# space and games

## February 5, 2007

### More on Hay Voting

Filed under: Voting — Peter de Blanc @ 10:36 pm

What has probably generated the most confusion about Hay Voting is that the original article does not give a formula for the amount of voting mass allocated to each candidate as a function of your utility function. I will derive such a formula here.

First of all, as indicated in the original article, each candidate starts out with √(n−1) units of voting mass on your ballot. The total amount of voting mass does not change; it is only reallocated. If we indicate the final mass allocated to a candidate i by Mi then the probability allocated to i is P(i) = Mi / n√(n−1).

Now, to transfer voting mass from a candidate i to a candidate k (or vice versa), you spend some voting credits. The amount that you transfer is equal to the square root of the amount of credit you spend. To put it another way, the amount you must spend is proportional to the square of the amount of mass you wish to transfer. Your utility density for transfers of mass from i to k is simply the difference in utility between the two candidates: Uk − Ui. As described in “The n-Substance Problem”, your optimal strategy is to transfer an amount of mass proportional to this utility density. So we’ll write ΔMi,k = cUkcUi, where c is some constant (which we will now solve for).

The amount of voting credit you spent on this transfer would be c2 (Uk − Ui)2. We transfer voting mass between each pair of candidates, so the total amount of credit spent is Σ{i, k} c2 (Uk − Ui)2 = 1, since you have one credit. Solving for c, we get c = 1 / √(Σ{i, k} (Uk − Ui)2).

The total mass allocated to candidate i is simply Mi = √(n−1) − ΣkΔMi,k. That is, the initial mass minus the amount transfered to each candidate. Note that the amount transfered can be negative, indicated that mass was transfered to candidate i from somebody else.

So our formulae are:

1. P(i) = Mi / n√(n−1), where
2. Mi = √(n−1) − ΣkΔMi,k, where
3. ΔMi,k = cUkcUi, where
4. c = 1 / √(Σ{j, k} (Uk − Uj)2)

Combining all the above formulae, we then have:

P(i) = Mi / n√(n−1)
= (√(n−1) − ΣkΔMi,k) / n√(n−1)
= (√(n−1) − cΣk(Uk − Ui)) / n√(n−1)
= (√(n−1) − Σk(Uk − Ui) / √(Σ{j, k} (Uk – Uj)2)) / n√(n−1)

Now why did I leave something this important out of the original article?

## 4 Comments »

1. Can you change the background and text to something that is more readable?

As far as this sight-impaired person can tell, your article doesn’t have anything in it.

Comment by JPK — February 6, 2007 @ 12:37 am

2. In my experience, light gray on black is very readable. If you’re in Firefox, you can hit ctrl-+ to make the text larger.

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