Posers and Puzzles
11 Aug 19
I have to arrange a small tournament (bowls if you're interested). There are 12 entrants and each player has to play 5 games, each against a different opponent. That wasn't difficult to arrange using random selection. But if there had been an odd number of entrants, would it be possible for them each to play an odd number of games? I can't see that it would. I think it would have to be an even number of games. Am I right?
@c-j-horse saidDo all of the games have to be played simultaneously?
I have to arrange a small tournament (bowls if you're interested). There are 12 entrants and each player has to play 5 games, each against a different opponent. That wasn't difficult to arrange using random selection. But if there had been an odd number of entrants, would it be possible for them each to play an odd number of games? I can't see that it would. I think it would have to be an even number of games. Am I right?
12 Aug 19
@c-j-horse saidSurely you have to arrange the tournament whether we are interested or not? 😉
I have to arrange a small tournament (bowls if you're interested).
@c-j-horse saidThat is correct.
But if there had been an odd number of entrants, would it be possible for them each to play an odd number of games? I can't see that it would. I think it would have to be an even number of games. Am I right?
An odd number of participants cannot all play an odd number of games.
The total number of games played is (participants)*(games)/2
@wolfgang59
Thank you for confirming that. And also for the quick formula to calculate the total number of games. I used a far more complicated method to get to the same thing.
@c-j-horse saidHave you looked at Berger Pairings charts? They might be of some help.
I have to arrange a small tournament (bowls if you're interested). There are 12 entrants and each player has to play 5 games, each against a different opponent. That wasn't difficult to arrange using random selection. But if there had been an odd number of entrants, would it be possible for them each to play an odd number of games? I can't see that it would. I think it would have to be an even number of games. Am I right?
Google it and you should be able to find details and sample charts.
Their charts cover odd and even group sizes.
The individual round match-ups are listed out for the full even number group size, but if the actual group size is one less, then in each round one player sits out for that round when matched up against the 'blank' player. (hope that makes sense)
You specified that each player was to play 5 games. Was that a total games limit or just a minimum limit for each player?
If you did 'round robins', you could divide your 12 players into 3 groups of 4 players, but then each player would end up playing 6 games each, but only against the 3 other players within their group. (2 games against each, one as black and one as white)
Another round robin option would be to have two groups of six, and each player would play just a single game against every other player in their group. You would need to try and balance out the games for playing black or white, but you would get your 5 games that way. Hopefully each player would end up with 3 playing one color and 2 as the other.
Hoping this may have helped in some way. 🙂
@wolfgang59 saidCould an odd number of players be accommodated if the requisite number of games was increased to six (13*6/2)?
That is correct.
An odd number of participants cannot all play an odd number of games.
The total number of games played is (participants)*(games)/2
@handyandy saidYes.
Could an odd number of players be accommodated if the requisite number of games was increased to six (13*6/2)?
It is only an odd number of playing an odd number of games which is not possible.
@wolfgang59 saidThe formula requires a number (players*games) divisible by two.
Yes.
It is only an odd number of playing an odd number of games which is not possible.