Steve Sachs Duke


Tuesday, May 04, 2004


Reflections on the Duty to Vote, Part I: Late last January, I sent in my absentee ballot for the Missouri primary. In the course of applying for the ballot, I remembered a discussion I had with a good friend a while back on whether citizens had a general duty to vote. My friend's argument was based on the well-known position that "my vote won't make a difference." She wasn't claiming that political process was irrelevant; the president of the United States has a vast amount of power, and putting the wrong person in office could seriously disrupt the lives of millions. Rather, she advanced the claim that voting couldn't be justified based solely on its effects. Ever since then, I've been working on a response, which has turned into this four-part series of blog posts.

Even in an important election, the argument goes, when the candidates' differences are pronounced, one vote will still be just a drop in the bucket. No one person's vote will change the outcome of a nationwide election. The same, of course, might be said of charity work; no one person's efforts will end world hunger. But in the case of charity, even a drop in the bucket can still be valuable in its own right -- it still represents a few more hot meals served, a few more sacks of grain, a few more warm winter coats. In an election, however, a single vote that does not decide the winner will have absolutely no consequences at all--it just changes the vote totals by one. And although one of the closest elections in American history recently came down to 500-odd votes in the state of Florida, given how many tens of millions of votes were cast, it's unlikely that an election will ever become any closer. Why should we vote, then, if the election is never really going to be tied?

Derek Parfit, in Reasons and Persons (summarized here), tries to answer this question with standard act-consequentialist analysis. On Parfit's argument, I should only vote if the expected costs of my voting are less than or equal to the expected benefit -- the benefit to the world of the superior candidate winning, multiplied by the chance that my ballot will decide the election. The costs of my voting are small (one half-hour on a given Tuesday), while the benefits of the superior candidate are large. In fact, if we look only at the economy (and ignore the other responsibilities of the office, like being commander-in-chief), assuming that a bad president would cost 0.5 percent of GDP means the benefit of a good one is something on the order of $50 billion.

The argument here crucially relies on the exact measure of the probability. Parfit estimates the chance of a split election at 1 in 100 million. By our benefit estimate above, though, that means the expected benefit from voting is $500--and that's ignoring most of what the president does. But is 100 million even low enough? How can we calculate the exact probability of a split election?

To make the calculations easier, we can start with some simplifying assumptions. First, let's assume that the election is entirely decided by the popular vote (goodbye, electoral college). Second, let's assume that there are two candidates, Bush and Gore, who are exactly tied in the polls (each at 50 percent support). As in the 2000 election, let's say there are 105,363,298 people who cast their votes before I do. What's the chance that the election would come out exactly split, and that my ballot will decide the election?

One way to estimate this chance is to pretend that each voter has a 50 percent chance, once in the voting booth, of casting a ballot for either Bush or Gore. Determining the chance of a split result is sort of like determining the chance that a fair coin, tossed a certain number of times, will come up exactly half heads and half tails. On the advice of Dr. Math, I used the binomial probability formula. If I flip a coin n times, the chance of r heads is given by P = C(n,r) · pr · q(n-r), where p is the chance of heads, q is the chance of tails, and C(n,r) = n! / (r! · (n-r)!). (Roughly, the pr and q(n-r) terms identify the chance of so many heads and so many tails, and C(n,r) identifies the different ways those heads and tails can be distributed in the order of tosses.) For instance, if we want to flip a coin 4 times and come up with 2 heads, then n is 4, r is 2, p = q = 0.5, and the final probability (calculating out all the factorials) is 0.375, or 3/8. In the election case, then, if we assume that n is 105,363,298, r is half that (or 52,681,649), and p = q = 0.5, what's the probability P?

This formula involves numbers far too large for most calculators--try taking the factorial of 100 million, and you'll see what I mean--so I eventually had to resort to programming. And after the numbers crunched, I found a truly surprising result: the probability figure was 7.7727·10-5. This is roughly one in 13,000, for an expected value of $3,886,350 per ballot -- a benefit far outweighing any conceivable costs.

