# Why your ballot isn't meaningless.

A mathematician's guide to the news.
Oct. 11 2004 1:29 PM

# Vote!

## Why your ballot isn't as meaningless as you think.

Let's say, for the sake of argument, you were among the 72,000 people who participated in the Guinness-certified world's largest chicken dance in Canfield, Ohio, in 1996. You probably feel pretty proud. But according to Slate's Stephen E. Landsburg, you shouldn't. After all, unless a previous chicken dance for 71,999 were on the books, your participation made no difference; the record would have fallen whether or not you'd shown up.

Landsburg is arguing against voting, not chicken-dancing: Your presidential vote, he says, "will never matter unless the election in your state is within one vote of a dead-even tie." That, of course, is extremely unlikely. So, the negligible chance of casting the deciding ballot is outweighed by the small but certain costs of voting, like the gas you'll use and the time you'll spend.

And yet people vote anyway, by the millions. Political scientists call this conundrum "the paradox of voting," and you could stay up half the night (I just did) reading research literature on the subject. Why do people vote when it's so unlikely to matter? Maybe because the pleasurable feeling of doing one's duty offsets the cost of gas. Maybe because people have an interest in their candidate not just winning but winning by as large a margin as possible. Maybe because we're motivated to avoid even small possibilities of regret—the regret that those Al Gore supporters who sat out Florida in 2000 surely feel, whether economists think they're being rational or not.

But let's stick to mathematics. Suppose we grant to Landsburg that voting carries a certain cost and that your vote should be considered worthwhile only if it decides the election. Everyone can agree that's unlikely—but how unlikely? Landsburg first proposes
modeling voters in a state, say Florida, as 6,000,000 coin-flippers, each choosing George Bush with some probability p and John Kerry with probability 1-p. For instance, if p is 1/2, 1-p is also 1/2; each voter has an equal chance of selecting Bush or Kerry. As you might expect, the odds of a tied outcome are not bad—about 1 in 3,100, as Landsburg computes.

But p might not be 1/2, and even a tiny bias in voter preference can make a tie exceedingly unlikely. For instance, if p = .51, the chance of a tie drops to 1 in 101046, a probability so small as to be effectively zero. (Here's Landsburg's computation.) Your vote is not going to count.

So, are we back to Landsburg's discouraging conclusion that voting is most often a waste of time? Not quite, because it's impossible to know in advance what proportion of your fellow Floridians are planning to vote for Bush. If you knew p was exactly 1/2, you'd be sure to get out and vote. If you knew Bush held a 51 percent advantage, you'd be foolish to bother. But you don't know, and without that knowledge you can't reason as Landsburg wants you to.

You don't know, but you can guess. A Sept. 29 poll of 704 Florida voters by CNN/USA has Bush leading Kerry 52-43. For simplicity, let's dump the still-undecideds and third-party enthusiasts and say that, among 669 randomly selected likely Florida voters, 366 supported Bush and 303 Kerry, a 55-45 margin in Bush's favor. If forced to make a guess, we might expect 55 percent of Florida voters to favor Bush. But how confident should we be that our guess is right? In particular, how likely is it that the real proportion of Bush votes in the state is very close to 50 percent?