Winning probability in a game with multiple players I do not come from a mathematical background, and hope you can answer this (probably very basic?) question.
I got a game group with friends where we play board games, normally 3-6 players each night. I have created an application which keeps track of game history, and keeps track of a Elo rating for each player. These Elo ratings can be used to determine the winning probability between two players.
However, how can I use these winning probabilities to get the probability of a player winning a multi-way game? 
Let's say we have a three way game, two players with a Elo rating of 1200 and one with 1400. The two with 1200 have 50% chance of winning against each other and the 1400-player has a 76% chance of winning in a heads up match against a 1200-player. How can I find the probabilities that each player wins the three-way game?
 A: If you assume that the weaker players won't gang up on the stronger player (a very strong assumption!), then a reasonable model would be the following.  (I'm following the notation of the "theory" section of the Wikipedia article on ELO.)


*

*let $R_A, R_B, R_C$ be the ratings of the three players.

*let $Q_A = 10^{R_A/400}$; define $Q_B, Q_C$ similarly.

*the probability of $A$ winning is $Q_A/(Q_A + Q_B + Q_C)$ and similarly for the other two players.


With the numbers you gave, $C$ has a probability of about $0.613$ of winning, and $A, B$ each have probability $0.194$ of winning.  
This seems like the "obvious" generalization of the Elo math.  The most obvious problem, to me, is that I wouldn't know how to update these ratings after a multi-player game is played.
A: Would the weaker players not band together against the stronger players? That is what happens in the truel (three player pistol fight with one shot per person). 
I observed that in monopoly and in three-player chess also. That said, if one  assumes the winning or losing of A over B and C are independent, then with the probabilities you mentioned
P(A wins against B and C) = P(A wins against B) * P(A wins against C) = .5 * .24 = .12
= P(B wins against A and C)
and P(C wins against both A and B)= .76 * .76 = .5776 .
The remainder is the probability of no-one winning. 
Note that in three-way chess, no-one winning because the situation "A wins against B, B wins against C and C wins against A" is not possible, as the first one to check-mate anyone is the overall winner, and the other two lose equally. So there the whole calculation would not hold. 
Summarizing I would say there is no unique answer to your question, you need to specify how ties are broken, if collaboration can occur, etc. This is not a statistics question, it is game theory. 
