Sub-page of SOS

RobertJasiek 2003-10-08: We agree that SOS can be used as a means to assist a program to make pairings during the tournament, although there is no consensus which pairing strategy should be considered the best. - You claim that SOS does not behave like a random variable. I disagree. It behaves differently from coin tossing, sure. However, there is no general description yet that would explain that difference, there are only empirical tests. With more empirical tests observations could differ. To get a general statement about the difference between SOS and coin tossing, one needs more than empirical tests: One needs probability theory that is not only some theory but also explains what one observes. - You claim that SOS is never worse than a random variable. I think you mean "coin tossing" as a concept for a random variable. I disagree: Suppose a 1 round Swiss tournament with 2 participants A and B of known equal strength and the game result A beats B. Then we always, i.e. also on average over many tournaments, have SOS(A) = 0 and SOS(B) = 1. For coin tossing it could be either Coin(A) = 1 and Coin(B) = 0 or Coin(A) = 0 and Coin(B) = 1, each with 50% probability. On average we get Coin(A) = 0.5 and Coin(B) = 0.5. Since we assumed equal strengths of both players, the tie-breaker Coin is much better than SOS in this tournament. Hence SOS can be much worse than coin tossing. - Concerning the early rounds of a tournament, one cannot assess a tie-breaker fairly since it is random how strong every player's opponent (among the pool of available opponents) is. The WAGC is a good (since extreme) example of that: There are, say, 3 very strong participants (Korean, China, Japan) and many weaker participants from intermediate to very weak. The 3 very strong players should win all their games until they meet each other. For the games before, SOS is not fair since very weak opponents will make fewer wins than intermediate opponents. Even a modified SOS is not fair (only slightly better on average); e.g., SOS-1 that throws out the weakest opponent or SOS-round1 that throws out the first round. Such does not solve the principle problem that even after eliminating extremes the remaining probabilistic expectations for one's opponents are not the same. - You claim that tiebreakers were the most needed to decide the first place. For which purposes? To distinguish final results that cannot be distinguished sufficiently meaningfully? The number of wins determines the strongest player, then tiebreakers make some of them luckier by giving titles, prizes, and honour only to them while Go was supposed to be a game of greater skill and not of greater luck.

LordOfPi: It seems there is a little mistake in that argumentation. If player A and player B are of equal strenght then the average SOS over many such tournaments will converge to 0.5 for both because both will win half of the games and lose half of them.

RobertJasiek: You are referring to my trivial example tournament? There is not a mistake in it. The tournament is defined so that it is player A that wins. (Maybe we could say that player A knows how to beat player B's style, even though generally they are of equal strength.) We can invent another example: Like that example, but the win is given due to a 50% winning probability of the player A in each one-game tournament. Then after an infinite number of such tournaments, SOS converges to 0.5 for each player. With a still finite number of such tournaments, however, often either player A or player B would win more. Say, player A wins more. Then SOS(A)<=0.5 and SOS(B)>=0.5. The tie-breaker Coin is not the game result coin but is still 0.5 for each player; it is thrown only as a tournament final result list tie-breaker. Interpret that type of tournament as you like...

tderz: Robert, common sense tells me that it makes no sense to discuss the best **tie**-breaker (SOS or coin) here, as there is none - **A won** ("game result A beats B").

IanDavis: *you claim that SOS is never worse than a random variable. I think you mean "coin tossing" as a concept for a random variable. I disagree: Suppose a 1 round Swiss tournament with 2 participants A and B of known equal strength and the game result A beats B. Then we always, i.e. also on average over many tournaments, have SOS(A) = 0 and SOS(B) = 1. For coin tossing it could be either Coin(A) = 1 and Coin(B) = 0 or Coin(A) = 0 and Coin(B) = 1, each with 50% probability. On average we get Coin(A) = 0.5 and Coin(B) = 0.5. Since we assumed equal strengths of both players, the tie-breaker Coin is much better than SOS in this tournament.* What sort of idiotic claim is this Robert? A beats B and wins the tournament, now we apply the tiebreaker. Hold on, did I just say now we apply the tiebreaker? Oh stupid me, I forgot, when two players are not tied we don't have a tiebreaker.