Odds of winning a race given the odds of each runner beating each opponent?

Reply Fri 15 Mar, 2013 04:12 am

Say we have three runners: A, B and C, and we have the probability of each runner beating each individual opponent:

A before B: 0.68
A before C: 0.42
B before A: 0.32
B before C: 0.30
C before A: 0.58
C before B: 0.70

Of course, the probability of A before B is = 1 - (probability B before A)

How would I go around calculating the odds of a given runner winning the race?

I thought I could consider the individual odds independent from each one, so I could just multiply them. Say:

Odds of A winning = P (A before B) * P (A before C) = 0.68 * 0.42 = 0.286

But then if I calculate the odds for B and C I get:
Odds of B winning = P (B before A) * P (B before C) = 0.32 * 0.30 = 0.096
Odds of C winning = P (C before A) * P (C before B) = 0.58 * 0.70 = 0.406

The three odds should come to 1, but it doesn't add up:
0.286 + 0.096 + 0.406 = 0.788 (not = 1)

What is it I am doing wrong?

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Reply Fri 15 Mar, 2013 01:32 pm
The events are not independent. Therefore:
P(A wins) = P(A beats B) * P(A beats C | A beats B)

So, the question becomes, "How do you calculate the conditional probability?"

I think that if you divide each result by their sum, you get the right values.
P(A wins) = .286/.788
P(B wins) = .096/.788
P(C wins) = .406/.788

To convince yourself that this makes some sense, set all probabilities to .5. Then each runner's probability of winning (prior to dividing by the sum) is 0.25. After dividing by the sum, the probabilities become 1/3, which is what you'd expect.
Reply Mon 18 Mar, 2013 02:47 am
Thanks for your reply

I've been pointed to this forum thread, which talks about the same issue. There's no way around integrating, it seems.

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