Pre-election polling and third party candidates

Abstract

I analyze voters’ incentives in responding to pre-election polls with a third party candidate. Third party supporters normally have an incentive to vote strategically in the election by voting for one of the major candidates. But these voters would vote third party if the third party candidate is doing surprisingly well in the polls. Because voters are more likely to vote third party if the third party candidate is doing well in polls, voters who like the third party candidate best have an incentive to claim they will vote third party in the polls so that more voters will ultimately vote third party in the election. The differing incentives faced during polls and elections accounts for why third party candidates do better in polls than in elections.

Appendix

Lemma 2

Suppose a voter prefers the third party candidate, has strict preferences between the two major party candidates, and believes the density at $(p_{A}, p_{B})$ is $f(p_{A}, p_{B})$ prior to the poll. Also suppose that $t$ of the voters in the pre-election poll like the third party candidate best and $m-t$ of the voters in the pre-election poll like one of the major party candidates best. Then if all other voters vote for their preferred candidate and $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$, it is a best response for this voter to vote third party for sufficiently large $n$. If $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx < \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$, then it is not a best response for such a voter to vote third party for sufficiently large $n$.

Proof

Note that for fixed values of $p_{A}$ and $p_{B}$, the probability that a voter likes the third party candidate best is $1 - p_{A} - p_{B}$, and the probability that a voter likes one of the major party candidates best is $p_{A} + p_{B}$. Thus the probability that $t$ voters in the poll like the third party candidate best and $m-t$ voters in the poll like one of the major party candidates best is $\frac{m!}{t!(m-t)!}(p_{A} + p_{B})^{m-t}(1 - p_{A} - p_{B})^{t}$. From this it follows that if $t$ voters like the third party candidate best and $m-t$ voters like one of the major party candidates best, then one believes the density at $(p_{A}, p_{B})$ is $f^{\prime }(p_{A}, p_{B}) = \frac{(p_{A} + p_{B})^{m-t}(1 - p_{A} - p_{B})^{t} f(p_{A}, p_{B})}{\int _{0}^{1} \int _{0}^{1-p_{B}} (p_{A} + p_{B})^{m-t}(1 - p_{A} - p_{B})^{t} f(p_{A}, p_{B}) \, dp_{A} \, dp_{B}}$.

Recall from Lemma 1 that if a third party supporter believes the density at $(p_{A}, p_{B})$ is $f^{\prime }(p_{A}, p_{B})$ and all other voters vote for their preferred candidate, then $\int _{1/3}^{1/2} f^{\prime }(x,1-2x) \, dx + \int _{1/3}^{1/2} f^{\prime }(1-2x,x) \, dx > \frac{1}{2} \int _{1/3}^{1/2} f^{\prime }(x,x) \, dx$ implies it is a best response for this voter to vote third party for sufficiently large $n$. Substituting $f^{\prime }(p_{A}, p_{B}) = k (p_{A} + p_{B})^{m-t}(1 - p_{A} - p_{B})^{t} f(p_{A}, p_{B})$, where $k = 1/\int _{0}^{1} \int _{0}^{1-p_{B}} (p_{A} + p_{B})^{m-t}(1 - p_{A} - p_{B})^{t} f(p_{A}, p_{B}) \, dp_{A} \, dp_{B}$, indicates that if $\int _{1/3}^{1/2} k (1-x)^{m-t} x^{t} f(x,1-2x) \, dx + \int _{1/3}^{1/2} k (1-x)^{m-t} x^{t} f(1-2x,x) \, dx > \frac{1}{2} \int _{1/3}^{1/2} k (2x)^{m-t} (1-2x)^{t} f(x,x) \, dx$, then it is a best response for this voter to vote third party for sufficiently large $n$. From this it follows that if $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$ and all other voters vote for their preferred candidate, then it is a best response for a voter whose favorite candidate is the third party candidate to vote third party for sufficiently large $n$. Similarly, if $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx < \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$, then it is not a best response for a voter whose favorite candidate is the third party candidate to vote third party for sufficiently large $n$ if the voter has strict preferences between the other candidates. $\square$

Proposition 1

Suppose it is an equilibrium for all voters to vote for their favorite candidate in the election if $t$ of the $m$ voters in the pre-election poll like the third party candidate best. Then for sufficiently large $n$ it is also an equilibrium for voters to vote for their favorite candidate in the election if more than $t$ of the $m$ voters in the pre-election poll like the third party candidate best.

