Blargh Blog

Tuesday, November 08, 2005

Rational Voting: The (Ir)relevance of Interference in Close Elections

An election is supposed to be simple. Voters cast their votes, those votes get counted, and the candidate with the most votes wins. As we all know, that's not how things work out in real life. There's fraud, intimidation, overly stringent restrictions that keep voters from having their votes counted, underly stringent restrictions that allow people who aren't eligible to vote to cast votes nonetheless, imperfect vote-counting, suspect voting machinery... And if the election is close enough so that any of these factors might have made a difference, then it will probably go to court, where a few people in robes can effectively hand the election to one candidate or the other.

This is bad for a variety of reasons. It can affect the outcome of elections. The lack of transparency undermines democratic values. Questions about legitimacy can lead to outrage and partisan divisiveness. It can make citizens feel disempowered, as the judges or the fraudsters seem to be determining the result.

This last effect intersects with a common blogosphere topic: why vote? The probability that a single vote will change the outcome of an election is tiny. Contra Frank, the question is not why so many people are voting against their economic self-interest, but why anyone would vote out of economic self-interest. Anyone's economic self-interest would be much better served by staying home and using that time productively. Levitt and Dubner (of Freakonomics fame) have an article in the NY Times Magazine giving their take on why people vote if the incentives don't seem to be there. After going through the standard argument for why an election is unlikely to be decided by a single vote - your vote - and providing some historical data on Congressional elections (or should I say "election") where the margin was a single vote, Levitt and Dubner go on to claim:
But there is a more important point: the closer an election is, the more likely that its outcome will be taken out of the voters' hands - most vividly exemplified, of course, by the 2000 presidential race.
But is this true? In the Crooked Timber thread I linked to before (and before), John Quiggan argues that your impact on the outcome does not change if close elections are decided by a "tortuous litigative and bureaucratic procedure":
In a close election, no vote is decisive, but every change in the (election-night) margin changes the odds. 1000 extra votes for Gore in Florida would have made a big difference, so every vote cast in Florida made (roughly) 0.001 of that big difference.
So does the intervention of judges and so forth reduce your chances of affecting the outcome of the election? Who's right, Quiggan or Levitt + Dubner?

I gave an answer in the comments at Freakonomics: Levitt and Dubner are wrong. I'd like to defend it here with somewhat more precise reasoning.

Let's say that there are two candidates, and let x be the number of votes that are cast for the candidate you prefer minus the number of votes cast for the other candidate. We define two functions in terms of x, both based on the best guesses that you can make before the election, when you are trying to decide whether to vote:

Let p(x) be the probability that your candidate will get x votes more than the other candidate from all of the voters besides you (so p(x) excludes your vote).

Let f(x) be the probability that your candidate will win the election, given that he received x votes more than the other candidate.

What is the probability that your candidate will win? Well, for each possible pattern of voting (which is represented by a particular value of x), we want to figure out how likely it is that this will be the voting pattern (that's p(x)), and then figure out how likely your candidate will win if this is the voting pattern (that's f(x)). We multiply these together and add up the values for all possible voting patterns and the result is the sum over all integers x of p(x)f(x).

What is the probability that your vote will make the difference in the election? Well, for each possible pattern of voting for all of the other voters (which is represented by a particular value of x), we want to figure out how likely it is that your candidate will win if this is the voting pattern and you do not vote (that's f(x)), and subtract that from how likely it is that your candidate will win if that is the voting pattern and you do vote (that's f(x+1), since your candidate is getting one more vote), and as before we multiply by the likelihood that this will be the voting pattern and sum over all possible voting patterns. The result is the sum over all integers of p(x)[f(x+1)-f(x)].

There is one assumption that we are making here, and that is that all votes are equal. Your vote is not less likely to be counted than anyone else's vote. If you're doing something reckless like casting a provisional ballot then this equation vastly overstates your chances of impacting the election. If you're more conscientious than the average voter at using the voting equipment, though, the probability that your vote will make a difference in the outcome of the election is slightly larger than what you'd get from this formula.

In an ideal election, you candidate would win if x>0
and he'd lose if x-1. That is, f(x)=1 for all x>0 and f(x)=0 for all x-1. Since f(x)=f(x+1) for all values of x but two, the infinite sum over all values of x of p(x)[f(x+1)-f(x)] reduces to p(-1)[f(0)-f(-1)]+p(0)[f(1)-f(0)], which equals p(-1)[f(0)]+p(0)[1-f(0)]. Assuming that p(-1)=p(0) (that is, a one-vote loss for your candidate is just as likely as tie, not an unreasonable assumption given that your predictions of other voters' behavior can hardly make much of such fine-grained distinctions), this simplifies to p(0). So the probability that your vote will make the difference in the election is equal to the probability that the election would result in a tie without your vote, a rather intuitive result.

