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As we are talking elsewhere about tournament mechanics, I would like to toss one other topic out there for discussion by the powers that be...

For State and higher can we up the minimum number of required rounds to be at least one more than is listed currently. Simply put, if you play only the minimum number of rounds listed the tournament is effectively a single-elimination event. The tournament host's guide speaks all about the advantages of Swiss-style tourneys compared to single elimination, but all of that is lost if you only host the current minimum number of required rounds.

Although the same holds true for local and district tournaments, I don't feel so strongly about those because there are a lot more chances to pick up district and local points. There are, however, limited opportunities to pick up State and Regional points, and it would stink to have your entire tournament decided by one horrendous draw.

I agree entirely. I also think the allotted time for each action is far too generous, so we could kill two birds with one stone by drastically shortening allotted time to act and requiring at least one more round.

By having the perfect number of rounds to determine a winner, you don't give anyone a chance to take first place except for the undefeated. This is true of elimination tournaments as well, the only first place possibility is the undefeated. By playing extra rounds you allow more people to have a shot at first place, people that lost or tied a game somewhere along the way.

At smaller tournaments, its nice when you get to play all the players in a category.   

@ROBM: Perhaps I should have clarified. All of us top three players had played each other.  

...why is it preferred to have more rounds...

Mainly because people came to the tournament to play Redemption.  The more rounds, the more they get to play.  I also like having an extra round where someone gets to try to knock off the undefeated person in 1st.  I think both of these are good reasons.

Mainly because people came to the tournament to play Redemption.  The more rounds, the more they get to play.

That is not a reason that Swiss is no better than elimination for determining a winner, nor does it answer the question of how Swiss automatically means an undefeated person in first.  People could play two official rounds of Redemption and then play the rest of the day and all night, as far as I'm concerned, it doesn't answer the question of going away from the statistical ideal.

I also like having an extra round where someone gets to try to knock off the undefeated person in 1st.

Again, a winner in a Swiss tournament is not necessarily undefeated; countless tournaments have been won with imperfect records.

Quote from: DaClock on June 29, 2009, 08:29:59 PM

By having the perfect number of rounds to determine a winner, you don't give anyone a chance to take first place except for the undefeated.

This actually is not statistically true.

It actually is statistically true when the number of players is a power of 2 or one less than that. At that point if you lose one game you are out of contention for first (barring multiple time out wins by the undefeated player). So, at the recent MN State tournament where we had 31 T1-2P players, playing the minimum number of rounds it would have been equivalent to a one and done for players interested only in taking first.

It is also true, as Gil noted, that mathematically a Swiss system with the minimum number have a substantial chances of messing up the ordering of second and third places.

Lastly, I'm not sure comparisons of Redemption to either chess or bridge are particularly useful in this regard. Redemption has a much higher degree of randomness than either of those two games. If you are interested in identifying the "best" player, this randomness needs to be compensated for and the easiest way to do that is to add rounds.

more rounds eaqual more time that most host just do not have for the tournament day.

However, Swiss style pits the best player and the 2nd best player against each other in the second round. The 2nd best player looses and then, without the additional swiss round, doesn't have a chance to place. I have strongly felt all along that swiss style is not complete and does not demonstrate 2nd place-3rd place accurately unless additional rounds are played.

How do you figure that the person playing either of the top two players in the second round placing are the best players in that category?  It's just a result of the pairing after one game.   That's the point of a Swiss style, is that players face different competitors every match, and the people at the top are ones who beat more and better players over the course of the tournament.

It actually is statistically true when the number of players is a power of 2 or one less than that. At that point if you lose one game you are out of contention for first (barring multiple time out wins by the undefeated player).

I don't see how this proves that only an undefeated player can win a category.  You seem to fall into the same trap as everybody else, assuming that one (in the interest of completeness - and only one) person will always be undefeated.  Seriously, has no one here ever attended a tournament that has ever had a winner that did not go undefeated?  Ever?

Obviously Nationals is not, because it has extra rounds.

... in T1-2P only.

And why District and up?  Have you never attended a Local, or have you had locals with one-loss winners?

Only at Nationals. Every other tournament I've been to, District and up, has had an undefeated winner. And I don't think Natz is the tournament in question here.

I know mine is only one example, but just to throw that in there.

Sorry, but that is an untruth...    Last year at SE Regionals which you attended, the winner of T1 2p, Ben Shadrick, had one loss.  I know that because I handed it to him.

Outside of that tournament, however, I believe I can make the same argument, concerning T1 2p.

Kirk

Stephen, set up a four player two-round tournament (or an eight player three-round tournament or...) and use Swiss pairings. Show me how a player with one or more losses can possibly win the tournament (barring multiple time out wins by the undefeated player).

A four-player category could just play round-robin.  Do not confuse the question of whether Swiss MUST ALWAYS result in an undefeated winner, with the question of whether Swiss is ideal in all situations.

You are correct that the winner will be undefeated IF the number of players exactly matches the binary algorithm.  But a). how often does that happen when you get beyond 8 players, and b). in order for the winner of that category to go undefeated, he must beat an undefeated player in every single round.  What we seem to be forgetting about Swiss is that it pushes players who win consistently to the top, and so the players defeated in later rounds are not going to be also-rans that just lucked into winning, but people who are themselves in contention to win.  Why should we NOT reward the player who beat everyone he faced, including all the people who rose up to the #2 slot in the tournament?

This happens all the time when you play more than the minimum number of rounds, or when you are just slightly over the threshold for the next round bump (say 9 players forcing four rounds).

But it's a threshhold for a reason.  A four-round ten player tournament is no different than one with 14 or 16 players, in that four rounds will determine a winner, in most circumstances going undefeated is not 100% required, and in the one case where it is, the winner defeated everyone in front of him, no one else was undefeated, and there are so many different scenarios once you have 16+ players that the number of rounds before every one-loss player gets one crack at the champion could as much as double the play time.  Don't you have to pull the plug at some point?  I can see the logic in recommending extra rounds in certain situations but I don't agree with the idea of changing the rule to fit the exception.