Proof

To prove this it suffices to show that if it is an equilibrium for all voters to vote for their favorite candidate in the election if $t$ of the $m$ voters in the pre-election poll like the third party candidate best, then for sufficiently large $n$ it is also an equilibrium for voters to vote for their favorite candidate in the election if $t+1$ of the $m$ voters in the pre-election poll like the third party candidate best.

From Lemma 2 we know that if $t$ of the $m$ voters in the pre-election poll like the third party candidate best, then for sufficiently large $n$ it can only be an equilibrium for all voters to vote for their favorite candidate in the election if $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx \ge \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$ for both $f = f_{A}$ and $f = f_{B}$. Similarly we know that if $t+1$ of the $m$ voters in the pre-election poll like the third party candidate best, then for sufficiently large $n$ it is an equilibrium for all voters to vote for their favorite candidate in the election if $\int _{1/3}^{1/2} (1-x)^{m-t-1}x^{t+1} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t-1}(1-2x)^{t+1} f(x,x) \, dx$ for both $f = f_{A}$ and $f = f_{B}$. To prove this proposition it therefore suffices to show that $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx \ge \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$ implies $\int _{1/3}^{1/2} (1-x)^{m-t-1}x^{t+1} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t-1}(1-2x)^{t+1} f(x,x) \, dx$ for both $f = f_{A}$ and $f = f_{B}$.

Note that $\frac{(1-x)^{m-t-1}x^{t+1}}{(1-x)^{m-t}x^{t}} = \frac{x}{1-x} > \frac{1}{2}$ for $x \in (\frac{1}{3}, \frac{1}{2}]$ and $\frac{(2x)^{m-t-1}(1-2x)^{t+1}}{(2x)^{m-t}(1-2x)^{t}} = \frac{1-2x}{2x} < \frac{1}{2}$ for $x \in (\frac{1}{3}, \frac{1}{2}]$. Thus for either $f = f_{A}$ or $f = f_{B}$, we have $\int _{1/3}^{1/2} (1-x)^{m-t-1}x^{t+1} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx$ and $\int _{1/3}^{1/2} (2x)^{m-t-1}(1-2x)^{t+1} f(x,x) \, dx < \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$. From this it follows that if $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx \ge \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$, then $\int _{1/3}^{1/2} (1-x)^{m-t-1}x^{t+1} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx \ge \frac{1}{4} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx > \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t-1}(1-2x)^{t+1} f(x,x) \, dx$. The result follows. $\square$

Proposition 2

If third party supporters with strict preferences between the major party candidates all vote for their preferred major party candidate, then for sufficiently large $n$, each of the major party candidates wins with strictly greater probability than they would have if these third party supporters had all voted third party.

Proof

As the number of voters goes to infinity, the difference between the probability $A$ wins if the third party supporters vote for their preferred major party candidate and the probability $A$ wins if the third party supporters vote third party approaches the difference between $Pr(p_{A} + p_{C}q_{A} > p_{B} + p_{C}q_{B}, p_{A} + p_{C}q_{A} > p_{C}q_{C})$ and $Pr(p_{A} > p_{B}, p_{A} > p_{C})$. To show the result holds for $A$, it thus suffices to show that this difference is strictly greater than zero.

Note that $p_{A} > p_{B}$ implies $p_{A} + p_{C}q_{A} > p_{B} + p_{C}q_{B}$ since $p_{A} > p_{B}$ implies $q_{A} \ge q_{B}$ or $p_{C}q_{A} \ge p_{C}q_{B}$ and $p_{A} + p_{C}q_{A} > p_{B} + p_{C}q_{B}$. Also note that $p_{A} > p_{C}$ implies $p_{A} + p_{C}q_{A} > p_{C}q_{C}$ since $p_{A} + p_{C}q_{A} \ge p_{A}$ and $p_{C} > p_{C}q_{C}$. Thus $p_{A} + p_{C}q_{A} > p_{B} + p_{C}q_{B}$ and $p_{A} + p_{C}q_{A} > p_{C}q_{C}$ must hold if $p_{A} > p_{B}$ and $p_{A} > p_{C}$ are satisfied. From this it follows that $Pr(p_{A} + p_{C}q_{A} > p_{B} + p_{C}q_{B}, p_{A} + p_{C}q_{A} > p_{C}q_{C}) \ge Pr(p_{A} > p_{B}, p_{A} > p_{C})$.

