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\begin{Large}
\begin{center}
{\large {\bf Should Candidates Flip a Coin if the Difference in Their
Votes
is Small? } \\[0pt]
} \bigskip  September 18, 2001 \\[0pt]
\bigskip Eric Rasmusen \\[0pt]
\bigskip {\it Abstract}
\end{center}

A coin flip can be a good way to settle an election if the margin of
victory
is small and it is known that there is a good chance of fraud by one
candidate. In that case, however, an even better rule is to award
victory
to
the apparent loser.

\noindent

{\small \hspace*{20pt}   Professor of Business Economics and Public
Policy
and
Sanjay Subhedar Faculty Fellow, Indiana University, Kelley School of
Business, BU 456, 1309 E. 10th Street, Bloomington, Indiana, 47405-
1701.
Office: (812) 855-9219. Fax: 812-855-3354. Erasmuse@indiana.edu.
Php.indiana.edu/$\sim$erasmuse. Keywords: Voting, Elections, Bias,
Estimation
JEL Classifications: C11, C44, D70, D72, D81.  Copies of this paper
can be
found at
Php.indiana.edu/$\sim$erasmuse/papers/coinflip.pdf. }



\noindent
 {\small I thank    Harvard Law School's
Olin Center   and the University of Tokyo's Center for International
Research on
the Japanese Economy for  their hospitality.   }

%%-----------------------------%----------------

\newpage



\bigskip

\noindent {\it 2. The Model}

Let us imagine that we are constructing rules for elections between a
dishonest candidate and an honest candidate. Denote the dishonest
candidate's margin of votes (votes for him minus votes for the honest
candidate) by $m$, his margin of legal votes by $x$, and the number of
illegal votes by $N$. Both $m$ and $x$ can be negative, indicating a
positive margin for the honest candidate, and $m= x+N$.

Let $x$ be distributed by density $f(x)$ with cumulative density $F(x)
$.
We will make $x$ is a continuous variable for neatness, so the
probability
of exact ties will be zero and will not need to clutter the analysis
with
special rules for tie-breaking.

	   \noindent
Assume:

\noindent (A1) The true winning margin density $f(x)$ is strictly
increasing in the range $[-2N,0]$.



\newpage
 \vspace*{-1in}



 \includegraphics[height= 1.932in, width=3.1626in]{coin1a.eps}

 	      \includegraphics[height= 1.932in, width=3.1626in]
{coin1b.eps}

	      \includegraphics[height= 1.932in, width=3.1626in]
{coin1c.eps}

	    \includegraphics[height= 1.932in, width=3.1626in]
{coin1d.eps}

\bigskip

Figure 2 shows two distributions that do not satisfy assumption A1.

 \includegraphics[height= 1.932in, width=3.1626in]{coin2a.eps}

  \includegraphics[height= 1.932in, width=3.1626in]{coin2b.eps}


\newpage

Our problem is to choose a
"victory rule": a rule which awards victory to one candidate or the
other. This rule depends on the observed margin of votes, and takes
the form $V(m)=p$, where $m$ is the dishonest candidate's margin and
$p$ is his probability of victory given that margin.


Assume society's objective is to maximize the probability of a
legitimate
victory, defined as the candidate with the most legal votes being
declared the victor. We will denote a legitimate victory by $L$, where
\begin{equation}  \label{e2a}
\begin{array}{ll}
L & =1\; if \; x \geq 0 \; and \; V=1 \\
& =1\; if \; x < 0 \; and \; V=0 \\
& =0 \;otherwise.
\end{array}
\end{equation}
If society knew which candidate was dishonest, which we have ruled
out,
the
optimal victory rule would simply replicate the objective by
subtracting
$N$
votes from the dishonest candidate's margin and declaring as winner
whoever
had the most legal votes, i.e.,

\noindent{\it The Full-Information Rule.} $V = 1$ if $m-N \geq 0$; and
V=
0
otherwise.

We will require, however, that any victory rule be symmetric, since we
do
not know the identity of the dishonest candidate in advance:

\noindent {\it Symmetry Requirement.} If $V(m) =p$, then $V (-m) = 1-
p. $

\newpage


 

\noindent{\it The Conventional Rule.} $V = 1$ if $m  \geq 0$; and V=0
otherwise.


