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# The Rational Voter Model Does Not Predict Zero Turnout

A famous Winston Churchill anecdote offers a good example of how conventional wisdom fails. Churchill approached his frequent antagonist, Lady Nancy Astor, and asked her whether she would sleep with him for a million pounds. She said of course she would. He then asked her whether she would sleep with him for five pounds. She responded, “What do you take me for?” Churchill response, “We’ve already established what you are, now we’re simply haggling over the price.”

Conventional wisdom also holds that the rational voter model implies voter turnout should be zero. By “the” rational choice model, the discussants typically mean the simplest model, but that’s still fair enough. Critics often proffer the claim as evidence of the manifest weakness of rational choice models of politics in general. After all, if rational choice theory cannot explain the most common political act in a democracy, then it can’t be of much use more generally, the thought goes.

I’ve heard the conventional wisdom repeated a hundred times, in formal research seminars and in informal conversations. Yet it’s wrong.

Contrary to conventional wisdom, even the basic rational choice model does not predict zero turnout.

To be sure, it’s difficult for the basic rational choice model to gin up the levels of voter turnout we commonly observe, but when we establish it predicts positive voter, to riff off of the Churchill quip, the principle is settled and we’re just haggling over the numbers.

A little brush clearing to begin with (with apologies to regular *L&L* readers for whom the brush has already been cleared). First, the bugaboo of “rationality.” It doesn’t mean what most people think it means. It only means people can choose among alternatives (“I prefer A to B, B to A, or “I am indifferent between A and B”) and people’s preferences are transitive (if a person prefers A to B and B to C, then that person prefers A to C, or if a person is indifferent between A and B, and B and C, then the person will be indifferent between A and C). Even certifiably crazy people can be rational according to the rationality postulate in the social sciences.

Nonetheless, behavior exists over which even this thin notion of rationality breaks down. Add one grain of sugar to a cup of coffee, and most people will report indifference between one cup and the next cup. But keeping adding sugar one grain at a time, and, over enough cups, people’s preferences over sugared-coffee will prove intransitive.

Critics often report results like this as though they threaten the foundations of rational choice theory. They don’t. I’m happy to concede the existence of intransitivities among some choices, or admit the fact of abnormal psychology, or systematic misperceptions, etc., etc., etc., I don’t mind sharing the explanatory world with theories that posit some, or even much, human behavior is inconsistent with the rationality postulates.

At the same time, I do think it noncontroversial to think most people are goal directed and that people often, but not always, respond in patterned ways to given incentive structures. If that’s conceded, then rational choice theory would seem to be one helpful tool to understand wide swaths of human behavior. It doesn’t need to explain everything.

But enough throat clearing.

The basic rational voter model contains three elements, the benefit, *B*, of one’s preferred candidate winning relative to the competitor, the cost, *c*, of voting, and the probability, *p*, that one’s vote will be pivotal, that is, will make the difference between one’s preferred candidate winning and the other guy winning.

Voters incur the cost of voting with certainty. But the benefit of voting is a collective good. So the possibility of free riding exists. We thus need to discount the benefit, *B*, by the probability that one’s vote is pivotal. The basic rational voter model holds people will vote when *pB* – *c* ≥ 0, and people will not vote when *pB* – *c* < 0. (We can assign the equality to either case.)

Depending on the election, the benefit of one’s preferred candidate winning can extend from pretty low (the candidates pretty much support the same policies) to really high (some voters fear one candidate will start a nuclear war, or not deter one). The cost of voting is typically modest. It’s the opportunity cost of registration and getting to the voting booth (whether for early voting or on Election Day). Usually a bit of time.

But in the vast majority of mass elections we observe, the probability one’s vote is pivotal – that is, that it makes a difference between who wins and who loses – is typically very low, approaching zero in very large elections.

We can play with this a bit, for example, by positing many people mini-max regret when they vote. I’ll chat about possible extensions to the basic model in another post. I want to consider only the basic model here, however. It’s the one critics typically have in mind when repeating the conventional wisdom that it implies voter turnout should be zero.

Here’s the intuition for the conventional wisdom: Even when the *B*s in an election are large, the *p*s are almost always so low that the turnout levels we see in the vast majority of mass elections seem inconsistent with the basic rational choice model.

I agree we need to add to the basic model to get turnout levels we commonly see in mass elections. That claim, however, is very different from the claim that the elementary rational voter model predicts zero turnout. Contrary to the all-too-conventional wisdom, it is easy to see the basic rational voter model does *not* predict zero voter turnout in equilibrium.

Let’s apply the rational choice model using the critic’s assumption that no one votes. If conventional wisdom is correct, then zero turnout would be an equilibrium for the basic model, that is, no one would have an incentive to deviate from rational non-participation to actually casting a vote.

If the conjecture is no one votes, does any non-voter, acting consistent with the rational choice model, have an incentive to deviate from the conjectured behavior? Of course. If the conjecture is that no one votes, then the probability one’s vote is pivotal equals 1. Your vote decides the election. Almost everyone would have an incentive to vote in this case, at least in elections of any consequence.

But we don’t need to go so far as to hang the model’s vindication on the formalism that it can gin up one voter turning out in an election. Let’s take the next step, assume the conjecture that everyone thinks only one voter will turn out vote. Well, in the vast number of elections it would seem that it would be rational, according to the elementary model, for a second person to turn out to vote. Then repeat.

Even with our simple model, turnout continues to climb until the conjectured “*p*” generates equality according to the equation.

To be sure, I’d concede that, without more, the basic rational voter model can’t get us up to turnout of 100,000,000 voters in U.S. presidential elections. But thinking of even modestly-sized *B*s, even this toy model gets us way past zero. I’m guessing, but I’d wager the basic model could get us to tens of thousands of voters in many elections, and perhaps hundreds of thousands.

Sure, it’s not what we observe in most modern elections. But the principle has been established, and now we’re just quibbling over the numbers.