Now note that if $p_{C} \ge p_{A} > p_{C}Q$, then $p_{A} > p_{C}$ does not hold, but $p_{A} + p_{C}q_{A} \ge p_{A} > p_{C}Q \ge p_{C}q_{C}$. Thus if $p_{A} > p_{B}$ and $p_{C} \ge p_{A} > p_{C}Q$, then $p_{A} + p_{C}q_{A} > p_{B} + p_{C}q_{B}$ and $p_{A} + p_{C}q_{A} > p_{C}q_{C}$ hold, but $p_{A} > p_{C}$ does not hold. So $Pr(p_{A} + p_{C}q_{A} > p_{B} + p_{C}q_{B}, p_{A} + p_{C}q_{A} > p_{C}q_{C})$ exceeds $Pr(p_{A} > p_{B}, p_{A} > p_{C})$ by at least $Pr(p_{A} > p_{B}, p_{C} \ge p_{A} > p_{C}Q)$. Thus it suffices to show that $Pr(p_{A} > p_{B}, p_{C} \ge p_{A} > p_{C}Q) > 0$.

Now note that $p_{C} \ge p_{A}$ holds if and only if $1 - p_{A} - p_{B} \ge p_{A}$ or $p_{B} \le 1 - 2p_{A}$ and $p_{A} > p_{C}Q$ holds if and only if $\frac{p_{A}}{Q} > 1 - p_{A} - p_{B}$ or $p_{B} > 1 - (1 + \frac{1}{Q})p_{A}$. Thus $p_{C} \ge p_{A} > p_{C}Q$ holds if and only if $1 - (1 + \frac{1}{Q})p_{A} < p_{B} \le 1 - 2p_{A}$.

Now if $p_{A} > \frac{1}{3}$, then $p_{B} \le 1 - 2p_{A}$ implies $p_{A} > p_{B}$. And if $p_{A} < \frac{1-\delta }{2}$, then $1 - 2p_{A} > \delta$. Also note that $Q < 1$ implies $1 - (1 + \frac{1}{Q})p_{A} < 1 - 2p_{A}$ for $p_{A} > 0$. Thus for any $p_{A} \in (\frac{1}{3}, \frac{1-\delta }{2})$, there is a continuum of values of $p_{B}$ which satisfy $p_{B} > \delta , p_{A} > p_{B}$, and $1 - (1 + \frac{1}{Q})p_{A} < p_{B} \le 1 - 2p_{A}$. Combining this with the fact that all voters believe the values of $p_{A}$ and $p_{B}$ are drawn from a distribution with a strictly positive density for such values of $p_{A}$ and $p_{B}$, it follows that all voters believe that $Pr(p_{A} > p_{B}, p_{C} \ge p_{A} > p_{C}Q) > 0$.

Thus for sufficiently large $n$, the probability that $A$ wins is strictly greater if third party supporters with strict preferences vote for their preferred major party candidate than if these voters vote third party. A virtually identical argument shows that the probability that $B$ wins is also strictly greater for sufficiently large $n$. $\square$

Proposition 3

For sufficiently large $n$, it is an equilibrium for voters who like a major party candidate best to claim they would vote for a major party candidate in the poll and for voters who like the third party candidate best to claim they would vote for the third party candidate in the poll.

Proof

The only situation in which a voter’s response to the poll affects the voter’s payoff is when exactly $t^{*}-1$ of the other voters in the poll claim they would vote third party and $m-t^{*}$ of the other voters in the poll claim they would vote for one of the major party candidates. In this case, if the voter claims he or she would vote third party in the poll, then third party supporters who have strict preferences between the major party candidates will vote third party. Otherwise such voters will vote for whichever of the major party candidates they prefer.

From Proposition 2 we know that, for sufficiently large $n$, if the third party supporters who have strict preferences between the major party candidates vote for their preferred major party candidate, then each of the major party candidates wins with strictly greater probability than they would have if these voters had voted third party. Thus in the one situation in which a voter’s response to the poll affects the voter’s payoff, claiming that one would vote for one of the major party candidates increases the probability that each of the major party candidates is elected for sufficiently large $n$. It thus suffices to show that voters who like one of the major party candidates best prefer a situation in which the probability that each of the major party candidates wins is increased and voters who like the third party candidate best prefer a situation in which the probability that each of the major party candidates wins is decreased.