The probability that the dishonest candidate wins under the
conventional
victory
rule is
\begin{equation}  \label{e3}
Prob(m>0) = Prob(x+N>0)= Prob (x > -N) = 1-F(-N).
\end{equation}
The probability that the dishonest candidate is the legitimate winner
is
\begin{equation}  \label{e4}
Prob(x >0)= 1-F(0).
\end{equation}
The probability that the honest candidate wins is
\begin{equation}  \label{e5}
Prob(m<0) =Prob(x+N<0) = Prob (x < -N) = F(-N).
\end{equation}

  The
probability of a legitimate victory is thus
\begin{equation}  \label{e6}
1-F(0)+ F(-N).
\end{equation}
The probability of a legitimate victory decreases in $N$, since bigger
$N$
means smaller $F(-N)$.

\newpage



The Conventional Victory Rule  is a special case, with $T=0$, of the
following.

\noindent{\it The Coin Flip Rule.} $V = 1$ if $  m  \geq T$; $V = 0$
if $m \leq -T$; and $V=.5$ otherwise.

  The probability that the
dishonest
candidate is the legitimate winner is not just the probability he is
legitimate, because sometimes, due to the coin toss, he fails to win
even
if
he is legitimate. The probability he is the legitimate winner and also
wins
under the victory rule is
\begin{equation}  \label{e8}
\begin{array}{l}
Prob (x>0,m>T) + .5 prob (x>0, -T<m <T) \\
=Prob (x>0, x+N>T) + .5 prob (x>0, -T<x+N <T) \\
= Prob (x>0, x >T-N) + .5 prob (x>0, -T-N<x <T-N)
\end{array}
\end{equation}

The probability that the honest candidate is the legitimate winner and
also
wins
under the victory rule is
\begin{equation}  \label{e9}
\begin{array}{l}
Prob (x<0, m<-T) + .5 prob (x<0, -T<m <T) \\
Prob (x<0, x+N<-T) + .5 prob (x<0, -T<x+N <T) \\
= Prob (x<0, x <-T-N) + .5 prob (x<0, -T-N<x <T-N)
\end{array}
\end{equation}

\newpage

We need to consider two cases: $T \geq N$,  and $T<N$.

(1) $T \geq N$ (threshold greater than the number of  dishonest votes)
.
The probability that the dishonest candidate is the legitimate winner
and
also wins under the victory rule is
\begin{equation}  \label{e10}
\begin{array}{l}
Prob(Dis.\; leg.\; win ) \\
 = Prob (x>0, x+N > T) + .5 prob (x>0, -
T<x+N
<T)
\\
  = Prob (x>0, x >T-N) + .5 prob (x>0, -T-N<x <T-N) \\
  = Prob (x >T-N) + .5 prob (0 <x <T-N) \\
  = [1- Prob(x <T-N)] + .5 [Prob(x<T-N) - Prob(x<0)] \\
  = 1- F(T-N) + .5 [F(T-N) - F(0)] \\
  = 1- .5F(T-N) - .5F(0)
\end{array}
\end{equation}

The probability that the honest candidate is the legitimate winner and
also wins under the victory rule is
\begin{equation}  \label{e11}
\begin{array}{l}
Prob (hon\; leg. \; win) \\
=   Prob (x<0, x+N<-T) + .5 prob (x<0, -T<x+N
<T)
\\
  = Prob (x<0, x <-T-N) + .5 prob (x<0, -T-N<x <T-N) \\
  = Prob ( x <-T-N) + .5 prob (-T-N<x <0) \\
  = F(-T-N) + .5 [F(0) -F(-T-N)] \\
  = .5 F(-T-N) + .5 F(0).
\end{array}
\end{equation}

\newpage

The probability of a legitimate victory is thus
\begin{equation}  \label{e12}
\begin{array}{l}
\pi= [1- .5F(T-N) - .5F(0)]+ [.5 F(-T-N) + .5F(0)]\\
 = 1-.5F(T-N) + .5 F(-
T- N)
\end{array}
\end{equation}

The optimal $T$ maximizes this. The first order condition is
\begin{equation}  \label{e13}
d\pi/dT= - .5 f(T-N) -.5 f(-T-N) =0.
\end{equation}

Expression (\ref{e13}) cannot be solved. The derivative is negative
for
all $T$ in
the
interval $[N, \infty]$ that we are considering so the smaller $T$ is,
the
better.
Thus, the optimum is $T^*=N$ if it is in this interval.

