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Imagine an athletics race where there are three runners A, B and C. They all have an equal opportunity of winning. So the chances of A winning are 1 in 3.

How would I get to this answer if I only knew that the chances of A beating B is 0.5 and A beating C is 0.5 ? At first I thought that A beating B and A beating C were independent results and therefore the probability would be 0.5 * 0.5 = 0.25, but I know the answer should be 0.33

If the chances of A beating B was 0.4 and the chances of A beating C was 0.35. What is the chance of A winning the race? How do I go about working this out.

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Why do you think that the probabilities you have are enough to work out the chance of $A$ winning in each case? –  Chris Eagle Sep 20 '12 at 21:42
    
Ok, if I knew the probabilities of all permutations for all of the runners. Would I then be in a position to work out the probability of A winning? If all runners had .5 probability of beating each other. Could I work out the chance that A wins? Excuse my lack of knowledge in this area, I haven't done much probability for a while. I've read on other posts that I could perform a pairwise comparison if I know all individual probabilities. Is this correct? If I can understand how to worn out the answer j can then move on to scenarios where runners have different probabilities of beating each othe –  Scott Sep 20 '12 at 22:05
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If all you know is that $A$ beats $B$ with probability $1/2$ and $A$ beats $C$ with probability $1/2$, you can’t determine the probability that $A$ wins the race. That information is consistent with the possibility that each runner is equally likely to win, but it’s consistent with other scenarios as well. Suppose, for instance, that $B$ always beats $C$. Then the only possible outcomes are $ABC,BAC$, and $BCA$. Suppose that these have probabilities $p,q$, and $r$, respectively. Then the probability that $A$ beats $B$ is $p$, so $p=1/2$. The probability that $A$ beats $C$ is $p+q$, but it’s also $1/2$, so $q=0$. Thus, in reality the only possible outcomes are $ABC$ and $BCA$, and $A$ wins with probability $1/2$, $B$ also wins with probability $1/2$, and $C$ wins with probability $0$.

This is a rather implausible scenario, but it’s possible in principle, and it’s consistent with the given information.

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