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 <G38p38.email@example.com>, > Peter L. Montgomery <Peter-Lawrence.Montgomery@cwi.nl> 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 firstname.lastname@example.org > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 >
-- Bob Wheeler --- (Reply to: email@example.com) ECHIP, Inc.