Below where I wrote "99% confidence" I should have specified the
confidence interval:
If n= 10000,the the standard deviation is 100*SQRT(P*Q) which is less
than 50, as long as P and Q are positive numbers that sum to unity. Here
P is the probability of one candidate being the preferred of the majority of
the voters (not just the ten thousand randomly chosen to vote on this
pairwise contest), and Q is the probability of the other candidate being the
preferred.
A one sided 99 percent confidence interval
corresponds to roughly 2.3 standard deviations, which, in this case would
correspond to about 115 voters. So if A beat B by as few as 115 out of ten
thousand votes (i.e. by a little more than one percent of the vote) then we
could be 99 percent confident that the same result would hold if
the entire population of voters (from whom the 10000 had been randomly selected) had
voted in that pairwise contest.
I submit that this is greater confidence than we now have in
typical large races with that close of a result.
Forest
I had written ...
My idea is that in a large enough election, the
individual pairwise contests could be farmed out at random to the
voters.
...
To be specific, suppose that you had twenty single
winner races with ten candidates each, and no ballot measures.
Each of the ten candidate races could be broken down
into 45 pairwise contests, so the total number of pairwise contests would be
20*45=900.
If there were nine hundred thousand voters, and each
of them received a random selection of ten pairwise contests to weigh in on,
then each pairwise defeat would be based on ten thousand
ballots, well above the statistical sample size requirement for 99%
confidence.
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