srnity: Your suggestion of resigning is probably the reason why some players abuse the vacation system: perhaps they hope to artificially inflate their win totals by goading their opponents into resigning out of frustration.

TarantinoFan: While I share your frustration with such unscrupulous behavior (autovacation is probably the most ill-conceived feature on this site), I think the best you can do is to grin and bear it. His vacation days will run out in a couple of weeks. After that, either he'll play and you'll have the game you signed up for, or he won't and you'll have a much-deserved victory.

tarcellius: There seems to be reason to believe that the one-tournament-per game restriction was removed without an announcement. Also, the fact that you are currently playing two Pente tournaments is strong evidence of this, unless you were a Rook at some point in the past. I would recommend against upgrading your membership, unless you want to do so for other reasons. It would be a shame if Fencer's habit of inadequate documentation led directly to an increase in his profits.

You're listed as being signed up for the tournament in question, and (while I could be mistaken) I believe that if the restriction were in place you would have been unable to sign up at all. I expect you'll be able to play without needing to upgrade.

BIG BAD WOLF: Really? That's a bizarre way to do things, but I guess it shouldn't surprise me.

This really needs to be stated clearly somewhere in BrainKing's woefully inadequate documentation. Pawns often need to plan their tournament choices carefully, and how are they to do that if they're not even told the requirements for entry?

forweg: Was it this one? If so, you would have needed at least 12 open slots.

Nothingness: It is possible precisely because the games are chosen randomly and independently. With 2 sections of 5 players, there are 20 games, each of which is equally likely to be of any given type, regardless of how the other 19 games are assigned.

If you tossed 2 coins, would you be surprised if they came up both heads or both tails? I hope not; they're as likely to be the same as different. What if you tossed 20 coins and they all came up the same? Of course that's much less likely, but it's still possible; in fact you should expect it to happen, on average, once in every set of 524288 sets of 20 tosses.

What if, instead of tossing 20 coins, you take 20 hats, put into each hat 11 cards marked 1 to 11, thoroughly mix the cards in each hat, and draw one card from each hat? Will you be surprised if you don't draw any cards marked 2, 5, or 9? Maybe such an outcome isn't very likely, but certainly it's possible; the only alternative would be for the cards in each hat to be influenced somehow by the cards in the other hats.

Of course that last example is equivalent to how the games in your tournament were assigned. The following table, if I've crunched the numbers correctly, shows the approximate probability of exactly n game types being represented when 20 games are assigned in a random tournament with 11 possible game types.

 n   Prob.
--------------------------------
 1   0.0000000000000000000163508
 2   0.0000000000000857251
 3   0.000000000854404
 4   0.000000532498
 5   0.0000617406
 6   0.00212913
 7   0.0275495
 8   0.150025
 9   0.356337
10   0.351113
11   0.112784

So, if you create many such tournaments, you should expect to see 9 or 10 game types in the first round most of the time. 15% of the time you'll get only 8 types, and all 11 types will be played only 11% of the time.

Rose: It's possible that tournament creation is more a game of skill than of luck.

Jim Dandy: The Sonneborn-Berger tiebreaker has nothing at all to do with ratings. It is calculated only from the results of the games played in this round of the tournament.

To compute your score in a tournament round, give yourself one point for each win, one half of a point for each draw, and zero points for each loss. In the present case, you played 6 games, winning 3 and losing 3, so your score is 1+1+1+0+0+0=3. The same is true for each of the other players.

To compute the S-B tiebreaker, multiply each other player's score by your score in games against that player, and add these up. In this case, each of the other players had a score of 3, and you won 1 game against each of them. Thus your S-B is (1*3)+(1*3)+(1*3)=9. The same is true for each of the other players.

Thus all players in this round finish with a score of 3 and an S-B of 9, and since no further tiebreakers are used, the result is a 4-way tie.

Tournament results are computed using only the results of games played in the tournament. No external factors (such as BKR) are considered. [The only exception to this in BrainKing's current tournament system is if a draw occurs in a single-elimination tournament, where, due to the structure of the tournament. it is necessary to have exactly one winner from each pair.]

Fencer: Yes, that is a good solution to be used by those who create tournaments. But it doesn't address the point I'm making, which perhaps I can clarify with a hypothetical situation.

Suppose there is a single-elimination chess tournament which uses normal games. Suppose it's a prize tournament, with a one-year Rook membership at stake. Player A is playing against player B in a semifinal round, and A's rating is slightly higher than B's. Their game reaches a position in which a draw is likely, and A becomes worried, because a draw would eliminate him from the tournament.

Now suppose that at the same time, A is also playing a game of chess against player C. This is just a regular game, not part of any tournament. Suppose A and C have both played well, and have reached a very interesting middlegame position in which any result is possible. Maybe A has a slight advantage.

But then A realizes that if C wins, it will cause A's rating to drop below B's, which would mean that in the event of a draw in the game between A and B, A would advance to the final round of the tournament. A considers the possibilities, and decides that he'll gladly give up a few ratings points to have a better chance at winning a valuable prize. He therefore begins to play poorly in his game against C. Maybe he even resigns prematurely.

Some might say that C should be pleased by this; after all, he won! But if C is like me, he will be upset, because a win isn't worth much if the opponent was trying to lose. C doesn't play games for such meaningless wins; he plays for the experience of trying always to find the best move, the best strategy, against a strong opponent who is doing the same. And C feels that he has been deprived of what could have been an excellent game, simply because the structure of a tournament (in which C was not participating!) gave A an incentive to throw away a game.

In my view, it's best if the rules of a tournament or other competition never give players an incentive to lose. Therefore I would prefer that the higher-rated player advance in the case of a draw.

Granted, situations like the one I've described probably will not occur very often, so I'm not inclined to worry about this much more. But I wanted to make this concern clear so that it could be considered for whatever it's worth.

Fencer: Agreed, but the point is that if the lower-rated player has the advantage in a tournament game, this creates an incentive for players to intentionally lose non-tournament games in order to improve their chances in the tournament.

Fencer: In light of Andre Faria's comment, I think it would be better to use the value from the start of the game. Otherwise there may be a strong incentive for players to lose games.

Luke Skywalker:
I would expect only one, or possibly two if a single-elimination tournament can be defined with two games between each pair of opponents.

But I've observed that the number of slots they tell you to need often has little or nothing to do with the number of slots you actually need. Caveat emptor.