# Score adjustment for trivia night pub quizes

## weblog/

I went to a trivia night where one of the other teams only had a single person. That person got 2nd place to a team with 6 people. It was quite impressive and got me thinking if there was a way to quantify how impressive it was. This seemed related to trying to figure out a scoring system that would account for team sizes.

Assume that the chance anybody knows any factoid is `p`. Also assume (somewhat unrealistically) that team members don't disagree witth each other. That is, if anybody on the team knows the factoid then the team will choose that correct answer.

For a team with `n` members, the probability they'll get the correct answer will be the same as not having having `n` people who don't know the answer:

``````1 - (1-p)ⁿ
``````

The normal scoring is to just sum up correct answers for `k` questions. So the expected score for the team will be:

``````score = k (1 - (1-p)ⁿ)
``````

Obviously a clear advantage to larger teams.

To account for this, we'd like that the expected adjusted score is independent of the team size `n`:

``````adjusted = k p
``````

To achieve this, solve the `score =` equation above for `p` and substitute into the `adjusted =` equation:

``````adjusted = k (1 - ⁿ√(1 - score/k))
``````

This successfully levels the playing field.

Of course, if your goal is to sell drinks at a bar, maybe you prefer big teams for a different reason :-).