Super Tuesday: campaign finance and the dynamics of sequential elections

Abstract

I develop a model of campaign finance in primary elections in which campaigns, which supply hard information about candidates’ electability, must be financed by strategic donors. I provide a rationale for Super Tuesday electoral calendars in which a block of voters vote simultaneously early in the election followed by other voters voting sequentially. For a range of campaign costs, such a calendar maximizes expected donations to nomination campaigns and, thus, the ex-ante probability of electing the best candidate over all possible electoral calendars.

Appendix

The vote count at a given history is $x$-$y$ if $A$ has received $x$ total votes ($\sum 1_{\left( v_{i}=A\right) }=x$), and $B$ has received $y$ total votes.

1.1 Proof of Theorem 1

Proof

Taking donations as given, $v^{*}$ selects the candidate with the highest posterior probability of being highly electable. Thus, there is no purely informational incentive for an informed voter to deviate from this voting strategy. However, voting against one’s signal could conceivably lead to more campaigns being funded. We must ensure that this possibility does not make deviation from $v^{*}$ profitable. We do so by considering the largest possible benefit which such a deviation could have and showing that it is outweighed by the damage done by casting a vote for the wrong candidate. This maximum benefit is only possible if the SIG can condition its future funding on the outcome of the relevant votes—an assumption we maintain throughout this proof. If this is not the case, the benefits of deviation are more limited, strengthening our results.

If the SIG will only fund one campaign, the voter’s decision is pivotal and it is clearly in his interest to vote according to his signal. By symmetry, whether the first vote goes to candidate A or B cannot influence the SIG’s best-response continuation strategy. Future votes, however, could plausibly do so.

Consider the case in which the SIG will fund at least two campaigns. The largest benefit from a voter’s deviation would come if voting according to his signal would mean the end of informative campaigns, while voting against his signal would mean that all five campaigns would be funded.

Case 1: If the first two signals are for the same candidate, the posterior probability that this candidate is highly electable is $\frac{q^{2}}{ q^{2}+(1-q)^{2}}$. Voting against his signal leads to a 1-1 vote count, so that the probability of electing the highly electable candidate is at most:

$$\begin{aligned}&\frac{q^{2}}{q^{2}+(1-q)^{2}}\left( q^{2}\left( 1+2(1-q)\right) \right) + \frac{(1-q)^{2}}{q^{2}+(1-q)^{2}}\left( q^{2}\left( 1+2(1-q)\right) \right) \\&\quad =\left( q^{2}\left( 1+2(1-q)\right) \right) =q^{2}\left( 3-2q\right) <\frac{q^{2}}{q^{2}+(1-q)^{2}} \end{aligned}$$

Thus, the deviation is not profitable for the voter.

Case 2: If the first two signals are for different candidates, the posterior probability that any candidate is highly electable is $\frac{1}{2}$. Thus, deviating and making the vote count 2-0 leads to nominating the highly electable candidate with probability no higher than:

$$\begin{aligned} \frac{1}{2}\left( q^{3}\right) +\frac{1}{2}\left( 1-\left( 1-q\right) ^{3}\right) \end{aligned}$$

Given that the SIG was willing to fund at least one campaign, it will also be willing to fund an additional campaign if the vote count is 1-1 (it must be that $q-\frac{1}{2}>c$). Thus, the probability of nominating the highly electable candidate will be at least:

$$\begin{aligned} q>\frac{1}{2}\left( q^{3}\right) +\frac{1}{2}\left( 1-\left( 1-q\right) ^{3}\right) \end{aligned}$$

Thus, the deviation is not profitable for the voter.

We have now shown that deviating by voting against one’s signal cannot be profitable for a voter in this model. We must also verify that abstaining when informed or voting when not informed is not profitable. This type of deviation is detectable by all other players, so that its profitability depends on out of equilibrium beliefs.

Detectable deviations

Because donations $d_{i}$ are observable, all players will know that a deviation has taken place when an informed voter abstains and vice-versa. Indeed, we specify off-equilibrium beliefs for the SIG such that all future voters will vote for the frontrunner regardless of their signal. Thus, the SIG has no motivation to fund future campaigns, removing the voter’s incentive to deviate from $v^{*}$. $\square$

1.2 Proof of Theorems 3 and 4

I begin by listing all possible electoral calendars in a five voter election. I then state two lemmas which simplify comparisons across calendars by establishing basic facts about the SIG’s equilibrium donation strategies. A series of claims narrows the set of calendars we must consider to three: Simultaneous, Super Tuesday and Sequential. Finally, I state Theorem 1 which gathers the results of Theorems 3 and 4 and prove it by explicitly calculating and comparing expected payoffs to campaign contributions. Given Theorem 1, I assume throughout that voting strategies are $v^{*}$.