If a voter’s favorite candidate is $A$, then the voter’s expected utility for the game is the probability that $A$ wins. Such a voter therefore prefers a situation in which the probability that each of the major party candidates wins is increased. Similarly, if a voter’s favorite candidate is $B$, then the voter also prefers a situation in which the probability that each of the major party candidates wins is increased. Finally, if a voter’s favorite candidate is $C$ and the voter is indifferent between the two major party candidates, then the voter’s expected utility for the game is the probability that $C$ wins or one minus the probability that a major party candidate wins. Such a voter therefore prefers a situation in which the probability that each of the major party candidates wins is decreased.

Now suppose a voter’s favorite candidate is $C$, and the voter strictly prefers $A$ to $B$. Then the voter’s expected utility for the game is $Pr(C \; \mathrm{wins}) + \frac{1}{2}Pr(A \; \mathrm{wins}) = 1 - \frac{1}{2}Pr(A \; \mathrm{wins}) - Pr(B \; \mathrm{wins})$. Such a voter therefore prefers a situation in which the probability that each of the major party candidates wins is decreased. Identical reasoning indicates that if a voter’s favorite candidate is $C$ and the voter strictly prefers $B$ to $A$, then the voter prefers a situation in which the probability that each of the major party candidates wins is decreased. The result then follows. $\square$

I now verify that the strategies in the paper continue to be an equilibrium as long as major party supporters are approximately indifferent between their least favorite candidates or if major party supporters have strong preferences between their least favorite candidates but they believe that $p_{A}$ and $p_{B}$ are drawn from a density $f(p_{A}, p_{B})$ that is symmetric in the sense that $f(x,y) = f(y,x)$ for all $x$ and $y$. First I show in Proposition 4 (not stated in main text) that major party supporters always vote for their favorite candidate in the election when they do not have strong preferences between their two least favorite candidates.

Proposition 4

Suppose a major party supporter obtains a utility of $1$ if his or her favorite candidate is elected, $0$ if his or her least favorite candidate is elected, and $\alpha \in (0, 1)$ if his or her second favorite candidate is elected. Also suppose the voter believes the density at $(p_{A}, p_{B})$ is $f(p_{A}, p_{B})$ prior to the poll. Then for any realization of the poll, if $\alpha < \beta \equiv \inf _{x \in [\frac{1}{3}, \frac{1-\delta }{2})} \min \left\{ \frac{f(1-2x,x)}{f(x,1-2x)}, \frac{f(x,1-2x)}{f(1-2x,x)}\right\}$, the voter will strictly prefer voting for the voter’s favorite candidate in the election for sufficiently large $n$ if all other voters in the election vote for their favorite candidate.

Proof

I prove the proposition for the case of a voter who likes $A$ best, $B$ second best, and $C$ least. The proofs of the other cases are virtually identical and thus omitted.

Suppose a voter derives utility of $1$ from the election of $A, \alpha \in (0, 1)$ from the election of $B$, and $0$ from the election $C$. Also suppose the voter believes the density at $(p_{A}, p_{B})$ is $f^{\prime }(p_{A}, p_{B})$. If all other voters are voting for their favorite candidate, then a similar argument to that used to prove Lemma 1 shows that if $2(1 - \alpha ) \int _{1/3}^{1/2} f^{\prime }(x,x) \, dx + \int _{1/3}^{1/2} f^{\prime }(x,1-2x) \, dx > \alpha \int _{1/3}^{1/2} f^{\prime }(1-2x,x) \, dx$, then this voter will strictly prefer voting $A$ to $B$ for sufficiently large $n$.

Suppose that $t$ voters in the pre-election poll like the third party candidate best and $m-t$ voters in the poll like one of the major party candidates best. Then we know from the proof of Lemma 2 that the voter believes the density at $(p_{A}, p_{B})$ is $f^{\prime }(p_{A}, p_{B}) = k (p_{A} + p_{B})^{m-t}(1 - p_{A} - p_{B})^{t} f(p_{A}, p_{B})$ for some constant $k > 0$. Thus the voter will strictly prefer voting $A$ to $B$ for sufficiently large $n$ if $2(1 - \alpha ) \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx + \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(x,1-2x) \, dx > \alpha \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(1-2x,x) \, dx$.

