Probability of win between two agents with known win rates

[kristinabui8]

Hey there,

Assuming that we have 2 agents playing the same game in the same environment of other players, with one of them having win rate of e.g. 56% and the other of e.g. 52%, can we actually say what the expected win chance would be in the matchup between them, without any further assumptions (e.g. without assumption of Elo rating distribution of "game skill" or anything like that)?

From common sense in the above scenario probability of the first player winning should be some p higher than 0.5 (as his win rate is higher than his opponent, suggesting him being better) and lower than 0.56 (as his enemy is better than average, with win rate >50%). Moreover, naturally, completely symmetrical but reverse matchup should give the second player winrate 1-p with the same equation.

Is there a closed form equation for the exact value of p?

[Guest]

If anyone else is interested, this is my solution. We basically want to "average" the two probabilities. The person with overall probability of winning of 0.56, against the person with overall probability of 0.52, will win with probability of [tex]\frac{0.56}{0.56+ 0.52}= \frac{0.56}{1.08}= 0.518518518...[/tex] and the other person will win with probability [tex]\frac{0.52}{0.56+ 0.52}= \frac{0.52}{1.08}= 0.481481...[/tex]

[Guest]

This is a really thoughtful question, and it highlights a subtle limitation of using aggregate statistics like win rates. At first glance, it feels like you should be able to infer the head-to-head probability directly, but that’s not actually the case without extra assumptions. Win rates summarize performance against a population, not against a specific opponent. Discussions around probability, gaming systems, and similar topics often come up in analytical communities and platforms like https://trueluck-global.com/el-gr/ , where different models and interpretations are explored in more depth.