Game Probability Problem Consider a game played by two people $X_1$ and $X_2$. Against the general population, $X_1$ wins with probability $0.51$ and $X_2$ wins with probability $0.49$. We have no knowledge of the distribution of their wins.
If we wanted to predict the probability that $X_1$ beats $X_2$, we could say that $X_1$ is $0.51-0.5 = 0.01$ better than average, $X_2$ is -0.01 worse than average, so the probability $X_1$ beats $X_2$ could be estimated as $$0.5 + 0.01 - (-0.01)=0.52$$
Clearly that are potential errors in this method i.e. if $X_1$ wins with probability $0.8$ against the general population and $X_2$ with probability $0.2$, then our estimated probability for $X_1$ beating $X_2$ is $1.4>1$. However for now we assume it is ok if the numbers are small.
My question is, how would this be generalised to 3+ people, where only one of them can win?
For example if $X_1$ wins against general population w.p. $2/3$, $X_2$ w.p. $1/3$ and $X_3$ w.p. $1/3$, then $X_2$ and $X_3$ are completely average, $X_1$ must therefore win with probability $2/3$ against them and as they are both equal against the average, their probability of winning would be $1/6$.
What about now if we have $(X_1, 2/3), (X_2, 2/3), (X_3, 1/3)$ where the $X_i$ represent the player and the number assigned to it represents their probability of winning against the average.
 A: You need a model for how the game works.  What follows is one possibility, but a different model would produce diffetrent results.
Assume that the overwealming majority of the population produce a standard normal random variable (mean $0$, standard deviation $1$) but a few individuals produce a normal random variable with a different mean but the same standard deviation, and the result of a particular match is that the winner is the largest value of the random variables produced.
Then to have a probability of $0.51$ of winning against a typical member of the population requires a mean of about $\sqrt{2}\Phi^{-1}(0.51) \approx 0.03545279$ and similarly to have a probability of $0.49$ requires a mean of about $-0.03545279$.  Put these two players against each other and the probability of the better one winning is about $\Phi(2 \times 0.03545279 / \sqrt{2}) \approx 0.5199937$, very close to the $0.52$ you thought sensible.
Similarly to have a probability of $0.8$ of winning against a typical member of the population requires a mean of about $\sqrt{2}\Phi^{-1}(0.8) \approx 1.190232$ and similarly to have a probability of $0.2$ requires a mean of about $-1.190232$. Put these two players against each other and the probability of the better one winning is about $\Phi(2 \times  1.190232 / \sqrt{2}) \approx 0.9538359$, rather more plausible as a probability than $1.4$. 
Moving on to your three-sided games, it might be possible to calculate the mean required to have a probability of $\frac23$ of winning against two typical members of the population in a three-sided game analytically, but empirically it seems to be something like $1.116$. So with two special individuals with this mean and a typical third with a mean of $0$, the chance that a particular one of the special individuals wins is empirically something like $0.4517$ and that the typical individual wins against the two special ones is about $0.0966$ - though the final digits may be wrong.  
To repeat my earlier warning, with a different model you would get different answers.
