Numbers Game

Ahhhh, the binomial distribution.

Let’s take the following values:

Bush: 2,794,346
Kerry: 2,658,125

Provisional Ballots: ~175,000

Now, if we make this a two-person race—hey, we’re approximating provisional ballots, so we can approximate the race as a two-person race—we can figure out Kerry’s odds of winning with enough provisional ballots as a binomial distribution.

binomial distribution function

Bush has a delta of 136,221 votes, meaning that Kerry would need 136,222 votes more than Bush amongst the ~175,000 provisional votes. This means that Kerry would need 155,612 votes of 175,000 remaining to be counted—if we again assume that every provisional will be counted. [The nature of provisionals is that some are going to be tossed out. If ~20,000 are tossed out, it’s game over. That may be a reasonable assumption alone, but I’ve been thinking about this entry all night, and I’m not stopping now.]

So, the P(155,612) in this case has the following conditions:

N = 175,000
r = 155,612
pi = … uhhh …

There are two ways you could look at pi. One is that pi should be approximated from the present totals, which assumes that the millions of votes we have now reflect the population as a whole, and what we have is a random sampling from that pool. The other approximation of pi could be 0.5, since one over an exact tie is what Sen. Kerry needs to be Pres. Kerry. Hell, let’s do both.

P(155,612) where pi = 0.49 [Kerry’s current percentage of the vote]: a number neither my calculator nor Excel will process.

Aha! Vassar to the rescue! Their calculator says: <0.000001, or in other words, less than 0.0001%.

Wow, I think my chances of being President are better.

[No point in running the other pi, I don’t guess.]