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Figured you math dudes (and dudettes) could knock this one out of the park!

I've got a website that stores users' game scenario results. Each scenario record can have one of three possible results: Side 1 wins, Draw, or Side 2 wins.

To this end, I'm currently representing balance by showing the simple percentage of total victories for each result. Example:

Result Totals
  Side 1 - 5
  Draw - 1
  Side 2 - 4

  Side 1    Draw    Side 2
   50%       10%     40%

but the astute reader will immediately note that using this methods means a scenario with 5 plays split 2-1-2 and a scenario with 100 plays split 40-20-40 will have the same apparent "balance".

What I would like to do is take into account the total plays for a scenario. Thereby, a scenario with a split of 41-20-39 will rank much higher than one splitting 2-1-2, even though the simple percentage is actually less balanced.

Does that make sense? Ideally, this would map to a scale of some sort, so scenarios could be given discrete balance ratings.

Thanks for any help you can provide.

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Do you consider a 50-0-50 split more or less balanced than a 0-100-0 split? – mhum Feb 24 '11 at 5:55
I had not considered it before. But after a minute's thought, my layman's brain seems content with putting 50-0-50 and 0-100-0 in the same pot. Is that wrong? – Andrew Heath Feb 24 '11 at 6:44
It's your model, you get to define it however you like. It helps clarify the problem a little bit for me. – mhum Feb 24 '11 at 6:52

In fact, your measurement of balance with 2-1-2 and 40-20-40 are the same-each player has the same chance of winning. What is different is the assurance that the measurement is reasonably accurate. Under the usual (worrisome) statistical assumptions you can calculate a standard deviation on the number of player 1 wins as $\sqrt{Np(1-p)}$ where $N$ is the number of games and $p$ is the fraction won by player 1. So for 2-1-2 the standard deviation on the number of player 1 wins is $\sqrt{5\cdot0.4\cdot0.6}\approx 1$ So you would say player 1 wins $(40 \pm 20)\%$. In the case of 40-20-40 you would have a standard deviation of about 5, so you would say player 1 wins $(40 \pm 5)\%$. It's not that the second measurement shows more balance, but you have better reason to think it is close.

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"assurance" or I suppose "confidence" was the angle I was looking for. thank you! – Andrew Heath Feb 24 '11 at 8:01

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