There are $2^{4}=16$ possible electoral calendars:

$$\begin{aligned} \Theta =\left\{ \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} {1.} \,\,\mathbf{Simultaneous},\{5\} &{} {5.\, \{3{\text {-}}2\}} &{} {9.\, \{2{\text {-}}1{\text {-}}2\}} &{} {13.\ \{1{\text {-}}1{\text {-}}1{\text {-}}2\}} \\ {2.} \,\, \mathbf{Sequential} \, \{1{\text {-}}1{\text {-}}1{\text {-}}1{\text {-}}1\} &{} {6.\, \{4{\text {-}}1\}} &{} { 10.\, \{1{\text {-}}2{\text {-}}2\}} &{} {14.\, \{1{\text {-}}1{\text {-}}2{\text {-}}1\}} \\ {3.} \, \mathbf{Super Tuesday} \, \{3{\text {-}}1{\text {-}}1\} &{} {7.\, \{1{\text {-}}4\}} &{} { 11.\, \{1{\text {-}}1{\text {-}}3\}} &{} {15.\, \{1{\text {-}}2{\text {-}}1{\text {-}}1\}} \\ {4. \, \{2{\text {-}}3\}} &{} {8.}\, \{2{\text {-}}2{\text {-}}1\} &{} {12.\, \{1{\text {-}}3{\text {-}}1\}} &{} { 16.\, \{2{\text {-}}1{\text {-}}1{\text {-}}1\}} \end{array} \right\} \end{aligned}$$

The following Lemma states that, if the SIG will fund some campaigns, without loss of generality we may assume that it will fund voter 1 and 2’s.

Lemma 1

If there is $d\in BR(v^{*},\theta )$ s.t. $\sum _{i=1}^{5}d_{i}\ge c$ for some $h_{5}\in H_{5}$, then $\sum _{i=1}^{5}d_{i}\ge c$ for all $h_{5}\in H_{5}$ and there is $d^{\prime }\in BR(v^{*},\theta )$ s.t. $d_{1}^{\prime }=c$.

If there is $d\in BR(v^{*},\theta )$ s.t. $\sum _{i=1}^{5}d_{i}\ge 2c$ for some $h_{5}\in H_{5}$, then $\sum _{i=1}^{5}d_{i}\ge 2c$ for all $h_{5}\in H_{5}$ and there is $d^{\prime }\in BR(v^{*},\theta )$ s.t. $d_{1}^{\prime }=d_{2}^{\prime }=c$.

Proof

Note that voters are identical except for the order they vote in. If the SIG will fund at least one campaign, its choice set is largest if it chooses to fund $V_{1}$’s. This implies that the SIG’s payoffs when it funds $V_{1}$’s campaign are weakly larger than when it waits and funds voter i(> 1)’s campaign first.

The SIG’s posterior belief about A’s type after observing $v_{1}$ will either be $q$ or $1-q$. By symmetry, the value of funding future campaigns must be the same in either case, so the SIG gains nothing by waiting until $v_{1}$ is observed. Thus, funding $V_{2}$’s campaign must weakly dominate waiting to fund a later campaign. $\square$

The following lemma states that, for any given electoral calendar and SIG donation strategy, the maximum number of donations which will be observed on the equilibrium path is odd or zero.

Lemma 2

Suppose $v=v^{*}$ and $d\in BR(v^{*},\theta )$. Then, $\max \limits _{h_{5}\in H_{5}}\{\sum _{i=1}^{5}1_{(d_{i}=c)}\}$ is either odd or zero.

Proof

Suppose a SIG has funded an odd number of campaigns n. The frontrunner has a lead of at least one vote. Thus, funding an additional campaign to get to an even n+1 will either confirm the frontrunner’s lead or tie the election. Tying the election does not change the identity of the candidate most likely to be highly electable. Thus, nothing is gained by funding campaign n+1. Therefore, the SIG will fund zero campaigns, one campaign, or fund until one candidate receives two out of three or three out of five informed votes. That is, an election will end with an even number of campaigns funded only if one candidate has a 2-0, 3-1, or 4-0 lead in informed votes. $\square$

The following series of claims narrow the set of calendars which we must consider. The first allows us to ignore calendars 7, 10, 11, 12, 13, 14, 15, and 16 and look only at 1, 2, 3, 4, 5, 6, 8 and 9 (or vice-versa).

