%vote.tex:  
%Plain TeX
%TeXShop, compile with XeTeX
%begin macros

\def\mod{\rm mod}
\def\Z{\bf Z}
\def\L{{\cal L}}
\baselineskip=14pt
\parskip=10pt
\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\W{{\cal W}}
\def\P{{\cal P}}
\def\Q{{\cal Q}}
\def\1{{\overline{1}}}
\def\2{{\overline{2}}}
\parindent=0pt
\overfullrule=0in
\def\Tilde{\char126\relax}
\def\frac#1#2{{#1 \over #2}}


%\headline={\rm  \ifodd\pageno  \RightHead  \else  \LeftHead  \fi}
%\def\RightHead{\centerline{
%Title
%}}
%\def\LeftHead{ \centerline{Doron Zeilberger}}
%end macros
\centerline
{
\bf
The Experimental Mathematics of Voting
}
\bigskip

\centerline
{\it By  Shalosh B. EKHAD and Doron ZEILBERGER}

{\bf Theorem 1} The weighted generating function of The Condorcet scenario $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ (with three candidates $1,2,3$ and
according to the weight 
$$
x_{123}^{NumberOf123} \cdots  x_{321}^{NumberOf321} t^{NumberOfVotes}
$$
is
$$
{\frac { \left( 1 \, + \, {t}^{3}{ x_{123}}\,{ x_{231}}\,{ x_{312}} \right) {t}^{3}{ x_{123}}\,{ x_{231}}{ x_{312}}\, }
{ \left( 1- {t}^{2}{ x_{123}}\,{ x_{321}} \right)  
\left( 1- {t}^{2}{ x_{213}}\,{ x_{312}} \right)  
\left( 1- {t}^{2}{ x_{312}}\,{ x_{123}} \right)  
\left( 1- {t}^{2}{ x_{132}}\,{ x_{231}} \right) 
 \left( 1- {t}^{2}{ x_{123}}\,{ x_{231}} \right)  
\left( 1- {t}^{2}{ x_{231}}\,{ x_{312}} \right) }} \quad .
$$

Setting all the the $x$'s to one, we get that the enumerating generating function for one of the Condorcet scenarios is
$$
{\frac {{t}^{3}+{t}^{6}}{ \left( 1-{t}^{2} \right) ^{6}}} \quad.
$$
Hence the number of such scenarios with $2n-1$ voters is the coefficient of $t^{2n-1}$ in ${\frac {{t}^{3}}{ \left( 1-{t}^{2} \right) ^{6}}}$,
i.e. the coefficient of $t^{2n-4}$ in $(1-t^2)^{-6}$, i.e.
the coefficient of $t^{n-2}$ in $(1-t^{-6}$, i.e. $(-1)^n{{-6} \choose {n-2}}= {{n+3} \choose {5}}$.
The total number of compositions is the coefficient of $t^{2n-1}$ in $(1-t)^{-6}$, i.e. ${{-6} \choose {2n-1}}={{2n+5} \choose {5}}$.
Hence assuming that all compositions are {\it equally likely} (a very wrong assumption), the probability of
one of the Condorcet scenarios, with $2n-1$ voters is
$$
{\frac {  \left( n-1 \right) n  \left( n+3 \right) }{  2 \left( 2\,n+1 \right) \left( 2\,n+3 \right)   \left( 2\,n+5 \right)  }} \quad .
$$

So it tends to $\frac{1}{16}$ as $n \rightarrow 16$.

What if each voter chooses one of the six alternatives with the same probability (i.e. $1/6$)?

{\bf Theorem 2}:

Suppose three candidates, $1$,$2$, and $3$, are running for office, and 
there are    $2n-1$ voters, each of them choosing one of the 3!=6
possible ranking with the same probability $(1/6)$. 
The votes are counted and  it turns out that in the vote of

                                 1 vs. 2, 1 won \quad 

                                 2 vs. 3, 2 won \quad 

                                 1 vs. 3, 3 won \quad 


In other words, there is a cycle $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$.

Let a(n) be the probability of that happening times $6^{2n-1}$ (i.e. the numerator)

The integer sequence, $a(n)$, satisfies the following linear recurrence equation with polynomial coefficients
$$
-1296\,{\frac {n \left( 2\,n+3 \right)  \left( 2\,n+1 \right) a \left( n \right) }{ \left( n+1 \right)  \left( n+2 \right) ^{2}}}
+36\,{\frac { \left( 2\,n+3 \right) \left( 22\,{n}^{2}+33\,n+12 \right) a \left( n+1 \right) }{ \left( n+1 \right)  \left( n+2 \right) ^{2}}}
$$
$$
-4\,{\frac { \left( 19\,{n}^{2}+57\,n+45
 \right) a \left( n+2 \right) }{ \left( n+2 \right) ^{2}}}+a \left( n+3 \right) =0 \quad .
$$

The asymptotic expression for $a(n)$ is $0.043869914022955- 0.021101164\,{n}^{-1} +O(n^{-2})$  .

By symmetry, this is also the probability of the scenario $1 \rightarrow 3 \rightarrow 2 \rightarrow 1$. 
So doubling, the asymptotic expression for the probability of a Condorcet scenario is
$$
 0.087739828045910- 0.042202328\,{n}^{-1} \quad .
$$

{\bf Corollary}: The probability for a strict Condorcet scenario with three candidates and $2n-1$ voters tends as $n \rightarrow \infty$ to
a constant, that is approximately equal to  $0.087739828045910\dots$, and that should be named the
Condorcet constant.


\end

