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I am curious to find out reasonable ways of dividing a prize among $n$ people in the following situation:

To make the example specific, we have $6$ people in total who are going to share a prize of $100$ dollars, and let us denote the amount received by each person $i$ as $p_i$. In addition, each person $i$ is given a score $s_i$, and we can think of $s_i$ as a way of measuring how well person $i$ deserves some portion of the prize. The intuition here is that we would like a higher-scoring person to receive a larger portion of the prize than a lower-scoring person, that is, $p_i\geqslant p_j$ if and only if $s_i\geqslant s_j$. The problematic thing here is that $s_i$ can be either negative or positive. For example, $s_1=1.3, s_2=2.1, s_3=-0.8, s_4=-3.7, s_5=0.7, s_6=5.2$.

So, what would be the proper ways of dividing the prize given these scores?

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    @BrianM.Scott, yes, the range of scores is bounded. For generality, let's say $s_{min} \leqslant s_i \leqslant s_{max}$.2012-01-10

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There is no clear mathematical answer. Certainly you can give the whole prize to the winner. You could say that the differences in scores are insignificant and give them all 1/6 (with small deltas if you want), or anywhere in between. You could add you favorite constant to all the scores to make them positive and share the purse in proportion. The larger the constant, the closer the prizes will be. So if you use 5, player 1 gets 6.3/34.8. If you use 1000, player 1 gets 1001.3/6004.8.

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    My examples are extreme cases, intended to show that there is a wide range of solutions. You need to think about what you want.2012-01-10
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One way is to distribute them proportionally to $e^{as_j}$ where $a$ is some carefully chosen positive constant to reflect some common sense belief of yours. For instance, if you think that a 5 point gap between the winner and the rest of the crowd justifies the "winner gets all" distribution, make $a=\frac 15\ln 100$.

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Use the softmax.

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    could you please elaborate some more about this seemly AI related method? thanks.2012-01-10