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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 $= 1 - \text{(probability B before A)}$

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:

$$\text{Odds of A winning }= \Bbb P(\text{A before B}) \cdot\Bbb P(\text{A before C}) = 0.68 \cdot 0.42 = 0.286$$

But then if I calculate the odds for B and C I get:

$$\text{Odds of B winning }=\Bbb P(\text{B before A})\cdot\Bbb P(\text{B before C}) = 0.32 \cdot 0.30 = 0.096$$

$$\text{Odds of C winning }=\Bbb P(\text{C before A})\cdot\Bbb P(\text{C before B}) = 0.58 \cdot 0.70 = 0.406$$

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?

Thanks!

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Thanks for the improved formatting, Brian M. Scott. –  Juan Mejías Gómez Mar 15 '13 at 11:27
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1 Answer

You don't have enough information to determine the answer: just knowing the probabilities for each pair is not enough.

Consider the following two possible situations: in situation $X$, all $6$ race outcomes are equally likely. In situation $Y$, runner $A$ cannot come second, but the remaining $4$ outcomes are equally likely. In both situations, the probability that any given runner beats another given runner is $1/2$. But in $X$, the probability that $A$ wins the race is $1/3$, while in $Y$, it is $1/2$.


Your numbers are rather messy, but here's an example to show they don't determine what you want (the first column lists possible outcomes of the race, the second lists the probability of the outcome):

Case 1
ABC 0.30
ACB 0.12
CBA 0.32
CAB 0.26

Case 2
BCA 0.30
CBA 0.02
CAB 0.26
ACB 0.42

Both cases satisfy all your conditions, but in case 1, B has probability 0 of winning, while in case 2 this probability is 0.30.


As to your own attempt: what you are doing wrong is assuming events are independent when you have no reason to think they are.

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Thanks, Chris Eagle, you're probably right that I don't have enough information, and that I am wrong in assuming they're independent events. I'm not 100% sure I understand your examples, though. In your first example, you say "the probability that any given runner beats another given runner is 1/2", but if A can't finish 2nd, is that still true? As for your second example, you say both cases satisfy my conditions. I guess that means they respect the individual odds I put on the list? But, then, how do you know the odds you are introducing are consistent with those? –  Juan Mejías Gómez Mar 15 '13 at 11:24
    
I actually have more information about each runner, although I did think what I mentioned would be enough. Each runner follows a normal distribution with different mus and sigmas. Using those, I calculated the list of probabilities I mentioned in my question. Would that be enough to find an answer? –  Juan Mejías Gómez Mar 15 '13 at 11:56
    
So, the way I see it now, my problem is limited to finding the probability than a normal distribution is greater than two others, when all of them are independent (because I consider times for a given runner to be independent from times of the others). Anyone can offer any help with that? –  Juan Mejías Gómez Mar 15 '13 at 12:08
    
I have created a new thread for the new question: link –  Juan Mejías Gómez Mar 15 '13 at 12:29
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