If $\alpha < \beta$, then $\alpha f(1-2x,x) < f(x,1-2x)$ for all $x \in [\frac{1}{3}, \frac{1-\delta }{2}), f(1-2x,x) = 0 = f(x,1-2x)$ for all $x \in (\frac{1-\delta }{2}, \frac{1}{2}]$, and $\alpha \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(1-2x,x) \, dx < \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(x,1-2x) \, dx$. But the inequality in the above paragraph holds if $\alpha \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(1-2x,x) \, dx < \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(x,1-2x) \, dx$. Thus if $\alpha < \beta$, such a voter strictly prefers voting for the voter’s favorite candidate for all realizations of the poll if all other voters are voting for their favorite candidate and $n$ is sufficiently large. $\square$

Next I show in Proposition 5 (again not stated in main text) that major party supporters always vote for their favorite candidate in the election if $f(p_{A}, p_{B})$ is symmetric in the sense that $f(x,y) = f(y,x)$ for all $x$ and $y$, even if they have strong preferences between their two least favorite candidates. This result will also hold if $f(p_{A}, p_{B})$ is close to a symmetric density.

Proposition 5

Suppose a major party supporter obtains a utility of $1$ if his or her favorite candidate is elected, $0$ if his or her least favorite candidate is elected, and $\alpha \in (0, 1)$ if his or her second favorite candidate is elected. Also suppose the voter believes the density at $(p_{A}, p_{B}), f(p_{A}, p_{B})$, satisfies $f(x,y) = f(y,x)$ for all $x$ and $y$ prior to the poll. Then for any realization of the poll, the voter will strictly prefer voting for the voter’s favorite candidate in the election for sufficiently large $n$ if all other voters in the election vote for their favorite candidate.

Proof

I prove the proposition for the case of a voter who likes $A$ best, $B$ second best, and $C$ least. The proofs of the other cases are virtually identical and thus omitted.

Suppose a voter derives utility of $1$ from the election of $A, \alpha \in (0, 1)$ from the election of $B$, and $0$ from the election $C$. Also suppose the voter believes the density at $(p_{A}, p_{B})$ is $f^{\prime }(p_{A}, p_{B})$. If all other voters are voting for their favorite candidate, then a similar argument to that used to prove Lemma 1 shows that if $2(1 - \alpha ) \int _{1/3}^{1/2} f^{\prime }(x,x) \, dx + \int _{1/3}^{1/2} f^{\prime }(x,1-2x) \, dx > \alpha \int _{1/3}^{1/2} f^{\prime }(1-2x,x) \, dx$, then this voter will strictly prefer voting $A$ to $B$ for sufficiently large $n$.

Suppose that $t$ voters in the pre-election poll like the third party candidate best and $m-t$ voters in the poll like one of the major party candidates best. Then we know from the proof of Lemma 2 that the voter believes the density at $(p_{A}, p_{B})$ is $f^{\prime }(p_{A}, p_{B}) = k (p_{A} + p_{B})^{m-t}(1 - p_{A} - p_{B})^{t} f(p_{A}, p_{B})$ for some constant $k > 0$. Thus the voter will strictly prefer voting $A$ to $B$ for sufficiently large $n$ if $2(1 - \alpha ) \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx + \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(x,1-2x) \, dx > \alpha \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(1-2x,x) \, dx$.

But since $f(x,1-2x) = f(1-2x,x)$ for all $x$, it follows that $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(x,1-2x) \, dx = \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(1-2x,x) \, dx$, meaning $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(x,1-2x) \, dx > \alpha \int _{1/3}^{1/2} (1-x)^{m-t}x^{t} f(1-2x,x) \, dx$, and the inequality in the previous paragraph is satisfied. Thus if $f(p_{A}, p_{B})$ satisfies $f(x,y) = f(y,x)$ for all $x$ and $y$, then the voter will strictly prefer voting for the voter’s favorite candidate in the election for sufficiently large $n$ if all other voters in the election vote for their favorite candidate. $\square$

Thus major party supporters will always vote for their favorite candidate in the election if they intensely prefer their favorite candidate over the other two candidates or if they have strong preferences between their two least favorite candidates but $f(p_{A}, p_{B})$ is symmetric in the sense that $f(x,y) = f(y,x)$ for all $x$ and $y$. Given this, note that identical proofs of Lemma 2 and Propositions 1 and 2 can be used to obtain these results when major party supporters are not indifferent between their two least favorite candidates. Also note that a virtually identical proof can be used to obtain Proposition 3 in this case. The only change needed is to make the minor adjustments needed to account for the fact that if a voter’s favorite candidate is $A$, then the voter’s expected utility is not exactly equal to the probability $A$ wins. For instance, if the voter prefers $A$ to both $B$ and $C$ sufficiently intensely, then the voter’s utility is approximately equal to the probability $A$ wins, and the voter will prefer the situation in which both major party candidates are more likely to win. A similar argument holds if the voter’s favorite candidate is $B$. Thus the results in the paper are robust to the assumption that the major party candidates are all indifferent between their two least favorite candidates.

