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Topic: Electoral college probablities
Replies: 9   Last Post: Nov 2, 2000 6:02 PM

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Bob Wheeler

Posts: 348
Registered: 12/7/04
Re: Electoral college probablities
Posted: Nov 1, 2000 9:59 AM
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Interesting, but rather complicated.

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.

Robert Israel wrote:
> In article <>,
> Peter L. Montgomery <> wrote:

> > In the US presidential election, the winner is determined
> >by electoral vote rather than by popular vote.
> >What is the chance that the two will give different outcomes
> >in an otherwise close race?

> > Make simplifying assumptions, such as an odd number
> >of equi-populous states.
> >In a two-candidate race, voters everywhere vote randomly
> >for one of the two candidates.
> >Every state has the same (odd) number of voters,
> >and the same number of electors --
> >the winning candidate within each state gets all of that state's electors.
> >What is the probability that the candidate winning
> >a majority of the overall vote will lose in a majority of the states?

> An interesting problem. Let's say there are S states, each with N voters,
> and S is fairly large. Let X_i be the (signed) difference between
> candidate 1 and candidate 2 in state #i, and E_i the event that candidate
> 1 wins state #i. Let M be the signed difference in popular vote, D
> the signed difference in states, and W the event that
> candidate 1 wins more states.
> Then E[X_i | E_i] = m = (N+1) (N choose (N+1)/2)/2^N.
> Of course E[X_i | not E_i] = -m, and
> E[X_i^2 | E_i] = E[X_i^2 | not E_i] = N.
> E[M | D] = ((D+N)/2) m + ((N-D)/2) (-m) = D m.
> E[D | W] = s = (S+1) (S choose (S+1)/2)/2^S
> (the same formula as for m), and
> E[D^2 | W] = E[D^2 | not W] = E[D^2] = S.
> Thus E[M | W] = m s
> E[M^2 | W] = E[M^2 | not W] = E[M^2] = SN
> Assuming S is large, we can probably use a normal approximation, so
> given W, M is approximately normal with mean m s and variance SN - (ms)^2.
> So the probability that the winning candidate loses in the popular vote
> is approximately F(-m s/sqrt(SN-(ms)^2)), where F is the standard normal
> cdf.
> For example, with S=51 and N=10^6 we would have s = 5.726033806
> and m = 797.8847608, and F(-m s/sqrt(SN-(ms)^2)) = F(-.8323716341)
> = .2025996041.
> Asymptotically, m = sqrt(2 N/Pi) and s = sqrt(2 S/Pi), so
> -m s/sqrt(SN - (ms)^2) -> -2/sqrt(Pi^2-4) = -.8255161606.
> Robert Israel
> Department of Mathematics
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2

Bob Wheeler --- (Reply to:

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