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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?

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  • $\begingroup$ I am guessing someone who wins both his matches wins the three-way game, but what happens if everybody gets one win and one loss? What's the tie-breaking rule? $\endgroup$
    – Nameless
    Commented Jul 3, 2013 at 12:43
  • $\begingroup$ It is not multiple matches, just multiple players in one match. Think three players playing Monopoly. You know the probability of each person winning over the other player in a heads up Monopoly match, and want to transform that into the probability of winning a three (or more) player game. $\endgroup$
    – ghdalum
    Commented Jul 3, 2013 at 12:46
  • $\begingroup$ Do you have data for the three way games? So, for example, can you see how often a player with an ELO rating of around 1400 won against two opponents with ELO ratings of around 1200? $\endgroup$
    – soakley
    Commented Jul 3, 2013 at 13:18
  • $\begingroup$ A,B,C are three players, A,B=1200 and C=1400, Then probability of A winning a game with B is 50% i.e P(A/B')=0.5. Similarly P(B/A')=0.5 and the probability of C winning is P(C/A'&B')=0.76 where A' (for example) is the complement of A. That is your question? $\endgroup$
    – SAAN
    Commented Jul 3, 2013 at 13:26
  • $\begingroup$ Hmm. Either you are stating the obvious, or your comment went over my head. What is my question? $\endgroup$
    – ghdalum
    Commented Jul 4, 2013 at 8:49

2 Answers 2

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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.

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    $\begingroup$ "The most obvious problem, to me, is that I wouldn't know how to update these ratings after a multi-player game is played." From what I understand, it seems that rankade algorithm manage this issue well: rankade.com/ree#ranking-system-comparison Unfortunately it's not open source at the moment. $\endgroup$
    – Aubrey
    Commented Nov 10, 2015 at 15:31
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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.

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  • $\begingroup$ Disregard concepts like allying and ties. Let's say all players are of equal skill, with a 50% chance of winning in a heads up game vs every other player. Logical thinking gives that each player got a 1/3 chance of winning in a three-way game. With this method you get a 1/4 chance of winning for each player, and 1/4 chance of getting no winner? $\endgroup$
    – ghdalum
    Commented Jul 4, 2013 at 8:52
  • $\begingroup$ Yes, under independence. The problem is that even if you don't allow two-player ties, the 1/4 probability of a three way tie arises out of the assumption of independence: A wins over B, B over C, and C over A has a positive probability of 1/8, and the opposite cycle also. If the game rules exclude ties, then this computation cannot hold. Do ties occur at all? $\endgroup$
    – vinnief
    Commented Jul 4, 2013 at 11:51

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