Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">Sorry, no offense intended...

n are the total number of players in the tourney
r are the number of players in the group (i.e. 3 for your case)

x! (spoken "x factorial") is defined as x! = x(x-1)(x-2)(x-3)...(3)(2)(1)

So 10! = 10x9x8x7x6x5x4x3x2x1

C(n,r) means C is a function of the variables n & r.

Bottom line:

The number of ways to group n people in Groups of r (i.e. the total number of games required to have everyone play everyone else in one and only one game) is C(n,r) = n!/[r!(n-r)!]

I seriously recommend you think up a better tourney. For 10 people, you need 120 games.

Slick.

Re: Math problem
 

Nevermind. Forget everything I said. C(n,r) is the wrong formula for this. Sorry for the confusion. http://forum.shrapnelgames.com/image...s/rolleyes.gif

Slick.

Re: Math problem
 

Outside of being an interesting mathematical series problem, the general question is void. It would be incredibly difficult to get 27 players in a round-robin style game. It would take months if not years for every combination to be played out. I'd recommend a bracket style where only the top one or two players advance to the next round. Which is how the NCAA tournament decides a winner in just 6 rounds out of a pool of 64 teams.

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">Not really. I suppose it depends on the players in the tourney though. Months I can see, but it should not take years.

27 players in games of three, where each player must play every other player once and only once. Each player only has to play 13 games total. That's a lot, but you can play them simultaneously, or at least 3 or 4 at a time. And 3 man games of SE4 go pretty quick.

But that would be a lot of games nonetheless. It would be hard to find 27 people willing to do that many probably.

Geoschmo

Re: Math problem
 

One of the good things about chess and its rating system (see my remark about one possible way to have a rating system in SE4 in the "New League" thread) is the wonderful way tournements can be run. Using the rating of the player, you just use Swiss-System pairing and get a tourney done in 4-5 rounds...considering anywhere from 15-30 people.

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">That's allright. I got Fyron to explain to me offline how that formula worked and it is interesting. It's close, but as you say not exactly right. It looks like it's counting all possible combinations, instead of each player facing every other player once and only once.

Maybe there is no forumla and LGM's way is the solution. Or maybe there is one and we'll all get the Nobel prize for mathematics for working it out. http://forum.shrapnelgames.com/images/icons/icon7.gif

Re: Math problem
 

There is certainly a formula for it, we just do not know it. Any professional mathematicians out there? http://forum.shrapnelgames.com/images/icons/icon10.gif

Re: Math problem
 

Choose 2 from N, since order is unimportant.
On my calculator, that's the "nCr" button
For N=27, that's 351 games to be played round-robin.

<font size="2" face="Verdana, Helvetica, sans-serif">No; Player 1 goes up against all 26 other players. Similarily for the others, so that's 26 games total for each player. http://forum.shrapnelgames.com/images/icons/icon12.gif

26games*27people = 702 plays
That makes for 351 two-player games.

[ August 06, 2003, 22:11: Message edited by: Suicide Junkie ]

Re: Math problem
 

But, each player only plays the others once each, so it does not work for larger games than 2 player ones.

Re: Math problem
 

More than two players would get you problems with teaming up, I imagine.