Good for you that it now works. It did seem strange that Mathematica couldn’t compute large values of n; the Pa(n) is just the expansion of (p + 1 - p)^ n minus the first ((n+1)/2)-1 terms, so Mathematica should be able to convert the formula to one that uses the binomial coefficients, which should be easily computable for very large values of n. (Pa(n)=SUM [upper bound = N, lower bound = (N+1)/2] BinomialCoefficien(N,i) * (.51^i) * (.49^(N-i)) I was just about to try to simplify it again using various identities for the binomial coefficients, but I think I will now defer to you and try to get back to my own work.

See also here.

I believe I have figured it out. For those keeping score at home, if you enter the summation formula with variables into Mathematica, it automatically converts it into a formula that uses the hypergeometric and gamma functions. It can then compute individual probabilities at N for high numbers of N, such as N= 100,001. I’ve checked this converted function against a bunch of known numbers, and it’s right.

If this is correct, then the marginal value of the 100,003rd voter in terms of his contribution to accuracy is 2.6×10^-14. In other words, he’s adding a trillionth of a percent of accuracy.

So, if the election is worth $10 trillion, his contribution is worth $26.

So, here’s the conclusion. If you defend democracy using Condorcet’s Jury Theorem, you should think mass elections are tremendously wasteful. Even if individual voters have a low probability of being right (51%), and even if the election is high stakes ($10 trillion difference between good and bad candidate) you only need about 100,000 voters before the marginal impact of additional voters becomes quite small ($26 bucks). When you’re around 100,000 voters, it’s a waste of their time to them vote, and they hardly do any good by voting. In contrast, the 1000th voter does millions of dollars worth of good.

The marginal value of the 25,001th voter is a 1.7 x 10-5 % increase in accuracy. So, if the right answer in election is worth $10 trillion more than the wrong answer, the 25,001th voter contributes $170 million worth of accuracy.

Let’s see if it can handle bigger numbers if I let it calculate for an hour.

Thanks for all your input. You’re right that I typed the formula in wrong.

What I’ve been doing for Pa, since it’s defined for odd numbers, is calculating Pa at N and N+2, taking the difference, and then dividing that by two. This slightly overstates the marginal contribution of the N+2th voter, since there are diminishing returns. Since I expect the contributions to become quite small, having an upper bound for them will do the work I need.

The idea of asking Mathematica to simplify Pa(N+2)-Pa(N) is a good one. I’ll try that when I get to work today.

I’ve got Young’s 1988 APSR paper as well as Black’s book and a few other sources. As far as I can tell, their proofs show that Pa has a limit of 1 at N–>infinity, but they don’t provide a different way to calculate the rate of change of Pa. Stink.

I’ve noticed that people who work on this are usually calculating Pa where N is 10,000 or less. So, I suspect their in the same predicament.

Should be

SUM [upper bound = N, lower bound = (N+1)/2] (N!/(N-i!)i!) * (.51^i) * (.49^(N-i))

No?

No, you didn’t misunderstand me. But I was using a quick back of the envelope pen-and-paper simplification, so I might be wholly wrong. I’ll recheck.

Have you tried defining a function Pa(N) and asking Mathematica to simplify the expression Pa(N+1)-Pa(N) and assign it to a function? Get Mathematica to do the symbolic manipulation for you, instead of calculating the actual value of Pa(N+1)-Pa(N)? Mathematica is usually good at coming up with simplifications of symbolic expressions. I’m away from a computer with actual symbolic manipulation capabilities, so I can’t check (does Wolfram Alpha serve as a full front end for Mathematica, by the way?). I haven’t used this stuff in a while, too, so take any of my suggestions with a grain of salt!

Another possibility that occurs to me is to look at the proof of the Theorem (presumably in Young 1988) to see how it shows that Pa(N) tends to 1 as N gets large; it may be that there is some obvious way of bounding Pa(N). I remember having read the proof some time ago, and I don’t remember it being especially difficult to understand, but unfortunately it’s late here and I can’t remember how it’s done.

I might be misunderstanding you, but I think your suggestions is that Pa(N) and Pa(N-1) have a bunch of common terms, so I can just cancel those terms out and calculate the remaining terms that they don’t have in common. However, as far as I can tell, Pa(N) and Pa(N-1) don’t have any common terms, though they have a bunch of terms that are pretty close.

For instance, the first term in Pa(101) is (101!/(101-51)!51!)(.51^51)(.49^(101-51)), while the first term in Pa(102) is (102!/(102-51.5)!51.5!)(.49^(102-51.5)). [Technically, the function Pa is supposed to be for odd sizes of N, but it more or less works for even jury sizes, too.]

However, I suspect I’m misunderstanding you. Summation functions were not my strong point back when I used to do more mathematics.

You’re right that I don’t need to know the accuracy of the jury/electorate for large numbers of N. I just need to know the difference in accuracy between a jury of size N and size N-1. I have tried asking Mathematic to calculate Pa(N)-Pa(N-1) for large values of N (such as N=10,001 or N=50,001), hoping that it had some algorithm for finding the difference without first calculating Pa(10,001) and Pa(10,000). Unfortunately, it times out. So does Wolfram Alpha online.

You can approximate the summation function Pa using indefinite integrals. Mathematica and other math programs can calculate Pa at higher values of N when you do that. However, they still can’t handle N> about 10,000.