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Go-Games
Posted: December 26th, 2006, 4:08 pm
by cublai
In a Go-Games between two teams, say Team C and Team J, the following rule holds:- Each team has exactly 7 members that are sorted in a predetermined order 1-7. Members and orders are not allowed to be changed during the games.- When the games starts, C1 plays against J1. Whoever loses is out. Then the next game will be the winner against the next member in the losing team. For example, C1 defeats J1, then J1 is out. The next game will be C1 playing against J2. And so on ...- The games are over when all members of one team are out.Question: Once the games starts, how many paths are there that leads to an end?
Go-Games
Posted: December 26th, 2006, 10:12 pm
by bhutes
Go-Games
Posted: December 26th, 2006, 10:36 pm
by BayAreaFan
I think it is 2 * SUM ( i = 0 to n -1) (n+i) C iThere can be cases where the losing team wins anything from 0 to n-1 games. So the number of games is between n & 2n - 1. In each situation either of the teams can win so there is a factor of 2. Then it just becomes a combination of two groups of (n) and (i) each.
Go-Games
Posted: December 27th, 2006, 8:20 pm
by cublai
You can try: When n=2, your formula gives 8 but there are actually only 6 paths.
Go-Games
Posted: December 27th, 2006, 10:32 pm
by gentinex
Suppose that side C loses all their members before side J. Then count separately the cases for the number of losses by side J. There is exactly one way for J to lose zero members. There are 7 ways for J to lose one member (before any of the seven members that C loses). There are exactly (8 choose 2) ways for J to lose two members (this is the number of ways of ordering the first six members of C, and the two losing members of J). Then (9 choose 3) ways for J to lose three members, and so on up to J losing six members. Thus, the total number of ways in which C can lose is(6 choose 0) + (7 choose 1) + (8 choose 2) + ... + (12 choose 6) = 1710Double this to also account for the number of ways in which J can lose, and you get 3420. It's easy how to see to extend this to n being something other than 7, and in particular n = 2 works out as 2((1 choose 0) + (2 choose 1)) = 6, as desired.
Go-Games
Posted: December 28th, 2006, 1:36 am
by cublai
Almost right except it should be 1716*2 = 3432.In fact this question has a simpler form of solution with a simpler method. (Note that 3432 is 14 choose 7.)The reason that I put 7 here is that these rules are actually real rules in the annual "LeiTai" Go-Games happening between China and Japan.
Go-Games
Posted: May 5th, 2007, 10:52 am
by giraffe
There can be a geometric interpretation here.Imagine Eucledean coordinate grid. Then it can be seen that the required number of go-game scenarios with $n$ players in each teamis the number of paths along the grid between points (0,0) and (n,n) that go in either LEFT or UPPER directionand can change it in the points with both integer coordinates only.Then it is clear that the number of scenarios is the binomial coefficientC_{2n}^n
Go-Games
Posted: May 5th, 2007, 10:55 am
by giraffe
The only correction.Paths can go in either RIGHT or UPPER directions