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If the conditions of the Condorcet Jury Theorem hold, then every additional jurist/voter adds some marginal amount of accuracy to the jury as a whole. However, this jury experiences diminishing marginal returns. If every juror has a 51% chance of being accurate, then the jury of 101 members has about a 57% chance of being accurate, a jury of 501 members has a 67% chance of being accurate, a jury of 1001 members has a 73% chance of being accurate, a jury of 5001 members has 92% chance of being accurate, and a jury of 10,000 members has a 99.99% chance of being accurate.I’d like to know what the marginal value (in terms of her contribution to accuracy of the jury) of the Nth voter is when N is rather large.
The accuracy of a jury of N members when each juror has a 51% chance of being accurate is given by the formula below:
Pa (probability the jury is accurate) = SUM [upper bound = N, lower bound = (N=1)/2] (N!/(N-i!)i!) * (.51^i) * (.49^(N-i))
Since that’s likely to be unclear, here’s a link to a nice print out of the formula:http://books.google.com/books?id=CdIOKZWc3oMC&lpg=PP1&dq=public%20choice%20iii&pg=PA129
It’s easy to calculate Pa using Mathematica for values where N < 6500. After that, Mathematica and other programs can’t handle it. So, what I’d ideally like to do is find some program that can calculate Pa for higher values of N, such as N=50,000, N=500,000, N= 1,000,000, etc.
Alternatively, if there is some way to find the first derivative of this function, that might be helpful as well. Does anyone know how to do this?
What I’d really like to know is what the optimal number of jurors/voters is when the conditions of the Condorcet Jury Theorem obtain. Even tiny increases in accuracy can have significant value if the value of being accurate is high enough. So, for example, the marginal value of the 5007th voters is about 0.002%. But if picking the better candidate is worth, let’s say, $1 trillion dollars, then the expected value of that vote is quite high. But what’s the marginal impact of the 50,000th vote? The 100,000th? The millionth?I’m wondering if you think that democracies are adequately modeled by the Condorcet Jury Theorem (you shouldn’t, by the way), what’s the optimum number of voters? Let’s say that the net value of being accurate is $1 trillion, and that adding additional voters is suboptimal once the marginal value of a vote goes below $1. In that case, the optimal number of jurors (N) is given by the formula:1,000,000,000,000 * [Pa(N+1) - Pa(N)] = 1
Alas, despite trying many things over the past week or so, I have no idea how to solve this without a supercomputer.
Another alternative would be to find some upper bound and prove the the actual number is below this (already low) upper bound. But I’ve been unsuccessful at that.
Yet another alternative is to calculate the real marginal value of votes at a bunch of Ns that Mathematica can handle, then run some regressions to find a function that models the marginal value well, and use that as substitute. I’ve done that with a few different functions, but the problem is that these functions are of questionable accuracy for high values of N.Any ideas?
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14 comments
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1 - Thursday, 25 June 2009 at 6:04 pm
Jason Brennan
By the way, in case it occurs to anyone:
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.
2 - Friday, 26 June 2009 at 4:34 am
Xavier Marquez
It looks to me that you only need to estimate Pa(N) - Pa(N-1), which is much easier to find. Try asking Mathematica to simplify the expression Pa(N)-Pa(N-1). Most of the terms in the sum cancel out, leaving only one term with factorials in it. I’m using pen and paper here, and I’m a bit rusty, but it looks to me that this should be something like p^n - ((n-1)!/((n/2)!)^2))*(p^(n/2))*((1-p)^(n/2 - 1))) for n even (check parenthesis - this is also a very quick pen and paper simplification, which may be wrong), which should be easy to find for very large values of n, even with the factorial terms. I’d be interested to know what you find!
3 - Friday, 26 June 2009 at 4:38 am
Xavier Marquez
Pa(N+1)-Pa(N) being precisely the marginal contribution of the additional voter to the accuracy of the jury, if I’m not mistaken - that’s the function you need, not the actual accuracy of the jury Pa(N).
4 - Friday, 26 June 2009 at 4:52 am
Jason Brennan
Xavier,
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.
5 - Friday, 26 June 2009 at 4:52 am
Jason Brennan
P.S: Thanks for your help! I appreciate that you’re taking a stab at this!
6 - Friday, 26 June 2009 at 6:35 am
Xavier Marquez
Jason,
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.
7 - Friday, 26 June 2009 at 6:42 am
Xavier Marquez
Also, it looks to me that your Mathematica formula SUM [upper bound = N, lower bound = (N=1)/2] (N!/(N-i!)i!) * (.51^i) * (.49^(N-i))
Should be
SUM [upper bound = N, lower bound = (N+1)/2] (N!/(N-i!)i!) * (.51^i) * (.49^(N-i))
No?
8 - Friday, 26 June 2009 at 6:46 am
Xavier Marquez
Sorry to keep adding what is probably only error (Pa is tending to zero here), but what we need is actually Pa(N+2)-Pa(N), since Pa(N) is only defined for N odd.
9 - Friday, 26 June 2009 at 12:37 pm
Jason Brennan
Hi Xavier,
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.
10 - Friday, 26 June 2009 at 3:35 pm
Jason Brennan
Mathematica can simplify Pa(N+2)-Pa(N). Its output is a complicated gamma function. Alas, it doesn’t seem to be able to calculate this at high values of N either… But I’m trying a few tricks to see if I can coax it into working.
11 - Friday, 26 June 2009 at 3:46 pm
Jason Brennan
I’m letting Mathematica crank away in the background.
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.
12 - Friday, 26 June 2009 at 4:03 pm
Jason Brennan
Oops, I meant 1.7 million not 170 million.
13 - Friday, 26 June 2009 at 8:02 pm
Jason Brennan
Sorry for the bad grammar earlier.
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.
14 - Friday, 26 June 2009 at 10:27 pm
Xavier Marquez
Jason,
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.