Finally I give a result which indicates that it may be reasonable to expect that there will be an equilibrium in which third party supporters vote third party if the third party candidate receives at least one-third of the votes in the poll.

Proposition 6

Suppose that $2[f(x,1-2x) + f(1-2x,x)] > f(x,x)$ for all $x \in [\frac{1}{3}, \frac{1}{2})$ and $t \ge \frac{m}{3}$. Then $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$.

Proof

Note that if $\frac{1}{3} \le x < \frac{1}{2}$, then $(\frac{1-x}{2x})^{2m/3} \ge (\frac{1-2x}{x})^{m/3}$ for any $m > 0$ since $(3x-1)^{2} \ge 0$ implies $1 - 6x + 9x^2 \ge 0$ or $1 - 2x + x^{2} \ge 4x - 8x^{2}$ or $(1-x)^{2} \ge 4x(1-2x)$ or $x(1-x)^{2} \ge (2x)^{2}(1-2x)$ or $(\frac{1-x}{2x})^{2} \ge \frac{1-2x}{x}$ or $(\frac{1-x}{2x})^{2m/3} \ge (\frac{1-2x}{x})^{m/3}$. Now if $\frac{1}{3} \le x < \frac{1}{2}$, then $0 < \frac{1-2x}{x} \le 1$ and $0 < \frac{1-x}{2x} \le 1$. Thus if $t \ge \frac{m}{3}$, then $m-t \le \frac{2m}{3}$ and $(\frac{1-x}{2x})^{m-t} \ge (\frac{1-x}{2x})^{2m/3}$ for all $x \in [\frac{1}{3}, \frac{1}{2})$. Similarly, $t \ge \frac{m}{3}$ implies $(\frac{1-2x}{x})^{t} \le (\frac{1-2x}{x})^{m/3}$ for all $x \in [\frac{1}{3}, \frac{1}{2})$. But then $(\frac{1-x}{2x})^{2m/3} \ge (\frac{1-2x}{x})^{m/3}$ implies $(\frac{1-x}{2x})^{m-t} \ge (\frac{1-2x}{x})^{t}$ for all $x \in [\frac{1}{3}, \frac{1}{2})$ if $t \ge \frac{m}{3}$.

From this it follows that if $2[f(x,1-2x) + f(1-2x,x)] > f(x,x)$ for all $x \in [\frac{1}{3}, \frac{1}{2})$, then $2(\frac{1-x}{2x})^{m-t}[f(x,1-2x) + f(1-2x,x)] > (\frac{1-2x}{x})^{t}f(x,x)$ for all $x \in [\frac{1}{3}, \frac{1}{2})$ if $t \ge \frac{m}{3}$. This implies that $(1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] > \frac{1}{2} (2x)^{m-t}(1-2x)^{t} f(x,x)$ for all $x \in [\frac{1}{3}, \frac{1}{2})$ if $t \ge \frac{m}{3}$. Thus if $2[f(x,1-2x) + f(1-2x,x)] > f(x,x)$ for all $x \in [\frac{1}{3}, \frac{1}{2})$ and $t \ge \frac{m}{3}$, then $\int _{1/3}^{1/2} (1-x)^{m-t}x^{t} [f(x,1-2x) + f(1-2x,x)] \, dx > \frac{1}{2} \int _{1/3}^{1/2} (2x)^{m-t}(1-2x)^{t} f(x,x) \, dx$. $\square$

From this it follows that if $f = f_{A}$ and $f = f_{B}$ both satisfy $2[f(x,1-2x) + f(1-2x,x)] > f(x,x)$ for all $x \in [\frac{1}{3}, \frac{1}{2})$, then it will be an equilibrium for all voters to vote for their favorite candidate if the third party candidate receives at least one-third of the votes in the poll. The third party supporters may not always have prior beliefs about the density $f(p_{A}, p_{B})$ which satisfy these assumptions. However, for moderately sized polls (say $m \approx 1000$), the information provided by the poll is normally far more important than any prior beliefs voters had about the fraction of voters that would like a given candidate best. Thus this result suggests that there will normally be an equilibrium in which third party supporters all vote for their preferred candidate if the third party candidate receives at least one-third of the votes in the pre-election poll.