Swiss automatic number of games

[koningco]

Can anyone help me to understand how many games there will be when a tournament creator set the number of games in a Swiss tournament to automatic? Here is an example tournament: https://boardgamearena.com/tournament?id=538458 (but I want to understand how it's calculated for smaller/bigger tournaments as well)

[koningco]

Can anyone answer my question? There is another tournament with automatic later today: https://boardgamearena.com/tournament?id=541409

[RazorOz]

I would imagine the reason you aren't getting a response is no-one knows for certain exactly how it works, unless there's a formula listed somewhere. It's a relatively new feature right, and we can guess how it roughly works, but an actual formula, I suspect no-one knows as it would require a lot of testing of wide ranging tournaments to be sure.

[Meeplelowda]

I would imagine the reason you aren't getting a response is no-one knows for certain exactly how it works, unless there's a formula listed somewhere. It's a relatively new feature right, and we can guess how it roughly works, but an actual formula, I suspect no-one knows as it would require a lot of testing of wide ranging tournaments to be sure.

There is a formula that could be used: log_2(P) rounded up to the nearest integer, where P is the number of players in the tournament, as discussed in this suggestion. For those who don't want to do the math, the way it is supposed to work is
5 to 8 players: 3 rounds
9 to 16 players: 4 rounds
17 to 32 players: 5 rounds
33 to 64 players: 6 rounds
etc.

Those would be the correct values for two-player games if ties are not possible because it would guarantee one undefeated player after that number of rounds.

However, the implementation is not strictly following that formula even for two-player games, as discussed in this thread. And I'm not sure anyone has looked at how it works for higher player counts.

[dschingis27]

5 to 8 players: 3 rounds
because it would guarantee one undefeated player after that number of rounds.

I haven't thought much about it or thought about the other counts. But I noticed that your claim already does not work with 6 players.

Swiss with 6 players:
after 1st round: 3 players with 1 win.
after 2nd round: either 1 player with 2 wins or 2 players with 2 wins.
after 3rd round: either 0 players with 3 wins or 1 player with 3 wins.

In case there are 0 players with 3 wins after 3 rounds, there must be at least 2 players with 2 wins.

[menettm]

There is a formula that could be used: log_2(P) rounded up to the nearest integer, where P is the number of players in the tournament, as discussed in this suggestion. For those who don't want to do the math, the way it is supposed to work is
5 to 8 players: 3 rounds
9 to 16 players: 4 rounds
17 to 32 players: 5 rounds
33 to 64 players: 6 rounds
etc.

Those would be the correct values for two-player games if ties are not possible because it would guarantee one undefeated player after that number of rounds.

However, the implementation is not strictly following that formula even for two-player games, as discussed in this thread. And I'm not sure anyone has looked at how it works for higher player counts.

Just to continue till 512 players (maximum in BGA ?)
65 to 128 players: 7 rounds
129 to 256 players: 8 rounds
257 to 512 players: 9 rounds

[tbhp]

Kinda defeats the point of making it "automatic" if no one can tell what it means.

So if I understand correctly, setting the tournament to automatic means the tournament will go on until there is one single winner?

I guess that would be relatively easy to check in this case, but it'd be nice to have the confirmation that this is how it works.

[mrdoctor]

Kinda defeats the point of making it "automatic" if no one can tell what it means.

Make a suggestion that they add an explanation to demonstrate how it works when people are setting up tournaments, and in the tournament main page once set-up? https://boardgamearena.com/bugs

[Meeplelowda]

5 to 8 players: 3 rounds
because it would guarantee one undefeated player after that number of rounds.

I haven't thought much about it or thought about the other counts. But I noticed that your claim already does not work with 6 players.

Swiss with 6 players:
after 1st round: 3 players with 1 win.
after 2nd round: either 1 player with 2 wins or 2 players with 2 wins.
after 3rd round: either 0 players with 3 wins or 1 player with 3 wins.

In case there are 0 players with 3 wins after 3 rounds, there must be at least 2 players with 2 wins.

I worded that badly (both now and previously I guess). It guarantees there will be at most one undefeated player, not that there will be one. Unless you hit exactly 8, 16, 32, etc. players you aren't guaranteed to have an undefeated player. However, it avoids the outcome of having two undefeated people who never get to play each other for the championship.

But it is true that the third winner from round 1 who obviously isn't paired with one of the other two winners might lose round 2, in which case there won't be a match between two 2-win players in round three.

All that said, I don't think anyone has reported the automatic numbers to be wildly off, as in a tournament with 20 players going 7 rounds or a tournament with 65 players only going 4 rounds. There never seems to be too few rounds and at worst there only seems to be at most one "extra" round. It could have to do with how they are trying to cobble this formula on 3+ player games, or because they are allowing for in-game ties or something like that and want a chance for the tie to be broken by a match instead of the tiebreakers.

[tbhp]

I worded that badly (both now and previously I guess). It guarantees there will be at most one undefeated player, not that there will be one. Unless you hit exactly 8, 16, 32, etc. players you aren't guaranteed to have an undefeated player. However, it avoids the outcome of having two undefeated people who never get to play each other for the championship.

But even if two players have only one loss each, you may still want the situation to be resolved by a confrontation instead of a tie breaker. That's why in my previous post I assumed that under this system, the tournament would go on until there is only one single winner by points (which for 5 to 8 players, would mean potentially 4 games).

This would also avoid declaring the winner based on the "opponents points" metric used in swiss system tournaments, which is quite frankly nonsensical, as it involves factors that are outside your control.

All that said, I don't think anyone has reported the automatic numbers to be wildly off, as in a tournament with 20 players going 7 rounds or a tournament with 65 players only going 4 rounds.

I don't think anyone is saying it's off, people are just bothered because they don't know how many remaining games there are.