# Probability of winning given pair probabilities

Given an analysis of every pair of competitors in a race, how may I determine the probability of any given competitor winning the race?

For example, what is the probability of competitor 2 winning the following race? P(Cx Win) means the probability of Competitor x winning.

 Cx   Cy   P(Cx Win)  P(Cy Win)
----------------------------------
1    2       0.3        0.7
1    3       0.4        0.6
1    4       0.9        0.1
2    3       0.8        0.2
2    4       0.7        0.3
3    4       0.9        0.1


I have tried to calculate a 'rating' for each competitor by adding their individual win probabilities. For example the rating for C1 would be 1.6. The rating for C2 would be 2.2 etc. I've tried different ways to use this rating to find the probability of the competitor winning, however my gut feel tells me something is wrong.

Is there a mathematical solution to this problem?

• How's the race constructed? Multiplying the probabilities will give a better rating, since a player with a zero probability versus a certain opponent willl never win, but can still have a high additive rating. – Marc Jun 24 '14 at 6:39
• This isn't enough information to solve the problem. As an example, suppose all the probabilities you're given are $50\%$. That could correspond to a race where every possible finish order is equally likely, or it could correspond to a race where one order and its exact reverse are the only possibilities, or it could correspond to any of infinitely many other possibilities. In this case, all you can say about the probability of any given racer winning is that it's no more than $50\%$. – user2357112 supports Monica Jun 24 '14 at 6:54

One possible type of model is that each person's time for any given race is a random variable with a distribution from some family of distributions, these random variables all being independent. Now if you knew the parameters, you could (in principle) determine the win probabilities in both two-person and four-person races. For example, suppose competitor $i$'s time $T_i$ is normal with mean $\mu_i$ and standard deviation $\sigma_i$. Then $T_i - T_j$ is normal with mean $\mu_{ij} = \mu_i - \mu_j$ and standard deviation $\sigma_{ij} = \sqrt{\sigma_i^2 + \sigma_j^2}$. The probability that player $i$ wins against $j$, i.e. that $T_i - T_j < 0$, is $\Phi(-\mu_{ij}/\sigma_{ij})$ where $\Phi$ is the standard normal CDF. This leads to six equations involving the eight parameters $\mu_i$, $\sigma_i$. However, there are two symmetry operations that preserve all probabilities: adding a constant to all $\mu_i$, and scaling all $\mu_i$ and $\sigma_i$ by a positive constant factor. So we can assume, say, that $\mu_1 = 0$ and $\sigma_1 = 1$, and hope that the six equations determine the other six parameters.
• In relation to your answer, can you please elaborate on how you go from the six equations i.e. $$P(T_i – T_j < 0) = \Phi( \frac{-\mu_{ij}}{\sigma_{ij}})$$ to determining the probability of Ci winning? – Jason Jun 25 '14 at 3:00
• Once you get the $\mu_j$ and $\sigma_j$, competitor $i$ wins if $T_i < T_j$ for all $j \ne i$. So if $f_j$ and $F_j$ are the pdf and cdf for $T_j$, $$P(i\text{ wins}) = \int_{-\infty}^\infty f_i(x) \prod_{j\ne i} (1 - F_j(x))\; dx$$ This won't have a closed form, but you can use numerical methods. – Robert Israel Jun 25 '14 at 7:51