Winners

[David.Hiskiyahu@BrinksGlo]

Assume a large collection of players in some kind of sport, e.g. professional backgammon.

The probability of win is determined by combination of luck and skill.

Assume player A which has an average win rate within this collection of players of 90%.
Assume player B which has an average win rate of 80%.

How does one calculate the average win rate of A in a long sequence of games between A and B?

 

[David.Hiskiyahu@BrinksGlo]

Might as well generalise the win rates to PA and PB, 90 and 80% are given for purpose of illustration.

 

[David.Hiskiyahu@BrinksGlo]

Intuitively it feels like the answer is PA/(PA+PB) but I cannot prove it.

In the illustrated case that would be 90/170=53%.

 

[Steven G]

I don't think this can be done.

Let's say there are 10 players numbered 1-10.

Suppose player 1 wins 100% of the games.
Suppose player 2 only loses to play 1
Player 3 only loses to player 1 and player 2
...
Player 9 only beats player 10
and player 10 always loses.

Assume no ties and that everyone plays everyone else the same number of times.

Then player 2 wins 90% of the games and player 3 wins 80% of the games.

In this situation p(player 2 beats player 3) = 1 while p(player 3 beats player 2)=0.

The problem is that it does not have to be the way as described above!

 

[David.Hiskiyahu@BrinksGlo]

Thanks Jomo, this is an interesting way to look at the problem.
The way you specified the situation, there is no random behavior in the system.
The conditions I discuss come from the real world of online games, specifically from www.itsyourturn.com.

In a specific kind of backgammon game I have an average win rate of 66% there, after a large number of games played.
Another player comes around let's say with 90% win rate.

The tendency to win is random with the above averages.
So, his chance against me seems to be 90 / (90 + 66) = 90 / 156 = 57.6%, while mine is 42.4%.
Just want to know if my formula makes sense.

 

[David.Hiskiyahu@BrinksGlo]

When I take the example to extreme, e.g. Player A tends to win 100% of the games, my formula does not work.

100/(100+66) <> 100.

 

[JeffM]

If player A has a win rate of 90% and player B has a win rate of 60% and a player of average skill playing against another player of average skill has a chance of winning of 50%, I'd think it is more likely that player A will win with probability of
(90 - 50)/{(90 - 50) + (60 - 50)} = 40/(40 + 10) = 80% than with probability
90/(90 + 60) = 90/150 = 60%.

That better reflects the differential in skill. But even this is probably a relatively crude estimate. A player with a win rate of 90% may not even play against players with a win rate of 10%.

There are so many variables that I doubt it is possible to do better than a relatively crude estimate, but I think measuring skill is best done by subtracting 50% from the win percentage rather simply looking at the win percentage itself.

I used to play bridge online, particularly as partner with my coach who had played a number of times for Britain. We attracted strong opponents who were friends of my coach. My win rate quickly fell below 50%. Your win rate does not reflect just your skill; it also reflects the skill of the opponents that you choose to play against.