----- Original Message ----- From: Kristofer Munsterhjelm Date: Saturday, June 25, 2011 2:26 pm Subject: Round robin tournament statistics To: EM Cc: Forest W Simmons
> Forest, > > You surely know statistics better than I do, so let me ask > something > I've been wondering about for some time. It even somewhat ties > into the > subject you've been discussing. > > Say you want to find out who's the best player (team, etc) in a > round > robin tournament. However, arranging matches is expensive, > mainly in > time. So you want to pair two players (teams, etc) against each > other > just enough to be able to decide who is best. > > How would you do this? > > It seems you could decide upon a confidence level and then have > a given > pair stop playing once you're confident that one of the players > in > question beats the other player. The level would then be picked > so that > one is reasonably sure that all pairwise contests "point the > right way" > (have the right winners). That would be conservative, since > methods > don't necessarily use all the information of every contest. > > It gets more difficult when one takes ties into account, though. > For > most games, no pair is exactly tied in the long run, but one > could > imagine a game where if both players cooperate, there's always a > tie > (such as two players in chess agreeing to always do a > grandmaster draw, > based on tit-for-tat reasoning). Then a long run of ties would > in itself > be significant: it means that neither player is (or chooses to > be) any > better than the other. Just eliminating ties from consideration, > as you > did in the winner calculation, wouldn't work because it could > take a > really long time before a non-tie result is granted. > That's where the "Independent Identically Distributed" proviso comes in. If there is any kind of mutual strategy, this condition cannot hold. What I was more concerned with, ultimately, was how equal rankings would affect the significance of the defeat. In other words, suppose there are 100 ballots, and W=40 support the winner, L=10 support the loser, and the other fifty rank them equally or truncate them both. Does this 40 to 10 defeat have the same significance as a 40 to 10 defeat in which there were only fifty ballots total? According to the above model (with the independent identical distribution condition) the answer is yes. That makes things nice for comparing pairwise defeat strengths in the case of sincere rankings. As I mentioned before, these sincere rankings are most likely in the case of informal polls before the actual election. In a poll on a yes/no question, this result says that you can just ignore the "no opinion" responses. ---- Election-Methods mailing list - see http://electorama.com/em for list info
