The Flathead Valley’s Leading Independent Journal of Observation, Analysis, & Opinion

6 November 2008

Will yet to be counted provisional ballots defeat the 911/OES bond?

Updated. It’s possible. Uncounted provisional ballots number approximately 400, while there are just 206 more Yes than No votes for the bond. But don’t count on it. I ran the numbers last night:

If we assume an under vote (elections professionals jargon for a contest in which a voter failed to make a choice) rate of approximately 7.7 percent, there are some 370 provisional votes to be allocated. The most likely outcome, significant at the .95 confidence level, is an even split plus or minus 20 votes. A reversal of electoral fortune for the 911/OES bond is as likely as an asteroid’s taking out Glacier High School at high noon on Thanksgiving.

The legislative district case. The probabilities are are almost as daunting on a smaller scale. Assume, for the sake of argument, that in a legislative district, Jones receives 2,010 votes to Smith’s 1,990, with 50 provisional votes remaining to be counted. If Jones gets 16 or more of the 50 provisional votes, he wins. What is the probability that Jones will get at least 16 of the provisional votes?

Updated paragraph. Students of statistics will recognize this problem as a classic example of the binominal distribution (highly technical discussion). Each vote is an independent event. The probability that a ballot records a vote for Jones (or Smith) is, given the closeness of the count thus far, 50 percent. The situation is identical to the classic coin toss problem taught in Statistics 101. After doing the math, we find there is a 99.2 percent chance that Jones will get the 16 or more votes he needs to win. You can try the calculation yourself at Stat Trek.

(If absolute statistical rigor is applied to this problem, the circumstances do not meet the four criteria necessary for the binomial distribution. Strictly speaking, the provisional ballots are not a random sample of the population of voters. They're a self-selected sample that may be skewed. But it’s the only sample we have, and common sense suggests it probably approximates a random sample. We therefore should perform the calculation, keeping in mind the constraints on precision imposed by the real world, and compare our calculated results to the actual count of the votes.)

We can also calculate the probability that the final margin will be more than ten votes, thereby avoiding one trigger for an automatic recount. Jones needs at least 21 of the provisional votes for that outcome. There’s an 83.9 percent chance that Jones will get at least 21 of the provisional votes.

Objective observers will conclude that there are pretty damned good odds that Jones with both win and not trigger the 10-vote-margin recount trigger. Human nature being what it is, Jones will, of course, conclude that he should stick with Valium until all the votes are counted.

Calculating the probability that a recount will produce an electoral reversal of fortune is trickier. A margin of 20 votes seems tantalizingly close, but unless a systemic error is skewing the outcome in only one direction, the most likely result is that errors — if any — will be distributed randomly in no one’s favor, and that the outcome will not change.

Candidates seeking a recount, of course, should demand a manual recount. That provides independent verification of the machine count, and avoids repeating counting machine errors (machines that count votes save time, but the trade-off is reduced accuracy compared to a well designed and conducted manual count).