But what happens when things get messy? What if a candidate could win the election despite garnering fewer votes, because of the crazy electoral system? What if f(x) is not a neat step function, but some complicated, smeared out function (possibly biased, maybe even non-monotonic!)? My claim is that your impact on the election is just about the same, as long as there exist integers a and b, with b>a, such that

1. f(x) is approximately 0 when x ≤ a
2. f(x) is approximately 1 when x ≥ b
3. p(x) is approximately constant for all x where a≤x≤b

In other words, even if the election procedure is messy for close elections (where the differential between a and b is small), it is reliable once either candidate gets a large enough lead. And, the range of voting patterns where the unpredictability happens is narrow enough so that it is hard to say beforehand which of the voting patterns within that range are most likely.

More precisely, we can write these conditions in terms of three errors, e1, e2, and e3, all of which are approximately 0:

1. f(a) = e1
2. f(b) = 1-e2
3. p(min)/p = 1-e3, where p is the maximum value of p(x) for a≤x≤b and p(min) the minimum

Then the probability that your vote will change the outcome of the election is no less than p(1-e1-e2)(1-e3). If p is approximately p(0) and e1, e2, and e3 are small, then this is about the same as p(0). Usually, p will be larger than p(0), since p is the maximum of p(x) for a≤x≤b.

This formula can be derived by evaluating the infinite sum over all x of p(x)[f(x+1)-f(x)]. Since this value is always positive, the sum over all x is greater than or equal to the sum from a to b-1 of p(x)[f(x+1)-f(x)], which is greater than or equal to the sum from a to b-1 of p(min)[f(x+1)-f(x)], which equals p(min)[f(b)-f(a)], which equals p(1-e3)[1-e2-e1].

Why can we say that the errors, e1, e2, and e3, are close to 0? We can make e1 and e2 small with values of a and b that are relatively close to 0 as long as the problems with the election procedure only arise in close elections. In an election with a million voters, we can assume that, as long as one candidate has at least a few thousand votes in his favor, it is very unlikely that the other candidate will be able to steal the election away. Since we get our information about how other voters are likely to vote from polls with margins of error of plus/minus a couple of percentage points (which would give us a 95% confidence interval with a length of 50,000 in our election with a million voters), our knowledge is not fine-grained enough to distinguish the likelihood of values of x that are within a few thousand of each other, as a and b are. So p(x) will be approximately constant in that range, and p(min)/p will be close to 1.

Even if these conditions do not hold very well, your chances of impacting the election will still be close to p. If there is a lot of variation in p(x) between a and b, then (p)(1-e1-e2)(1-e3) is likely to significantly underestimate the true value of your impact, since we are using the minimum value of p(x) on that domain. Unless the procedure for determining the winning candidate was really wacky, it seems unlikely that your probable impact on the election would ever drop below something like three-fourths of p(0).

Given the uncertainty in everything else that you might want to measure (the benefit of making your candidate win, the cost of voting), the probability of a tie among the rest of the voters seems like a very good (and intuitive) estimate of the probability that your vote will be decisive in favor of your candidate. The precise procedure by which the election will be decided is much less important in influencing the rationality of voting than the other factors that determine your incentives. And there is no reason to feel disempowered when the outcome of the election is determined by judges, or bad voting machines, or ... . Angry, maybe, or outraged, or disgusted, or betrayed, or ashamed, or bemused, or frustrated, or vengeful. But not disempowered. As long as your vote was cast, it had just about as much of an expected impact as always. Not that its impact is very large.

UPDATE (11/11/05): Here are some more intuitive ways to think of this result, two from correspondents and one from me:

- In both ideal elections and real elections where judges can get involved, a few thousand votes will swing the election from one candidate to the other. The average impact of those few thousand voters is the same in both cases, and, since we can't predict vote totals precisely enough to distinguish possibilities that are within a few thousand votes of each other, the more fine-grained differences between the two cases are irrelevant. The expected impact of a single voter is the same in both cases.

- Any loss in marginal vote power attributable to "stolen elections" is counterbalanced by the gain in marginal vote power from (a) being able to push a potentially stolen election into the sure-win range and (b) being able to push a sure-loss into the potentially stolen range.

- What judges do is spread out the probability that your vote is the pivotal one, rather than concentrating it on one marginal guy.


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