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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

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    $\begingroup$ It depends on how the tournament is conducted and the rules for resolving ties: please explain. $\endgroup$
    – whuber
    Mar 29, 2022 at 18:52
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    $\begingroup$ How are matches structured? Is it a round-robin tournament where everyone plays everyone and players are ranked on the total number of wins? Or is it a single-elimination bracket tournament? If it's the latter, that formula won't work - A might always lose to B, but if B gets eliminated by C in an earlier round, A can still win the tournament. In an elimination tournament with N players, the winner only plays log2(N) games, not N games. Who they play will make a big difference - do we assume random matchups? $\endgroup$ Mar 29, 2022 at 18:52
  • $\begingroup$ @NuclearHoagie added some additional context info. Please let me know if you need anything else. $\endgroup$
    – Jage
    Mar 29, 2022 at 19:08
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    $\begingroup$ You need to assume something about the relationship between this "all on the field at once" contest and the "head to head matchups" for which you actually have data. Why not explain what's really going on, rather than asking us to guess based on this abstract description of the problem? $\endgroup$
    – whuber
    Mar 29, 2022 at 19:11
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    $\begingroup$ @whuber Sorry, I figured being general would be best and make answers the most helpful to others in the future. 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. $\endgroup$
    – Jage
    Mar 29, 2022 at 19:14

1 Answer 1

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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.

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  • $\begingroup$ I really like the concept of intransitive dice. Seems very interesting and applicable here. A question I have: The lower variance of C resulting in it losing more frequently than A and B is not strictly true in all cases right? Whether higher or lower variance is good or bad will depend on the distribution of head to head predictions correct? $\endgroup$
    – Jage
    Mar 31, 2022 at 5:04
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    $\begingroup$ I believe that's correct: comparing variances is just a little too crude to arrive at definite results. But this does provide insight into why some dice emerge as better and others as worse when all are competing with each other, compared to head-to-head matchups. $\endgroup$
    – whuber
    Mar 31, 2022 at 13:26

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