Claim 1

Let $\theta$ be any calendar s.t. $\theta (V_{1})<$ $\theta (V_{2})$ and $\theta ^{\prime }$ be a calendar s.t. $\theta ^{\prime }(V_{1})=\theta ^{\prime }(V_{2})$ and $\theta ^{\prime }(V_{i})=\theta (V_{i})-1$ for $i\in \{3,4,5\}$. Then, $\theta$ weakly dominates $\theta ^{\prime }$ and vice-versa.

Proof

The donor knows that the election will either be 1-0 or 0-1 after one informative vote. Because of symmetry ($\Pr (A)=\Pr (B)=\frac{1}{2}$), continuation strategies lead to the same expected utility in either case. Therefore, the optimal $d_{2}$ will be the same regardless of whether the donor can condition on the outcome of $v_{1}$. $\square$

The following claim eliminates calendars 5 and 9 from consideration, leaving 1, 2, 3, 4, 6, and 8.

Claim 2

Any calendar in which $\theta (V_{3})<$ $\theta (V_{4})=\theta (V_{5})$ (calendars 5, 9, 10, and 13), is weakly dominated by $\theta ^{\prime }$ s.t. $\theta ^{\prime }(V_{4})<\theta ^{\prime }(V_{5})$ and $\theta ^{\prime }(V_{i})=\theta (V_{i})$ for $i\in \{1,2,3\}$ (calendars 2, 3, 15, and 16).

Proof

If $d_{1}=d_{2}=d_{3}=c$, the election will either be 2-1 or 3-0. Only the first case is relevant since the election is over if it is 3-0. If the last two voters vote simultaneously, the SIG will either fund both or neither since only two votes against the front-runner can change the result. Whenever $c$ is such that $d_{4}=d_{5}=c$ under $\theta (V_{4})=\theta (V_{5})$, $d_{4}=c$ if $\theta (V_{4})<\theta (V_{5})$, and $d_{5}=c$ if the election is still undecided. This is because the expected benefit of both continuation strategies is the same, but the expected cost is strictly lower in the Sequential case.

By Lemma 1, $d_{1}=c$ whenever $\sum _{i=1}^{5}d_{i}>0$ for some $h_{5}$, and $d_{2}=c$ whenever $\sum _{i=1}^{5}d_{i}>c$ for some $h_{5}$. If $d_{3}=0$, either $d_{4}=c$ or $d_{5}=c$ is possible only if the informative vote count is 1-1 (a 2-0 lead cannot be overcome by two votes).

The only remaining question is whether $d_{3}$ could be adversely affected by having $V_{4}$ and $V_{5}$ vote sequentially ($\theta (V_{4})<\theta (V_{5})$) rather than simultaneously ($\theta (V_{4})=\theta (V_{5})$). Suppose that the informative vote count is 1-1. It will be 2-1 after an informative $v_{3}$. Applying Lemma 2, if $d_{3}=0$ it will fund at most one more campaign, but in this case it may as well fund $V_{3}$ ’s. If the election is 2-0, it is only optimal to fund further campaigns if the SIG is prepared to fund campaigns until a candidate reaches a 3 informative votes. This continuation strategy must include $d_{3}=c$. $\square$

Next we eliminate calendar 6 from contention, leaving 1, 2, 3, 4, and 8.

Claim 3

The calendar {4-1} (no. 6) is weakly dominated by Super Tuesday {3-1-1}.

Proof

If all four campaigns in the first block of {4-1} are funded, it must be that the donor would fund the fourth campaign conditional on the election being 2-1 since it will either be 2-1 or 3-0, in which case the election is over. Therefore, if $d_{4}=c$ under {4-1}, then $d_{4}=c$ if $\theta =$ {3-1-1} and the election is 2-1.

