Isn’t the conventional “economic” argument that it’s not worthwhile for any individual to vote because they’ll almost never be the one vote that puts the better candidate over the line? Whereas this analysis indicates that (in many cases, but obviously depending on the distribution of q) no matter how many other people are voting, your participation actually increases the odds of a socially optimal outcome, even if every other voter is smarter than you?

If you tweak the formula to assume *everyone* has some definite, better than even chance of voting for what’s good for society, so that q = q1*(r^n)+c, where q1,n,c are constant and r is the ranking, no matter how small c is, you eventually get more voters improving the odds of selecting the socially optimal candidate again. For q1=0.1, n=1 and c=0.001 (thus giving a minimum probability of 50.05% of selecting the best candidate), the best voter has a 55.05% chance of getting the right outcome, then it’s downhill until you have 475 voters and a 52.10367% chance, then up-hill after that, getting back to 55.0503% at 14013 voters and continuing to increase thereafter. This is when only the top 100 voters have a better than 50.1% chance of picking the socially optimal candidate, everyone else varies between that and 50.05%, but at least they vary independently.

Self-interest probably works fine as a proxy for q too: the socially optimal candidate is socially optimal because he or she benefits lots of people, so they’re likely to benefit any particular individual — so if you have a signal q’ for a candidate that helps you, you probably have a signal q=q’/?? for the socially optimal candidate.

A simple solution would be to suppose that some voters have a less than 50 percent chance of being right. But I don’t think that would save this model, because it isn’t modeling the fact that politics is an adaptive system. Whatever the dispositions of the voters are, political parties adjust the alternatives offered until the expected vote is again split 50/50. Voters are polarized into 2 groups (in the US), which self-adjust to each claim about 50% of the voters. And if you believe that your group is right, then it is always rational to vote – even on issues you don’t understand, or for candidates you’ve never heard of.

Or possibly ‘Vote and Die!’

Also, when I spoke of using a log-normal distribution for IQ, this is misleading. While it may be true that you can fit some of the data better with a log-normal distribution, this log normal distribution will be so close to a normal distribution, that if you plotted it and showed it to a statistician, he would call it a normal distribution.

Skills usually have approximately normal distributions, because they are the combination of a large number of random factors. You can sometimes use a log-normal distribution to account for a skew caused when the distribtion is bounded below but not above.

Now, how does a skill, with a nearly normal distribution, map into a probability of voting correctly? Looking at our usual data, such as times in running the 100m dash, might be problematic because it isn’t clear whether to consider times bounded or unbounded.

So I plotted the 1273 scores in the Netflix competition. These scores are in root-mean square error of guessed movie ratings, and theoretically range from 0 to about 1.05 (what you get if you guess the average value for each rating). In practice, the lower bound can be approximated by having a person try to guess their own ratings for movies that they already rated in the past; based on one experiment, this lower bound is about .79.

This is a pretty good substitute for a probability, because the RMSE is closely related to the probability of guessing the right rating. (It is sqrt(9*p(off by 3) + 4*p(off by 2) + p(off by 1)).)

Sadly, I can’t post the picture of the plot here, but I can tell you what the histogram looks like. It occupies the range .864 to .951, with a mode of about .905. Most of the mass is between .90 and .93, with a rather sharp drop-off down towards .86. The end with the lowest-RMSE scores (corresponding to our highest-ranked voters) is nearly flat in the histogram. In other words, it looks vaguely normal, but heavy in the range .9 to .945. As with all skill-based scores, in this distribution, there are a few people who are very bad, and a few (fewer) who are very good, with most being in the middle.

This is in contrast to Robin’s proposed distribution, in which almost everyone is very bad. That distribution is extremely sharp; it looks like an L if you plot p(correct) vs. rank, for all parameters I have tried. That is not a realistic distribution. And it is that unrealistic distribution (and possibly some issue with how Robin handles variance) that leads directly to the conclusion that almost everyone shouldn’t vote.