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Originally Posted by DeltaLover
Before we continue the interesting conversation, we need to clarify that I can detect a pitfall in the way you are stating the problem.
Your { P1, P2, .., Pn} is not a probability distribution and this is of course why the tuple does not add up to 1 but to 2. The logical reason of course is that Pi is not independent from Pj.
I will continue later with a more detailed view
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Yeah but ...
Suppose I know the win probabilities {W1,W2,W3 ...}.
Those are pdf definitely and W1 + W2 + W3 ... = 1.
Now you say suppose the 2 won, what is the probability of 1 winning among the rest, come first that is in the depleted by one field.
You could say S2 = W1 / (W1 + W3 ...) + W1 / (W1 + W4 + ...) + ...
So P1 = W1 + S1.
This is valid is n't it ?
I don't like it because it's not the best way of doing it, but essentially that's how to get to the P values (the better formula is a bit more complicated but follows the same logic).
Or suppose that only speed is taken into account and in this case there is a closed integral solution to give you the probabilities:
http://www.untruth.org/~josh/math/normal-min.pdf
With this integral you compute the probabilities of 1sts, 2nds, 3ds ...
If you are not happy again with the results and it does n't make money it's yet another story, but it is a closed solution.
Nonetheless you will have the same thing with place probabilities, they add up to 2.