Under {4-1}, the SIG will never fund only 3 date-1 campaigns. Funding three voters in the first block of a {4-1} means that $d_{5}=0$ because $d_{5}=c$ only if the informative vote count is tied, which is impossible when an odd number of informative votes have been cast thus far. Moreover, if the SIG funds 2 date-1 campaigns, he can make the funding decision for the third ($V_{5}$) after conditioning on the outcome of the first two (i.e. fund it only if the informative vote count is 1-1 and not 2-0). Therefore, it is strictly better for the SIG to fund two campaigns on date 1 and then fund voter 5 if the informative vote count is tied, thus giving the same probability of success at a strictly lower expected cost.

If only two campaigns are funded in the first block of {4-1}, at least two will be funded if $\theta =${3-1-1}. In both cases, only one additional campaign may be funded: if the informative vote count is 2-0 after the first block, the lead cannot be overcome; if the informative vote count is 1-1, one additional signal will make it 2-1 and that lead cannot be overcome. Therefore, if it is optimal to fund the first two voters when $\theta =$ {4-1}, it is also optimal when $\theta =${3-1-1}. $\square$

The following result shows that calendars 4 and 8 are weakly dominated by the Sequential calendar and narrows our set of contenders to 1, 2 and 3.

Claim 4

Any calendar in which $\theta (V_{2})<\theta (V_{3})$ is weakly dominated by the Sequential calendar {1-1-1-1-1}.

Proof

By Lemma 1, $d_{1}=d_{2}=c$ whenever the comparison of these calendars is in question. Therefore, I compare calendars conditional on two informative votes having been cast.

Suppose a candidate has a 2-0 lead. Then, the SIG will only fund further campaigns if it is willing to fund campaigns until one candidate has received three favorable informed votes. This may be done at a lower expected cost when the final three voters vote sequentially because the SIG can choose to stop funding as soon as one candidate reaches 3 votes. Therefore, having the final three voters vote sequentially weakly dominates all other arrangements of the last three voters conditional on the first two voters voting informatively for the same candidate.

Now suppose the election is tied 1-1. The SIG will fund voter 3’s campaign since the cost of previous campaigns is sunk and it was willing to fund the campaign of voter 1 (it must be that $q-\frac{1}{2}>c$). One candidate will have a 2-1 lead after voter 3’s vote. Because of the symmetry of the game, it does not matter which candidate it is for the SIG’s funding decision and therefore a calendar in which voters 3 and 4 vote simultaneously is strategically identical to one in which they vote sequentially. By the second claim in this section, if the first three campaigns have been funded, the calendar with voters 4 and 5 voting sequentially weakly dominates the one in which they vote simultaneously. $\square$

These claims leave us with three contenders for the voter-optimal electoral calendar: Sequential, Simultaneous, and Super Tuesday. As is proven in Theorem 1 below, the Simultaneous calendar is dominated by the Super Tuesday calendar. However, because of its special role in the literature, I will examine it more closely than other dominated calendars. I include the proof of this dominance relation in the proof of the following theorem.

Rather than prove Theorems 3 and 4 separately, we present the results contained in both into Theorem 1.

Theorem 1

There exist values $c_{all}<c_{sim}<c_{st}<c_{\mathrm{seq}}$ such that:

• The Super Tuesday calendar weakly dominates all other calendars, and strictly dominates the Sequential calendar, when $c\in \left( c_{all},c_{st}\right)$.

• The Sequential calendar weakly dominates all other calendars, and strictly dominates the Super Tuesday calendar, when $c\in \left( c_{st},c_{ \mathrm{seq}}\right)$.

• The Simultaneous calendar weakly dominates all other calendars when $c<c_{sim}$ or $c>c_{\mathrm{seq}}$.

Proof

Funding one (or two) campaigns results in the h-type winning the nomination with probability $p(1)=q$. The first voter will be funded under any electoral calendar if $\Delta (1)=q-\frac{1}{2}>c$.

Simultaneous calendar

In a Simultaneous election, funding three (or four) campaigns leads to selecting the correct candidate with probability:

$$\begin{aligned} q^{2}\left( q+3\left( 1-q\right) \right) =q^{2}\left( 3-2q\right) =p(3) \end{aligned}$$

The increase in the probability of selecting the h-type resulting from funding three campaigns rather than one is:

$$\begin{aligned} q^{2}\left( 3-2q\right) -q&= -q\left( 2q^{2}-3q+1\right) \\&= -2q^{3}+3q^{2}-q=\Delta (3)>0 \end{aligned}$$

