# Probability Calculation in Games – Determining Player Winning Probability in Head-to-Head Matchups

combinatoricsprobabilityranking

I have predicted the probability that player A beats player B in a head to head matchup in a game, for all combinations of players. I would like to convert this into the probability of a player winning the entire match, based on head to head predictions, and rank them.

Is this as simple as P(A beats B) * P(A beats C) * … * P(A beats N – 1) for the N players?

Thanks for your help, much appreciated.

EDIT: Additional info: There is no elimination. This is a single trial/event. All players are competing against all other players at the same time, that is, all n players are on the "field" at once. In a tie, all t players that tied are ranked equally. There is no tiebreaker, they split the "winnings".

EDIT 2: My specific case is a race. The head to head probabilities are the probabilities that a player finishes ahead of the other player. So if P(A > B) = 0.6, there is a 60% chance Player A beats player B, but does not make a mention of their actual finishing position. Based on this info, I would like to rank players by probability of winning the entire race

This question is interesting but unanswerable. I found it instructive to think about Efron's intransitive dice for the insight they give us into some subtle issues.

Let's model the "races" (head-to-head contests) between individuals $$A,$$ $$B,$$ and $$C$$ with ordinary fair dice that have been relabeled. When two dice are thrown, the one with the larger value wins. When all three dice are thrown, the one with the largest value wins. To avoid dealing with ties, we will make sure that no two dice share a value.

• Die $$A$$ has the numbers $$2,2,4,4,9,9.$$

• Die $$B$$ has the numbers $$1,1,6,6,8,8.$$

• Die $$C$$ has the numbers $$3,3,5,5,7,7.$$

Straightforward calculations establish that in any head-to-head matchup, die $$A$$ beats die $$B$$ with a probability of $$5/9.$$ We can do this by brute force with a table, if you like. Since, in effect, each player is producing one of just three values (with equal probabilities), we only need to tabulate all $$3\times 3 = 9$$ possibilities. Here are the winners of all possible in a head-to-head matchups of $$A$$ (columns) against $$B$$ (rows):

$$\begin{array}{r|ccc} \text{B\A:}&2 & 4 & 9 \\ \hline 1 & A & A & A \\ 6 & B & B & A \\ 8 & B & B & A \end{array}$$

Similar calculations establish that $$B$$ beats $$C$$ with a chance of $$5/9,$$ and also $$C$$ beats $$A$$ with the same chance. This is a probabilistic rock-paper-scissors situation where each player tends to beat the next in a cyclic list $$A\to B\to C\to A.$$

A more complicated calculation, but still straightforward, tabulates all $$3\times3\times 3 = 27$$ (equiprobable) outcomes when all three players are "in the field together." It turns out that $$A$$ and $$B$$ each have a $$10/27$$ chance of being the overall winner, while $$C$$ only has a $$7/27$$ chance of winning.

What differentiates $$C$$ from $$A$$ and $$B$$ to make $$C$$ inferior in this collective sense? All three dice have the same mean value of $$5,$$ showing they get the same "score" on average. (To continue the race metaphor, they are all equally strong runners.) But the variance of $$A$$ is $$((2-5)^2 + (4-5)^2 + (9-5)^2)/3 = 26/3$$ and the variance of $$B$$ is the same (because the values of $$B$$ are $$10$$ minus the values of $$A,$$ which leaves the variance unchanged). However, the variance of $$C$$ is only $$((3-5)^2 + (5-5)^2 + (7-5)^2)/3 = 8/3,$$ considerably smaller.

In short, $$C$$ gets far more consistent results. Contestants $$A$$ and $$B$$ get their wins by being inconsistent: sometimes they greatly exceed their average and sometimes they fall far short.

Based on this little example, then, we might suppose that the shape of the distribution of head-to-head results -- especially its spread -- likely is an important factor in estimating the chances of any competitor being the overall winner. It's not enough to suppose that all three outcomes are independent, nor does it suffice to know just the head-to-head probabilities.