# Is it possible?

C J Horse
Posers and Puzzles 11 Aug '19 08:34
1. C J Horse
A stable personality
11 Aug '19 08:34
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?
2. HandyAndy
Non sum qualis eram
12 Aug '19 01:31
@c-j-horse said
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?
Do all of the games have to be played simultaneously?
3. wolfgang59
Mr. Wolf
12 Aug '19 03:52
@c-j-horse said
I have to arrange a small tournament (bowls if you're interested).
Surely you have to arrange the tournament whether we are interested or not? ðŸ˜‰
4. wolfgang59
Mr. Wolf
12 Aug '19 03:55
@c-j-horse said
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?
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
5. C J Horse
A stable personality
12 Aug '19 08:43
@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.
6. mwmiller
RHP Member No.16
12 Aug '19 13:211 edit
@c-j-horse said
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?
Have you looked at Berger Pairings charts? They might be of some help.
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. ðŸ™‚
7. HandyAndy
Non sum qualis eram
12 Aug '19 16:27
@wolfgang59 said
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
Could an odd number of players be accommodated if the requisite number of games was increased to six (13*6/2)?
8. wolfgang59
Mr. Wolf
12 Aug '19 21:19
@handyandy said
Could an odd number of players be accommodated if the requisite number of games was increased to six (13*6/2)?
Yes.
It is only an odd number of playing an odd number of games which is not possible.
9. HandyAndy
Non sum qualis eram
12 Aug '19 21:43
@wolfgang59 said
Yes.
It is only an odd number of playing an odd number of games which is not possible.
The formula requires a number (players*games) divisible by two.