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Elo and Glicko are simply wrong

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luckisK

What is the correct rating of each of the soccer teams when they have completed the championship of the year? (wins-losses)/(number of games) which is equivalent to ((wins)+0.5(draws))/(number of games). AND NOT Elo or Glicko rating. You will say, yes but Elo and Glicko are solutions for the problem that players have not faced opponents of  equal strength. Yes but these solutions are wrong and the above argument proves that they are wrong. I HAVE found a modification of ((wins)+0.5(draws))/(number of games) which is a solution for the problem, but it is so complicated that it is preferable not to implement it, but instead to force all players to confront players of all ranges of strengths, like a soccer championship of 100000 teams that each opponent is chosen randomly. The only modification of ((wins)+0.5(draws))/(number of games) that is needed then, is to take in account the number of games a player has played, as if one played only one game and won, he cannot have the maximum possible rating of 1. I have found such a modification and it is very simple to implement.

tygxc

#1
No, your reasoning is simply wrong and the elo / glicko is right.
Winning 9 games against 10 beginners is easy.
Winning 1 game against 10 grandmasters is hard.
Chess championships are decided like you say: wins + 0.5 * draws
However to rank a player glicko or elo are right.