# What is the probability your vote will make a difference?

Nate Silver, Aaron Edlin, and I estimated the probability that a single vote in any state will be decisive in the presidential election. It was Aaron’s idea, Nate supplied the simulations, and I calculated the probabilities and made the graphs.
Here’s our article describing what we did, here’s the abstract:

One of the motivations for voting is that one vote can make a difference. In a presidential election, the probability that your vote is decisive is equal to the probability that your state is necessary for an electoral college win, times the probability the vote in your state is tied in that event. We compute these probabilities for each state in the 2008 presidential election, using state-by-state election forecasts based on the latest polls. The states where a single vote is most likely to matter are New Mexico, Virginia, New Hampshire, and Colorado, where your vote has an approximate 1 in 10 million chance of determining the national election outcome. On average, a voter in America has a 1 in 60 million chance of being decisive in the presidential election.

and here are some graphs:

There are more graphs if you follow the link to the article.

As Aaron, Noah, and I have discussed, it can be rational to vote even when the probability of decisive vote is 1 in 10 million.

P.S. Typo in Figure 1 caption above fixed.

P.P.S. See here for more discussion of why we can compute the probability of a decisive vote, even though the election might be decided by a recount

## 20 thoughts on “What is the probability your vote will make a difference?”

1. Actually, the chance that your vote would be decisive is zero. Even if the “true” vote were exactly 50/50, limitations in verifying such the count would never allow for a single vote to be decisive. Recount after recount would ensue, and none would be definitive to that degree of accuracy–not to mention the legal battles that would ensue.

2. This is kind of an obnoxious question since I haven’t really taken the time to understand your methodology but…

If I use a simple indpendent binomial model for voting in California, starting with a uniform distribution on the percentage of Obama voters and then conditioning on the last four poll results (giving McCain everyone who didn’t respond for Obama), I seem to be getting the probability of a tied vote in California as around 10^-19. This is a bit lower than the roughly 10^-8 you have in your paper. Obviously there might be some problems with the independent binomial model but I think I was fairly generous in my calculation in other ways (e.g. the uniform prior and giving McCain all undecided/third-party votes). Is there a simple explanation for why I should believe that the probability of a tied election in California is closer to 10^-8/10^-9 than to 10^-19?

3. Again no breakdown of Nebraska by Congressional districts.

Omaha – the only city in the USA with its own electoral vote, and a very even split between Obama and McCain, should certainly be up near the top of this chart.

We’ll split NE’s vote and bring Electoral College reform to the rest of America.

4. Abe: See recent blog entry for an explanation.

Nick: Read the linked article. Our calculations come directly from the uncertainty distributions derived from Nate’s simulations. The binomial distribution is irrelevant to the question.

Tom0063: Well, we do mention Nebraska and Maine in a footnote . . .

5. Andrew: I had read the paper, though not as thoroughly as I perhaps should have. Clearly you didn’t actually use the independent binomial model but I didn’t understand why it should give such wildly different results. I’m guessing that the answer to that, though, is wrapped up inside the simulations which aren’t described in this paper, hence the lack of enlightenment on my part. (Reading around a bit, it looks to me like perhaps a lot of the difference comes down to putting too much confidence in the polls.)