7
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Jul 3 '13 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 Jul 3 '13 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 Jul 3 '13 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 Jul 3 '13 at 13:26
  • $\begingroup$ Hmm. Either you are stating the obvious, or your comment went over my head. What is my question? $\endgroup$ – ghdalum Jul 4 '13 at 8:49
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ "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 Nov 10 '15 at 15:31
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jul 4 '13 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 Jul 4 '13 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.