\newpage

(2) $T<N$ (threshold less than the number of  dishonest votes). The
probability that the criminal candidate is the legitimate
winner
and also wins under the victory rule is
\begin{equation}  \label{e14}
\begin{array}{l}
Prob (dis \;legit \;winner )  \\
 = Prob (x>0, x+N>T) + .5 Prob (x>0, -
T<x+N
<T) \\
  = Prob (x>0, x >T-N) + .5 prob (x>0, -T-N<x <T-N) \\
  = Prob (x >0) + 0 \\
  = [1- Prob(x <0)] \\
  = 1- F(0)
\end{array}
\end{equation}

The probability that the honest candidate is the legitimate winner and
also
wins
under the victory rule is
\begin{equation}  \label{e15}
\begin{array}{l}
Prob (Honest \;legit \;winner )  \\
 = Prob (x<0, x+N<-T) + .5 prob (x<0,
-T<x+N <T) \\
  = Prob (x<0, x <-T-N) + .5 prob (x<0, -T-N<x <T-N) \\
 = Prob ( x <-T-N) + .5 prob (-T-N<x <T-N) \\
  = F(-T-N) + .5 [F(T-N) - F(-T-N)] \\
 = .5 F(-T-N) + .5 F(T-N).
\end{array}
\end{equation}

\newpage

The probability of a legitimate victory is thus
\begin{equation}  \label{e16}
\pi = 1- F(0)+ .5 F(-T-N) + .5 F(T-N).
\end{equation}

The optimal $T$ maximizes this. The first order condition with respect
to
$N$
is
\begin{equation}  \label{e17}
d\pi/dT= -.5 f(-T-N) + .5 f(T-N) =0.
\end{equation}

The derivative in (\ref{e17}) is always positive, because $f(-T-N)$ is
always less
than
$f(T-N)$, as shown in Figure 3. Both winning margins $x$ are negative
numbers in
this
case, but $T-N$ is closer to 0, where the density is greater under our
assumptions.

Thus, $T^*=N$ is the optimum.

\newpage

 \includegraphics[height= 3.0796in, width=4.9943in]{coin3.eps}

We can conclude that when we think that one candidate will have $N$
illegal votes,
the
optimal coin flip rule flips a coin if the margin of victory is less
than
$N$.

\newpage

\vspace *{-.5in}

\noindent
{\it A Bayesian Approach.}


If
society observes margin $m$, what should its posterior belief be of
the
probability
that the legal margin $x$ is also positive?

 On observing
$m=
m^{\prime}
$, society knows that either (a) $ x=m^{\prime}-N$ or (b) $x= -
m^{\prime}-
N$,
depending on which candidate is the dishonest one. If $m^{\prime}>N$,
then
in case
(a), $x> 0$, and in case (b), $x^{\prime}<0$, so the posterior should
be
that with
probability 1 the apparent winner is the legitimate winner. This is
why
$T^*$
should
not exceed $N$.

If $m^{\prime}\in [-N,N]$, then society cannot deduce with certainty
who
was the
legitimate winner.   The posterior probability that the apparent
winner is
the
legitimate winner is, by Bayes's Rule,
  \begin{equation}  \label{e18}
    P(m^{\prime}) =
\frac{f(-m^{\prime}-N)} {f(m^{\prime}-N)+f(-m^{\prime}- N)}.
 \end{equation}
If $P(m)$
is greater than .5-- i.e., if $f(-m^{\prime}-N)>f(m^{\prime}-N) $--
then
victory
ought
to be awarded to the apparent winner.  
Assumption
A1 tells us that that is false, however, because both $-m^{\prime}-N$
and
$f(m^{\prime}-
N)$
are in the interval $[-2N, 0]$ over which the density is increasing.
Thus,
for
margins
between 0 and $N$,   the apparent winner is probably
{\it not}
legitimate!

\newpage

  Suppose $N=500$, and the winning
margin is
100. 

If the dishonest candidate is the apparent winner, with $m=100 $,
then $x= -
400$,
and we would like a rule that reverses his victory.

 If the honest
candidate is the
apparent winner, with $m=-100$, then $x= -600$, and we want a rule
that
confirms
the
apparent winner.