The additional cost, given that one campaign is being funded, is $2c$. Therefore, the SIG will fund three campaigns if:

$$\begin{aligned} c<\frac{1}{2}\Delta (3) \end{aligned}$$

Funding five campaigns leads to selecting the correct candidate with probability:

$$\begin{aligned} q^{3}\left( q^{2}+5q\left( 1-q\right) +10\left( 1-q\right) ^{2}\right) =q^{3}\left( 10-15q+6q^{2}\right) =p(5) \end{aligned}$$

The increase in probability of success from funding voters 4 and 5 is:

$$\begin{aligned} p(5)-p(3)&= q^{3}\left( 10-15q+6q^{2}\right) -q^{2}\left( 3-2q\right) =3q^{2}\left( 2q-1\right) \left( q-1\right) ^{2}\\&= 6q^{5}-15q^{4}+12q^{3}-3q^{2}=\Delta (5)>0 \end{aligned}$$

The difference in cost between the two funding strategies is $2c$, so the SIG will fund all five campaigns if:

$$\begin{aligned} c<\frac{1}{2}\Delta (5)=c_{sim} \end{aligned}$$

When this inequality holds, the Simultaneous electoral calendar weakly dominates all others.

Super Tuesday calendar

In a Super Tuesday election, funding two campaigns in the first block leads to selecting the right candidate with probability $p(3)$ at an expected cost of:

$$\begin{aligned} 2c+c\left( 1-q^{2}-\left( 1-q\right) ^{2}\right) = 2c\left( 1+q\left( 1-q\right) \right) <3c \end{aligned}$$

The increase in cost from funding only one campaign to following this strategy is $c+2cq\left( 1-q\right)$,Therefore, at least two date-1 campaigns will be funded in a Super Tuesday election if:

$$\begin{aligned} c<\frac{\Delta (3)}{1+2q\left( 1-q\right) }=c_{\mathrm{seq}} \end{aligned}$$

Note that the SIG will be willing to fund this strategy for higher $c$ than it is to fund three campaigns in a Simultaneous election because $1+2q\left( 1-q\right) <2$.

In a Super Tuesday election, funding all three period-1 campaigns leads to selecting the right candidate with probability $p(5)$at an expected cost of:

$$\begin{aligned} 3c+c\left( 1-q^{3}-\left( 1-q\right) ^{3}\right) +c\left( 6q^{2}\left( 1-q\right) ^{2}\right) =3c\left( 1+q+q^{2}-4q^{3}+2q^{4}\right) \end{aligned}$$

The increase in the probability of nominating an h-type from funding voter 3’s campaign is $\Delta (5)$. Subtracting the expected cost of the “fund 2” strategy from that of the “fund 3” I find the difference in expected cost:

$$\begin{aligned}&3c\left( 1+q+q^{2}-4q^{3}+2q^{4}\right) -c\left( 1+q\left( 1-q\right) \right) =c\left( 6q^{4}-12q^{3}+5q^{2}+q+1\right) \\&\quad =c\left( 1+q\left( 1-q\right) \left( 1+6q\left( 1-q\right) \right) \right) \end{aligned}$$

Therefore, the SIG will fund all 3 date-1 campaigns in a Super Tuesday election if:

$$\begin{aligned} c<\frac{\Delta (5)}{1+q\left( 1-q\right) \left( 1+6q\left( 1-q\right) \right) }=c_{st} \end{aligned}$$

When this inequality holds, the Super Tuesday calendar weakly dominates all others.

Because $q\left( 1-q\right)$ reaches a maximum for $q\!\in \! (0,1)$ at $\frac{ 1}{4}$, $q\left( 1-q\right) \left( 1\!+\!6q\left( 1-q\right) \right)$ $\le \frac{5 }{8}$ and therefore $c_{st}>c_{sim}$.

Sequential calendar

Suppose that one of the candidates has a 2-0 lead in informed votes in a Sequential election. The SIG’s posterior belief about the probability that the leading candidate is the correct choice is $\frac{q^{2}}{q^{2}+\left( 1-q\right) ^{2}}$. That is also the probability of the correct candidate winning the election if the SIG funds no further elections. If the SIG does continue to fund campaigns, it must be willing to do so until a candidate reaches 3 votes. This increases the probability of electing the correct candidate to:

$$\begin{aligned}&\frac{q^{2}}{q^{2}+\left( 1-q\right) ^{2}}\left( q+\left( 1-q\right) q+\left( 1-q\right) ^{2}q\right) +\frac{\left( 1-q\right) ^{2}}{q^{2}+\left( 1-q\right) ^{2}}q^{3}\\&\quad =\frac{q^{3}}{2q^{2}-2q+1}\left( 2q^{2}-5q+4\right) \end{aligned}$$

