Math problem

[geoschmo]

Ok, I like math but my brain doesn't work well when you get past two dimensions.

I am trying to figure something out. Assume a hypothetical tournament which are all 3 player games. In the tournament everybody has to play everybody else once, but only once.

If there are three players it's easy, A,B,C. What's the next number of players that this format will work for? Can it be done with 9 or do you need 27, or is it some other number completely?

Is there an easy way to figure this out including who has to play who?

[geoschmo]

Ok, I figured it out the long way using graph paper.

So I guess then it's x squared, where x is the number of players per game. Not x to the power of x. So with 4 player games it would take 16 people in the tourney and 5 player games would take 25? Does that sound right? Is there a way to verify that without using graph paper?

[BBegemott]

Originally posted by Geoschmo:
I am trying to figure something out. Assume a hypothetical tournament which are all 3 player games. In the tournament everybody has to play everybody else once, but only once.

IMHO it can't be done with 3 player games with any number of players (except 3 of course).

Assume 9 players tournament. Nine players will be marked with numbers 1,2,...,9.
I could make only legal 10 triplets:

1) 123
2) 145
3) 167
4) 189
5) 246
6) 278
7) 259
8) 347
9) 369
10)358
11)48?
12)49?

There is no legal substitution for 11th & 12th triplet, because 4th player has already played 1,5,2,6,3,7. Putting any number instead of '?' contradicts to the rule 'everybody has to play everybody else once, but only once'.

Or have I missed some kind of hidden trick? I am really curious to see your solution

[tesco samoa]

hey geo i use that for figuring out how many tickets i need to do in betting on sport games

3 of 3 rotations ( 3 teams to 25) right up to 12 of 12 rotations

3 of 4 is 4
3 of 5 is 10
3 of 6 is 20
3 of 7 is 35
3 of 8 is 56
3 of 9 is 84
3 of 10 is 120
4 of 5 is 5
4 of 6 is 15
4 of 7 is 35
4 of 8 is 70
4 of 9 is 126
4 of 10 is 210
5 of 6 is 6
5 of 7 is 21
5 of 8 is 56
5 of 9 is 126
5 of 10 is 252
6 of 7 is 7
6 of 8 is 28
6 of 9 is 84
6 of 10 is 210
7 of 8 is 8
7 of 9 is 36
7 of 10 is 120
8 of 9 is 9
8 of 10 is 45

many different rotations...

[Z.e.r.o]

The number of matches you need is n!
where n! = 1*2*3*4*...*(n-1)*(n)
and is called factorial and can be used to determine the numbers of permutations of n tokens.

In a 3 players match where anyone is playing a role only once is 3! (1*2*3 = 6)

Empirical solution:
123
213
312
321
132
231

If I'm not right you can spank me with a crowbar

[tesco samoa]

umm mine was total games that would be needed to play each other in every combination...

ignore