# Notes on a randomized variant of majority judgement

I like randomized voting. I like majority judgement. Naturally I’ve wondered for a while whether there is a natural randomized variant of majority judgement.

One option is to do this just by resampling majority judgement, but this doesn’t seem like a good option. You have to decide how many people to sample, and if it’s all of them then the answer is just going to asymptotically approach deterministic majority judgement.

I came up with a solution recently that I think holds a certain amount of appeal as a design, but after some analysis I think it doesn’t hold much appeal as a voting system. These are some notes on it for posterity.

The idea is as follows: People vote as in majority judgement, assigning a grades to each candidate.

We then  run the following process:

1. If there is only one remaining candidate, they win
2. If there are multiple remaining candidates but they all have exactly the same grades, pick one at random
3. From each candidate, independently pick a voter and assign that candidate the grade that voter gave them (so if 90% of people ranked a candidate excellent we give them the grade excellent with 90% probability). Of the grades assigned this way, pick the highest. Any candidate which got less than that grade drops out.

When I came up with this I thought it was obvious and natural and clearly the randomized way to do MJ. In reality it doesn’t track the behaviour of MJ well at all.

In particular it’s very easy to show that the relation “A beats B in a two candidate race” is non-transitive. How? You just construct grades which constitute non-transitive dice.

Further, as a result of this, it lacks the property of majority judgement that adding a new candidate to the list doesn’t affect whether A beats B (because you can add a candidate which knocks A out of the race earlier than B).

It also doesn’t track random-ballot very well I think. Consider the case with three candidates, A, B and C and three grades. Suppose candidate A has 50% good and B and C have 25% each. In random ballot that would also be the probability of them winning. In this system:

The possible outcomes for the second round are {A}, {B}, {C}, {A, B}, {B, C}, {A, C}, {A, B, C}. Let WA be the event that A wins and let $$p = P(WA)$$. Then:

\begin{align*} P(WA|{A}) &= 1 \\ P(WA|{B}) &= P(WA|{C}) = P(WA|{B, C})= 0\\ P(WA|{A, B}) &= \frac{3}{4}\\ P(WA|{A, C}) &= \frac{3}{4}\\ P(WA|{A, B, C}) &= p \\ \end{align*}

So
\begin{align*} p &= \frac{1}{2}*\frac{3}{4}*\frac{3}{4} \\ & + \frac{3}{4}(2 * \frac{1}{2} * \frac{1}{4} * \frac{3}{4}) \\ & + p ( \frac{1}{2} * \frac{1}{4} * \frac{1}{4} + \frac{1}{2} * \frac{3}{4} * \frac{3}{4})\\ p &= \frac{9}{16} + \fraction{5}{16}p\\ p &= \frac{9}{11} \\ \end{align*}

I made multiple arithmetic mistakes in doing that, so although I’ve double checked it it could still be wrong, but it’s at least in the plausible direction: In most likely scenarios, A gets several chances to win it can either win immediately by getting a high grade and the other two getting a low grade (which happens with about a 28% chance), or it can be put into a head to head which it is likely to win (which happens with about 19% chance) or it can restart and get those chances again. So this system significantly distorts random ballot in favour of the majority.

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