So that the increase in the probability of electing the correct candidate is:

$$\begin{aligned} \frac{\left( 1-q\right) ^{2}}{q^{2}+\left( 1-q\right) ^{2}}q^{3}-\frac{q^{2} }{q^{2}+\left( 1-q\right) ^{2}}\left( 1-q\right) ^{3}= q^{2}\left( 2q-1\right) \frac{\left( q-1\right) ^{2}}{2q^{2}-2q+1} \end{aligned}$$

This strategy brings with it an additional cost of:

$$\begin{aligned}&c+c\left( \frac{q^{2}}{q^{2}+\left( 1-q\right) ^{2}}\left( 1-q\right) + \frac{\left( 1-q\right) ^{2}}{q^{2}+\left( 1-q\right) ^{2}}q\right) \\&\quad +c\left( \frac{q^{2}}{q^{2}+\left( 1-q\right) ^{2}}\left( 1-q\right) ^{2}+\frac{ \left( 1-q\right) ^{2}}{q^{2}+\left( 1-q\right) ^{2}}q^{2}\right) \\&\quad = \frac{2c}{q^{2}+\left( 1-q\right) ^{2}}\left( 2q^{4}-4q^{3}+3q^{2}-q+1\right) \end{aligned}$$

Therefore, the SIG will fund voter 3 after a 2-0 start if:

$$\begin{aligned} c&< \frac{q^{2}\left( 2q-1\right) \frac{\left( q-1\right) ^{2}}{q^{2}+\left( 1-q\right) ^{2}}}{\frac{1}{q^{2}+\left( 1-q\right) ^{2}}\left( 2q^{4}-4q^{3}+3q^{2}-q+1\right) }\\&= q^{2}\left( 2q-1\right) \frac{ \left( q-1\right) ^{2}}{2q^{4}-4q^{3}+3q^{2}-q+1}=c_{all} \end{aligned}$$

Note that $c_{all}<c_{sim}<c_{st}$, so that the Simultaneous and Super Tuesday calendars strictly dominate the Sequential when $c\in (c_{all},c_{sim})$, a result in line with the findings in Battaglini (2005). Moreover, the Super Tuesday calendar strictly dominates the Sequential calendar over the campaign cost interval $(c_{sim},c_{st})$. I verify $c_{all}<c_{sim}$ numerically:

$$\begin{aligned} \left( 6q^{5}-15q^{4}+12q^{3}-3q^{2}\right) -2q^{2}\left( 2q-1\right) \frac{ \left( q-1\right) ^{2}}{2q^{4}-4q^{3}+3q^{2}-q+1}>0 \end{aligned}$$

In a Sequential election, if three campaigns have been financed leading to a 2-1 vote lead by one of the candidates, continuing to fund campaigns makes sense for the SIG only if it is willing to fund until one candidate has three votes. This leads to electing the correct candidate with probability $p(3)$, while stopping funding now means the frontrunner will win the election, which is the correct choice with probability $q$. The increase in the probability of selecting the correct candidate is therefore $\Delta (3)$. This strategy leads to additional expected costs of $c\left( 1+2q\left( 1-q\right) \right)$. Therefore, the SIG will continue this funding if:

$$\begin{aligned} c<\frac{\Delta (3)}{1+2q\left( 1-q\right) }=c_{\mathrm{seq}} \end{aligned}$$

Note that the SIG’s problem following a 2-1 vote and following a 1-0 vote while expecting not to fund more than three campaigns are identical. Therefore, when $c>c_{\mathrm{seq}}$, the SIG will fund at most one campaign. If this is the case, the voter is indifferent among all electoral calendars.