Obviously, this figure is far too large. We've had a good number of elections in the U.S. since the republic was founded, and the idea that one in 13,000 would result in an even split (with 100 million people voting!) is simply incredible. What could have gone wrong?

One possibility is simple programming error; but given that the programs seemed to work well at lower numbers, giving consistent results using theoretically equivalent algorithms, this is unlikely to be the problem. (If you're interested in checking the math, I've posted the GAWK source code. I originally used the simple binomial program here, but when the numbers got too large, I was forced to switch to adding up logarithms (here) and then expanding the sum at the end. A third program provided a more efficient means of adding the logarithms (here), since many of the terms in the factorials cancel out, and produced the same results as the first two.)

A second possibility is inaccuracy in the description of the problem. "Of course," one might argue, "actual voters don't flip a coin when they go into the voting booth. There's a set number of people who are going to vote for Bush, and a set number of people who are going to vote for Gore. So your 50-50 probability has nothing to do with which candidates people actually choose." This objection can be countered by looking at the situation from the perspective of the voting machine, which doesn't know which candidate the person who just entered the booth supports. All it knows is that the person will pull the lever either for Bush or for Gore, and that roughly 50 percent have been pulling the lever for each. (Alternatively, imagine that there is a pre-set population of people who support Bush or Gore, but that not all of them will end up voting. Those who actually do go out and vote are randomly selected from this 50-50 group--or, more precisely, the process whereby certain individuals end up voting doesn't bias the election in favor of one candidate or the other, so it seems like we've taken a random sample. In this case, we really are taking a 50 percent chance that the next voter we select, or just the next voter to enter the booth, will vote for a given candidate. This assumption that the two candidates' supporters vote at equal rates isn't often true in actual elections, since one candidate's supporters may be more committed than the other's; in that case, assume that the poll data has been weighted to reflect the how likely a voter the respondent is, and that the final figures are still 50-50.) From these perspectives, the election does seem like a matter of probability. The issue here isn't that the election result is actually random and unpredictable, like quantum mechanics, but rather that we can use probability theory to produce a reasonable ex ante estimate when we don't know who the actual voters will be.

A third possibility, and the one that seems most likely to me, is that the assumption of exact 50-50 support is unrealistic. This formula, especially with large number of voters, is incredibly sensitive to minor variations in the level of support. If Bush were receiving 50.1 percent support in the polls, for instance, instead of being 50-50, the chance of a split election would drop to roughly 1 in 1.3·1096, which is tiny beyond comprehension. (1096 is much larger than the number of atoms in the universe.) And since polls generally have a margin of error of ±3 percent, the chance that the level of support is exactly 50-50 is also miniscule.

These extraordinarily low probabilities make it seem far more likely that the math is right, since it accords with our intuition that evenly split elections are extraordinarily rare. So we really need to conduct our analysis at two levels. First, what is the actual breakdown of Bush and Gore supporters in the country? This question probably won't be answered with a single number (it could be, but we don't know it); instead, it's a probability distribution over various values within the margins of error for our poll data. Plus, given that polls themselves aren't always right, the distribution could be a lot wider than that.

Second, once we have the distribution of possible support breakdowns, what is the probability that, given a random sampling of that population (so we'd have to weight the support breakdown to reflect the fact that not everyone is equally likely to go to the polls), we get a split election? To my mind, if differences that minor can produce huge changes in the probabilities, this means that the actual chance of getting a split election is vanishingly small. If the polls are showing a statistical dead heat, there may be some minute chance that a vote would make a difference. (How small? The chance of a split election in a 50.01-49.99 race is roughly 1 in 100,000; if we assume--inaccurately, but conservatively--that a 50-50 poll means there's a 1-in-1000 chance that the actual level of support is between 50.01 and 49.99 percent, then we're back at Parfit's 1-in-100-million figure.) But if the polls are showing a 53-47 race, or a 56-44 race, the probability that my vote would make a difference is too small even to consider. I might as well stay home.

But should I stay home? Do these probability calculations invalidate any duty we have to vote? To find out, tune in for tomorrow's post.

UPDATE: See further post above.




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