LOS calculation: Does the same result is always the same?

[Rémi Coulom]

Besides, there is the chess* intuitive explanation that should be really convincing. (being stronger at chess is equivalent to being stronger at chess*, where the game is restarted whenever it ends in a draw, so that no draw is possible in chess*).

I must admit this does sound very good (although it is not a 100% proof since not every sequence of chess games is a sequence of chess* games).

I believe if we use the hypothesis of independence of game results, we can re-order games in order to make each sequence of chess games a sequence of chess* game.

Hypothesis: iid games, probabilities p_1,p_0.5,p_0.

p_1 = probability that A wins one game of chess
p_0 = probability that B wins one game of chess
p_0.5 = probability of a draw in one game of chess

A stronger than B at chess <=> p_1 > p_0
A stronger than B at chess* <=> p_1 > p_0
(So A stronger than B at chess <=> A stronger than B at chess*)

If we consider a sequence of chess games, for instance with result wwlddwld:

P(A stronger than B at chess | wwlddwld) = P(A stronger than B at chess | dddwwwll) = P(A stronger than B at chess* | dddwwwll) = P(A stronger than B at chess* | w*w*w*l*l*)

(w* is a win at chess*, w is a win at chess...)

The last = might need some thinking, though.

Rémi