Timminz wrote:joriki wrote:However, it still has a major disadvantage -- among good players, it favours those who tend to play more opponents. In the extreme case of a player who always wins, their score would be 2 if they only play two-player games, but 8 if they only play 8-player games.
I really like your ideas, but I disagree with the quoted bit. I wouldn't call that a disadvantage to the system. I would say that someone who plays only 8 player games, and wins every game, is better than someone who wins 100% of the time, but only plays 2 player games. Beating one opponent is a whole lot easier than seven.
That's true. Sorry, I was being a bit too compact there; I should elaborate on that argument.
We want equal scores to indicate similar strength independent of the number of players a player typically plays. Using #win/#games * #players/#games, a hypothetical perfect player playing only two-player games and winning them all would get a score of 2. A player playing only eight-player games would only need to win one game out of four to get that score, and by winning more than that, they could get a score that the player playing only two-player games could never attain.
Let's call a player's probability to win an n-player game p (n). Ideally we want the score the player is expected to get if they play only n-player games to be independent of n. Using #win/#games * #players/#games, the expected score is p(n) * n, so setting that equal to a constant c independent of n, we get p (n) = c / n. That is, for this formula to reflect a player's strength independent of n, the player's probability of winning a game would need to drop inversely with the number of players. For an average player with c = 1, this is probably roughly realistic; for a very good player with c = 2 it seems unrealistic, since it means certain wins for n = 2 and only one out of four wins for n = 8; and for p (8) > 1/4 and hence c > 2, it's impossible, since it would imply p (2) > 1.
On the other hand, if we use #opponents beaten / #games lost, the expected score is p (n) * k / (1 - p (n)), where k = n - 1 is the number of opponents per game. Setting this expression to a constant c independent of n yields p (n) = c / (k + c). For an average player with c = 1 this yields p (n) = 1/n, like the other formula. But for high c, this result behaves more nicely. We have p (n) <= 1 for all n (with p (n) = 1 only if c is infinity). We can power-expand p (n) for c much larger than k and get p (n) = 1 - k/c + ..., which makes sense: If an extremely good player has a small 1/c chance of encountering an opponent they can't beat in a two-player game (where k = 1), then they have roughly a k/c chance of encountering such a player in a game with k opponents.
So the behaviour of these formulas is similar for scores near 1, but the formula #opponents beaten/#games lost seems to be more realistic for high scores.
