Tricky probability problem about voting

Question

Parliament contains a proportion p of Labour members, who are incapable of changing their minds about anything, and a proportion 1 − p of Conservative members who change their minds completely at random (with probability r) between successive votes on the same issue. A randomly chosen member is noticed to have voted twice in succession in the same way.

What is the probability that this member will vote in the same way next time?

Answer 1

Let's say there are L Labour members and C Conservative members, then p=L/(L+C) (probability of randomly choosing a Labour member) and 1-p=C/(L+C) (probability of choosing a Conservative member).

On the next vote rC/(L+C) Conservative members will change their minds while (1-r)C/(L+C) will vote the same as they did before. The respective probabilities are r(1-p) and (1-r)(1-p).

But on the next vote all Labour members (probability=1) will vote the same again. The combined probability is p.

So p+(1-r)(1-p)=p+1-p-r+pr=1-r+pr.

EXAMPLE

Let L=200, C=400, so L+C=600, p=⅓, 1-p=⅔.

Let r=⅗, so 1-r=⅖.

Since Labour members don't change their minds on a particular issue, 200 will vote the same on successive votes. ⅖ of the 400=160 Conservative members will also vote the same. Total voting the same=360. So 360/600=0.6, ⅗ or 60% is the proportion and the probability that a random member will vote the same on successive votes.

The formula also predicts 1-r-pr=1-⅗+⅕=⅗.