[EM] typo

[MIKE OSSIPOFF]

To correct what I just said in my "correction" message, when
individual frontrunner probabilities are judged from the win
probabilities, A is 95/2 times as likely to be one of the 2 frontrunners
as B is.

If the win probabilities are A: .9, B:.04, C: .04, and D: .01,
those don't add up to 1. Of course the sum could be more than 1 in
a small election where ties are possible, but in a public election
that wouldn't be so. But, using those win probabilities, and calculating
for D the way I did for A, B, & C, it's D: .1

Those are just numbers estimated to be proportional to the candidates'
individual frontrunner probabilities.

If we multiply those numbers together for each pair, we get numbers
that are estimates of numbers proportional to each pair's frontrunner
probability. If we then add those up, and then divide each pair's
product by that sum, we get estimates for each pair's frontrunner
probability.

Multiplying, for each pair, the numbers calculated for the candidates:

A & B: .95 X .2 = .19
A & C: .95 X .2 = .19
A & D: .95 X .1 = .095
B & D: .2  X .1 = .02
C & D: .2  X .1 = .02
B & C: .2  X .2 = .04

These add up to .555

So the various pairs' frontrunner probabilities can be estimated by
dividing that pair's product by .555

Mike Ossipoff

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