I'll walk you through the calculations below. Let me apologize in advance for not breaking down each step in detail, but typing equations on a computer is a bitch. I assure you that I have not cheated and that my calculations are accurate, but if you really don't believe me, I can send you a pdf with more detail.My reply, and a further expansion of this topic, later today...
Anyway, in order for a single person's vote to "count" in an election (i.e. influence the outcome), that person's vote must break an exact tie. The probability of an exact tie can be calculated using the binomial distribution formula:
Probability of exactly n = x = [N! / ((n!)(N-n)!)] * [(p^n)(1-p)^(N-n)]
where N = the total number of people voting
n = the number people voting for a candidate Joe Blow
p = the probability of a someone voting for a candidate Joe Blow
In the case of an tied election, N = 2n and N - n = n
In which case the formula simplifies to:
x = [(2n)! / (n!)(n!)] * [((p)^n )(1-p)^n )]
Since factorials for large n cannot be calculated numerically by any normal calculator, use the Stirling's approximation for n!:
n! ~ ((2*pi*n)^.5) * (n/e)^n , which is extremely accurate for large n.
Plugging that in yields:
x = [(4(p - p^2))^n] / [(pi*n)^.5] (Eq. 1)
Now if the probability of someone voting for Joe Blow is exactly p = .5, then the formula simplifies to:
x = 1 / [(pi*n)^.5] (Eq. 2)
So let's say that you live in a relatively small state in which 1,000,000 people vote and the probability of someone voting for Joe Blow is exactly p = .5. In that case, the probability of an exact tie is about 1/1,253.
That seems like a pretty decent chance, right? Yes, but unfortunately it only holds true if p is exactly .5. If p varies even slightly, the probability of a tie drops dramatically. For example, if p = .51, the probability of a tie is about 1x10^(-90). (Note: I calculated that number by taking the log of both sides of Eq.1 before plugging in the numbers, which takes advantage of the fact that log (a^b) = b * log (a), and avoids the problem of trying to calculate something to the 500,000th power.)
The situation is even worse if you live in a large state like NY (number of voters ~ 7,500,000) in which the population strongly favors one candidate (I'll be conservative and say p = .55). In that case, the probability of a tie is about 1x10^(-16371), or in layman's terms, a decimal followed by 16371 zeros!
And the above calculations have not even taken into account (a) the probability of a particular state's electoral votes being decisive or (b) the probability, in the event of a very close election, of vote recounters or the Supreme Court screwing with the numbers and rendering your decisive vote not so decisive.
Wednesday, October 15, 2008
Not surprisingly, my argument for voting for Obama -- or perhaps, simply voting -- has prompted some replies. Here's one that specifically goes over the math for why it is essentially idiotic to ever vote:
at 6:37 AM