 Which is more probable, $m=100$ or $m=-100$? 

It is $m=
100$ that
is
more probable, because it arises when $x= -400$, which is more
probably
than $x=-
600$
given assumption (A1).

 In short: if a candidate wins by too few votes, the
most
likely
explanation is that he actually lost the legal vote and only flipped
the
result by
virtue of illegal votes.

\newpage

This suggests that the following victory rule is superior to the coin
flip
rule.

\noindent{\it The Reversal Rule.} $V = 1$ if $m \in [-N, 0]$ or $m> N$
;
$V
= 0$ otherwise.

We have seen that the optimal rule has $T^*=N$. Let us compare the
optimal
Coin
Flip
Rule with the Reversal Rule using the following general rule (called
``general''
only
for convenience; note that it takes the threshold $ T=N$ as given).

\noindent{\it The General Rule.} $V = z$ if $m \in [-N, 0]$; $V = 1$
if $m>N$ ;$V = 1- z$ if $m \in [0, N]$; $V = 0$ if $m < -N$ .

If $z=.5$, the General Rule is identical to the optimal Coinflip Rule;
if
$ z=0$,
it
is identical to the Reversal Rule. Let us determine the optimal level
of
$z$.

The probability that the dishonest candidate is the legitimate winner
and
wins
under
this
victory rule is
\begin{equation}  \label{e19}
\begin{array}{l}
z Prob( x>0, -N<m <0) + (1-z) Prob( x>0, 0<m <N)+\\
  Prob (x>0,m>N) \\
= Prob (x+N>N) \\
= Prob (x >0)
\end{array}
\end{equation}
Equation (\ref{e19}) is telling us that if the dishonest candidate
wins
legitimately, the General Rule always awards him victory, so $z$ is
irrelevant to his probability of being the legitimate winner and also
winning under this victory rule.

The probability that the honest candidate is the legitimate winner and
also
wins
under the victory rule is
\begin{equation}  \label{e20}
\begin{array}{l}
Prob (x<0,m<-N)+ (1-z) Prob( x<0, -N<m <0) \\
 + z Prob( x>0, 0<m <N) \\
= Prob ( x+N<-N)+ (1-z) Prob( -N<x+N <0) \\
= Prob ( x <-2N)+ (1-z) Prob( -2N<x <0)
\end{array}
\end{equation}

Thus, the probability of the legitimate winner winning under the
General
Rule is
\begin{equation}  \label{e21}
\begin{array}{l}
Prob (x >0)+ Prob ( x <-2N)+ (1-z) Prob( -2N<x <0),
\end{array}
\end{equation}
which is   maximized by setting $z=0$ and using the Reversal Rule.

\newpage

 
The 
objective function in this problem is out of the ordinary. 

Voting is a
winner-take-all tournament, not an attempt to measure the winning
legal
margin with minimal mean squared error. 

  Suppose we have a scale that we know is either 40 or
-40
milligrams off, with equal probability, and we are measuring an object
from
a population whose weights are unimodally and symmetrically
distributed
with
mean 5000 milligrams. Our measurement is 5010 milligrams. We deduce
that
the
true weight is therefore either 5050 or 4070 milligrams. Typically,
our
objective is to come up with an estimate for the weight which is
unbiased
with minimum variance, or perhaps which might be biased but has
minimum
mean
squared error. In both cases, the estimate would be somewhere between
4070
and 5000 milligrams, since 4070 is more probable than 5050 as the true
weight, but 5050 also has positive probability.

 If, however, our
objective
was to maximize the probability of estimating the weight absolutely
correctly, or to maximize the probability of choosing an estimate in
the
correct interval $[0,5000]$ or $[5000, \infty]$ our best estimate
would
be
4070.  



\newpage

\noindent {\it 3. Conjectures}

 

\noindent {\it Conjecture 1.} If the dishonest candidate chooses how
many
illegal votes $N$ to buy after the victory rule is chosen, that could
result
in either higher or lower $T^*$ and $N$, depending on functional form
and
parameter values.

 
\noindent {\it Conjecture 2.} If the dishonest candidate chooses how
many
illegal votes $N$ to buy before the victory rule is chosen, that could
result in either higher or lower $T^*$ and $N$, depending on
functional
form
and parameter values.

 



\end{Large}

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