Abstract
In order to assess this question, we can consider what
kinds of voting statistics would be generated by such a random model, and
then ascertain whether the observed patterns in the 2000 presidential election
differ significantly from the behavior of the random model.
f(X)=binomial(N,X) * PAX * PB(N-X)
where binomial(N,X) is the binomial coefficient defined by N!/(X! * (N-X)!) and N is the number of voters.
Note that this formulation is statistically equivalent to one in which the system consists of a randomly-generated population of deterministic voters, each of which reliably votes for one party or the other with 100% probability.
As is well-known, the mean of a binomial distribution is M=N*PA and the variance is V=N*PA*PB. The standard deviation, which is defined as the positive square root of the variance, would be S=sqrt(N*PA*PB), or sqrt(N)*P if PA=PB=P. Thus, if we were to hold many such elections (holding the parameters fixed), we would expect that the number of votes for candidate A would tend toward M, with deviations from M generally not exceeding 2*S in absolute magnitude.
In fact, the empirical rule tells us that for any mound-shaped
distribution (such as the binomial distribution), we can expect roughly
95% of all observations to fall within the interval [M-2*S,M+2*S].
Thus, any observation outside this interval suggests that the system under
study does not follow a binomial distribution with the given parameters,
so a sufficiently large deviation from the mean would indicate that the
real system was not simply casting votes uniformly at random.
For the national vote, with approximately 97,759,658 ballots divided between the two dominant parties, the appropriate binomial distribution would have a mean of N*P = 97,759,658 * 0.5 = 48,879,829 (assuming PA=PB=P) and a standard deviation of sqrt(N)*P = 9,887 * 0.5 = 4,943. The actual results, 48,783,510 Republican and 48,976,148 Democrat, deviate from the mean by 96,319, a number far greater than the standard deviation. The fact that this deviation is more than 19 times as large as the standard deviation makes this a highly statistically significant result, leading us to reject the null (random) model.
For the Florida vote, with 5,816,744 ballots divided between the two dominant parties, the appropriate binomial distribution would have a mean of N*P = 5,816,744 * 0.5 = 2,908,372 and a standard deviation of sqrt(N)*P = 2412 * 0.5 = 1206. The actual results, 2,909,260 Republican and 2,907,484 Democrat, deviate from the mean by 888 (producing a margin of 1,776), thereby falling soundly within one standard deviation and offering no objective evidence for rejecting the null model.
Given this interpretation, it would be rather alarming if the candidate favored by the national vote was defeated on the basis of the Florida count. It appears that this is precisely what will occur. If we consider that the small numerical advantage (888) of the one candidate could easily have been reversed through the many random contingencies involved in deciding not only how votes are cast but also how many and which people even cast their votes (i.e., the existence of traffic congestion, an overabundance of voters shortly before a polling station closes, etc.), it would seem that selecting a candidate based on such a close count, given the clear preference of the overall population for the competing candidate, would be highly undesirable behavior for any electoral system.
An alternate electoral system might account for these
considerations by rendering inadmissible counts from any state which did
not differ significantly from random, or more simply by selecting candidates
directly through the popular vote. In the case of a close count in
the popular vote, there does not seem to be any principled alternative
to random selection, so there would presumably be no harm in leaving the
matter to the whim of statistical fluctuations.