This has been posted as a reply in another thread.
Go here for state Z-scores and corresponding probabilities:
http://www.democraticunderground.com/discuss/duboard.php?az=view_all&address=203x361617
Prior analysis has focused on the probability that 16 states
would exceed the margin of error for Bush, using the standard
95% (actually one-tail 97.5%) confidence level for polling).
This analysis focuses on the Z-score, which is the number of
standard deviations that the actual vote is displaced from the
polling mean, based on the normal distribution (bell-curve).
The Z-score exceeded one (1) standard deviation (a 68.26%
confidence level, 84.13% for one-tail) in 35 states for Bush
and just 5 states for Kerry.
Calculating the probabilities for each occurrence:
Prob (Bush Z >1 in 35 states): 1/4500 trillion (?);
Prob (Kerry Z >1 in 5 states): 91% or 9 out of 10 (very
plausible)
To calculate the probability of the Z-score, we use the Excel
function:
Prob P = 1-NORMSDIST(Z),
where Z = the number of standard deviations, as above.
To calculate the probability of the Z-score exceeding 1 in at
least N states, we use the Excel function:
Prob P(N>Z) =1-BINOMDIST(N-1,50,1/P,TRUE)
Prob Favor N Z Pr(Z) Pr(N>=Z)
15.866% Bush 35 1.00 1 in 6 1 in 4500 trillion
15.866% Kerry 5 1.00 1 in 6 90.9%
Probabilities for other Z-score levels:
1.390% Bush 16 2.20 72 1 in 160,000 trillion
0.466% Bush 10 2.60 215 23 trillion
0.040% Bush 4 3.35 2475 165 million
0.009% Bush 3 3.75 11307 74 million
0.002% Bush 2 4.10 48382 1.9 million
Using confidence levels of 95% (as in the previous analysis)
and 99% for the probability calculations, we have:
95% Confidence (Z=1.96):
2.50% Bush 16 1.96 40 1 in 19 trillion
99% Confidence (Z=2.58):
0.50% Bush 10 2.58 202 1 in 13 trillion