On the other hand, if the SIG funds only two date-1 campaigns in a Super Tuesday election it will fund a third if the election is tied 1-1 in informed votes after the first block has voted, but will never fund more than that because a 2-1 lead which would ensue could never be overcome by a single informed vote. Therefore, there is a cost interval $c\in \left( c_{st},c_{\mathrm{seq}}\right)$ over which the Sequential calendar strictly dominates the Super Tuesday calendar. This interval in non-empty if:

$$\begin{aligned} c_{\mathrm{seq}}-c_{st}=\left( \frac{-q\left( 2q^{2}-3q+1\right) }{1+2q\left( 1-q\right) }-\frac{6q^{5}-15q^{4}+12q^{3}-3q^{2}}{1+q\left( 1-q\right) \left( 1+6q\left( 1-q\right) \right) }\right) >0 \end{aligned}$$

Again, I verify this inequality numerically. $\square$

1.3 Proof of Theorem 5

Each voter must determine whether he is willing to invest $c$ in order to receive an informative signal about candidate quality. Note that, in contrast to the model presented in Section 3, information acquisition decisions here do not depend on the expected cost of future expenditures on information. This makes the proof of Theorem 5 somewhat simpler than the proof of Theorems 3 and 4. To avoid repetition, we will use notation and probability calculations developed in the proof of Theorems 3 and 4; in particular, $\Delta (3)$, $\Delta (5)$ and $p(3)$ are defined in that proof, and the value of pursuing a best out of five strategy when after a 2-0 start is calculated there.

Whenever the electoral calendar allows it, early voters will not acquire information, choosing instead to free-ride on later voters. In some cases, such as Simultaneous elections, the ordering of voters is not meaningful and there may be multiple equilibria in which the identity of the voters who vote varies.

Voters are most likely to acquire information if they believe that other voters will not compensate for their failure to do so by acquiring information themselves. Thus, in efficient pure strategy equilibria voters who acquire information compare their cost of doing so, c, with the expected social benefit of doing so.

We proceed by analyzing several information acquisition cost intervals separately and checking whether any voter could have the incentive to acquire information that would lead to a one signal, three signal or five signal equivalent outcome.

If $c>q-\frac{1}{2}$ the informational value of the first signal is less than its cost, so that no voter will choose to acquire information.

If $q-\frac{1}{2}\ge c>\Delta (3)$ the informational value of the first signal is more than its cost, so that voter 5 will acquire information if no other voter has or is expected to. However, the incremental value of eliciting three signals rather than one, $\Delta (3)$, is lower than the cost of acquiring a signal so that no voter will be willing to invest c in order to increase the probability of nominating the high type from $q$ to $p(3)$. Thus, the efficient equilibrium in this case involves only one signal being acquired.

If $\Delta (3)\ge c>\Delta (5)$ the value of eliciting three signals rather than one is higher than c, so that any individual voter who believes he is pivotal for that outcome will be willing to acquire information. However, going from a three signal outcome to a five signal outcome does not justify the expense $c$, so that equilibria involving more than three voters acquiring information are not feasible. Furthermore, in an equilibrium in which a maximum of three voters are expected to acquire information, being able to condition one’s vote on prior votes or signals does not affect the voter’s decision, except in the case in which the election has already been decided and is 2-0. The only alternative is that the election is 1-1 and acquiring information brings benefits of $q-\frac{1}{2}<c$. Therefore, the probability of selecting a high-type nominee in an efficient pure strategy equilibrium of the information acquisition game when $c\in [\Delta (3),\Delta (5))$ is independent of the electoral calendar.

In these first three cases, the electoral calendar plays no role in determining the efficient pure strategy equilibrium. However, in part of the remaining interval of information costs $c$, $[0,\Delta (5)]$, electoral calendars affect the amount of information acquisition in the set of efficient pure strategy equilibria.

If $\Delta (5)=c_{2}\ge c$ and voter 3 can condition his information acquisition decision on voter 1 and 2’s votes or signals, he will not acquire information after a 2-0 start if the benefit of doing so $\frac{ q^{2}(2q-1)(q-1)^{2}\ }{2q^{2}-2q+1}=c_{1}$ is less than its cost $c$. However, if voter 3 cannot condition on the first two votes or signals, the expected value of acquiring information is $\Delta (5)\ge c$ so that he will choose to do so. Therefore, for the cost interval $(c_{1},c_{2}]$, calendars which allow voter 3 to condition his decision on the outcome of the first two votes, that is calendars in which $\theta (V_{2})<\theta (V_{3})$, underperform other calendars.

The class of electoral calendars which satisfy $\theta (V_{2})=\theta (V_{3})$ includes the Simultaneous and the Super Tuesday calendars while the Sequential calendar satisfies $\theta (V_{2})<\theta (V_{3})$.

Finally, if $c\le c_{1}$ all voters will be willing to acquire information regardless of the electoral calendar.