In article <3A002FEC.4AEE86C3@echip.com>, Bob Wheeler <firstname.lastname@example.org> wrote:
>If p is the popular vote proportion and if p is >the same for all states, then the expected >electoral total is 540p with a variance of >540p(1-p). Assuming normality, this gives the >probability of erroneous electoral result as 0.32 >when p=0.51, 0.085 when p=0.52, and 0.03 when >p=0.53. Hence, a popular-electoral mismatch seems >unlikely when the popular vote exceeds 50% by two >or more points.
Unless I misunderstand your assumptions here, this is wrong. You seem to be saying that each state has probability p of a "correct" electoral vote. This would work if the state's electoral vote was decided by one randomly chosen voter, but that's not what happens.
The problem as I interpreted it had p = 0.5, i.e. each voter votes "randomly" with equal probabilities for the two candidates. And the conclusion is that the probability of a mismatch is about 0.2.
Robert Israel email@example.com Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2