\
&=m^2+\frac{1}{\binom{m+n}{m}}\sum_{k> m}(2k-1)
\binom{m+n}{k}\\
&=m^2+\frac{2(m+n)}{\binom{m+n}{m}}\sum_{k> m}
\binom{m+n-1}{k-1}
-\frac{1}{\binom{m+n}{m}}\sum_{k> m}
\binom{m+n}{k}\\
&=m^2+\frac{m+n}{\binom{m+n}{m}}\sum_{k> m}
\binom{m+n-1}{k-1}
+\frac{m+n}{\binom{m+n}{m}}\sum_{k\ge m}
\binom{m+n-1}{k}
-\frac{1}{\binom{m+n}{m}}\sum_{k> m}
\binom{m+n}{k}\\
&=m^2+n+\frac{m+n}{\binom{m+n}{m}}\sum_{k> m}
\binom{m+n}{k}
-\frac{1}{\binom{m+n}{m}}\sum_{k> m}
\binom{m+n}{k}\\
&=m^2+n+\frac{m+n-1}{\binom{m+n}{m}}\sum_{k=0}^{n-1}
\binom{m+n}{k}.
\end{align*}


Subtracting the square of the expectation, we get:

\begin{theorem} The variance of the number of correct guesses is
given by
\begin{align*}
n-\frac{m+1-n}{\binom{m+n}{m}}\sum_{k=0}^{n-1}
\binom{m+n}{k}-\frac{1}{\binom{m+n}{m}^2}\Bigg(\sum_{k=0}^{n-1}
\binom{m+n}{k}\Bigg)^2. \qed
\end{align*}

\end{theorem}



In the symmetric case $m=n$, this can be simplified to
\begin{equation*}
m+\frac14-\frac{4^{2m-1}}{\binom{2m}{m}^2},
\end{equation*}
in agreement with the earlier result of Sulanke \cite{Sulanke95}.























\section{How often will you beat a naive player?}

Zagier \cite{Zagier90} considered the probability that
 you beat the naive player (with your superior strategy) and gave the
asymptotic answer 
\begin{equation*}   
\frac12+\frac{1}{\sqrt{8}}+\mathcal{O}(m^{-1/2})
\end{equation*}
for the symmetric case $m=n$. 

Now we can give  an exact result:
Recall that  the probability that  you score $k$ by
 guessing the more likely outcome
is given by




\begin{equation*}   
A(m,k)=\frac{1}{\binom{2m}{m}}\bigg[
\binom{2m}{k}-\binom{2m}{k+1}
\bigg],\qquad m\le k\le 2m.
\end{equation*}
The  distribution for the naive player is given by
\begin{equation*}   
C(m,k)=\frac{1}{4^m}
\binom{2m}{k},\qquad 0\le k\le 2m.
\end{equation*}
Therefore the probability that you win is
\begin{equation*}   
\sum_{0\le k<m}C(m,k)+\sum_{m\le k<j\le 2m}C(m,k)A(m,j).
\end{equation*}
The first, easy sum is
\begin{equation*}   
\sum_{0\le k<m}C(m,k)=\frac12\bigg(1-\frac{\binom{2m}{m}}{4^m}\bigg).
\end{equation*}

The other sum is
\begin{align*}   
\frac{1}{\binom{2m}{m}4^m}
&\sum_{m\le k<j\le 2m}\binom{2m}{k}\bigg[
\binom{2m}{j}-\binom{2m}{j+1}
\bigg]\\
&=\frac{1}{\binom{2m}{m}4^m}
\sum_{m\le k< 2m}\binom{2m}{k}
\binom{2m}{k+1}\\
&=\frac{1}{2\binom{2m}{m}4^m}
\sum_{0\le k< 2m}\binom{2m}{k}
\binom{2m}{k+1}\\
&=\frac{1}{2\binom{2m}{m}4^m}
\sum_{0\le k< 2m}\binom{2m}{k}
\binom{2m}{2m-k-1}\\
&=\frac{1}{2\binom{2m}{m}4^m}
\binom{4m}{2m-1}.
\end{align*}




\begin{theorem} The probability that
the player guessing the more likely colour beats the one
guessing naively is (exactly) given by
\begin{equation*}   
\frac{1}{2}+\frac12\frac{\binom{4m}{2m+1}}{\binom{2m}{m}4^m}		
-\frac{1}{2}\frac{\binom{2m}{m}}{4^m}.
\end{equation*}

\end{theorem}

Stirling's formula readily gives  the result

\begin{equation*}   
\frac12+\frac{1}{\sqrt{8}}-{\frac{1}{2\sqrt{\pi m}}}
-\frac{7\sqrt{2}}{64m}+\frac{1}{16\sqrt{\pi}m^{3/2}}+
\mathcal{O}(m^{-2}).
\end{equation*}

