Thursday, November 09, 2006

Counting votes

As Howard Rodman points out, the media that called Florida for Bush in 2000 refused to call the Virginia race, even though Webb's lead (7000) was ten times what Bush had. And Allen has refused to concede. He wants a recount.

What is the likely error in this vote? There were 2.3 million voters. Lets say voters were equally likely to vote for either candidate, and every vote was a possible error. What would be the expected difference in votes? About 1500. (This is the standard deviation of a binomial distribution.) Even with these extreme assumptions, the margin (7000) is nearly five standard deviations away. The probability of getting that by chance is vanishingly small.

And, in fact, voters did not vote randomly. Very few votes (lets say 5%, to be generous?) would have been cast wrongly. Very few (let's say another 5%) would have been counted wrongly. That makes 115000. If these votes are assumed to be random, and equally distributed between the candidates (a fair assumption given the narrow margin of victory), the expected error is about 340. Webb's margin of victory is 20 standard deviations away.

Only if the miscounted votes were skewed in their distribution (because of voter intimidation, fraud, whatever), and only if there were many, many such miscounted votes, would the possible error begin to approach Webb's margin. I don't expect "Macaca" Allen, who wants a recount, to understand statistics (and if there was any voter intimidation, Allen probably already benefited). But what about the craven TV channels?


Tabula Rasa said...

craven is exactly right. i can't believe people are sitting around and not saying "let's move on" like they did in florida.

Rahul said...

I just love this line from the NYT: "The Associated Press, a widely accepted authority for calling elections... (declared) Mr. Allen, a Republican, the loser." The NYT doesn't have the courage to do their own calling, so AP to the rescue...