Very similarly, we can consider the probability that the
game against the naive player is a draw, which turns out to be


\begin{align*}   
\sum_{m\le k\le 2m}&C(m,k)A(m,k)=
\frac{1}{\binom{2m}{m}4^m}
\sum_{m\le k\le 2m}\binom{2m}{k}\bigg[
\binom{2m}{k}-\binom{2m}{k+1}
\bigg]\\
&=
\frac{1}{2\binom{2m}{m}4^m}
\Bigg[\sum_{0\le k\le 2m}\binom{2m}{k}^2+\binom{2m}{m}^2\Bigg]
-\frac{1}{2\binom{2m}{m}4^m}
\binom{4m}{2m+1}\\
&=
\frac{1}{2\binom{2m}{m}4^m}
\binom{4m}{2m}+
\frac{1}{2\,4^m}\binom{2m}{m}
-\frac{1}{2\binom{2m}{m}4^m}
\binom{4m}{2m+1}\\
&=
\frac{1}{2(2m+1)\binom{2m}{m}4^m}
\binom{4m}{2m}+
\frac{1}{2\,4^m}\binom{2m}{m}\\
&=\frac{1}{2\sqrt{\pi m}}+\frac{\sqrt2}{8m}+\mathcal{O}(m^{-3/2}).
\end{align*}





\section{A variation}

Instead of guessing at the same time as the naive player, 
suppose that
the game is
now played alternately. The naive player guesses, then the 
clever player, then the
naive player again,  and so on. 
We will see that is makes a (small) difference whether
the clever or the naive player starts. We will discuss again the average score
of the clever player and start by considering the case where
the clever player plays first.




\begin{align*}
T(m,n)&=[\![\text{$m+n$ even}]\!]\frac{m}{m+n}+
\frac{m}{m+n}T(m-1,n)+\frac{n}{m+n}T(m,n-1),\ 0\le n<m,\\
T(m,m)&=\frac12+2T(m,m-1), \qquad m\ge1, \ \ T(0,0)=0.
\end{align*}
($[\![P]\!]$ is 1 if condition $P$ is true, 0 otherwise.)
It is quite easy to guess the solution and prove it by induction:



\begin{align*}
T(2m,2n)&=m+\frac{1}{\binom{2m+2n}{2n}}\sum_{k=0}^{n-1}
\binom{2m+2n}{2k}, 
\\
T(2m,2n+1)&=m+\frac{1}{\binom{2m+2n+1}{2n+1}}\sum_{k=0}^{n-1}
\binom{2m+2n+1}{2k+1}-\frac{m}{2m+2n+1},\\
T(2m+1,2n)&=m+\frac{1}{\binom{2m+2n+1}{2n}}\sum_{k=0}^{n-1}
\binom{2m+2n+1}{2k+1}-\frac{n}{2m+2n+1},\\
T(2m+1,2n+1)&=m+\frac{1}{\binom{2m+2n+2}{2n+1}}\sum_{k=0}^{n-1}
\binom{2m+2n+2}{2k+1}+\frac12.
\end{align*}
Therefore 
\begin{equation*}
T(m,m)=\frac{m-1}{2}+\frac{4^{m-1}}{\binom{2m}{m}}.
\end{equation*}


Now the case when the naive  player starts  is briefly sketched.

\begin{align*}
T(m,n)&=[\![\text{$m+n$ odd}]\!]\frac{m}{m+n}+
\frac{m}{m+n}T(m-1,n)+\frac{n}{m+n}T(m,n-1),\ 0\le n<m,\\
T(m,m)&=2T(m,m-1), \qquad m\ge1, \ \ T(0,0)=0.
\end{align*}

And the solution is



\begin{align*}
T(2m,2n)&=m+\frac{1}{\binom{2m+2n}{2n}}\sum_{k=0}^{n-1}
\binom{2m+2n}{2k+1},\\
T(2m,2n+1)&=m+\frac{1}{2\binom{2m+2n+1}{2n+1}}\sum_{k=0}^{n}
\binom{2m+2n+2}{2k+1},\\
T(2m+1,2n)&=m+\frac{1}{\binom{2m+2n+1}{2n}}\sum_{k=0}^{n}
\binom{2m+2n+1}{2k}+\frac{n}{2m+2n+1},\\
T(2m+1,2n+1)&=m+\frac{1}{\binom{2m+2n+2}{2n+1}}\sum_{k=0}^{n-1}
\binom{2m+2n+2}{2k+1}+\frac{m+3n+2}{2(m+n+1)}.
\end{align*}


Therefore 
\begin{equation*}
T(m,m)=\frac{m}{2}+\frac{4^{m-1}}{\binom{2m}{m}}.
\end{equation*}

So one does slightly better (by $\frac12$) when one is allowed to play
after the naive player. And the sum of the scores in both instances is

\begin{equation*}
T(m,m)=m-\frac{1}{2}+\frac{4^{m}}{2\binom{2m}{m}},
\end{equation*}
as it should be, since the score is only divided into the score obtained at
even (resp. odd) steps. 

\bigskip
{\bf Acknowledgment}.
The insightful comments of two anonymous referees are gratefully
acknowledged.

\bibliographystyle{plain}

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\end{thebibliography}


\end{document}
