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Topic: PLAYERS WINNERS GAME (Read 388 times) |
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coolnfundu
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PLAYERS WINNERS GAME
« on: Nov 9th, 2004, 11:45pm » |
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With n number of players winning a game and a player being out if he looses m amount of games (applies universally), how many minimum games are needed to decide on the winner? Whats the maximum games which can be held to decide on the winner without any redundancy?
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Grimbal
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Re: PLAYERS WINNERS GAME
« Reply #1 on: Nov 10th, 2004, 1:04am » |
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::There must be at least m*(n-1) losses to eliminate (n-1) persons.
There can be at most (m*n)-1 losses or everybody gets eliminated.
And there is exactly one looser per game.::
This assumes that the relation A wins over B is not transitive. If it is, exactly one person will never loose, so whatever the value of m, he will never be eliminated and the question becomes how to find the "maximum" in a transitive relation.
Of course, if m>(n-1), the players must play multiple times against the same player. Is that what you call redundancy?
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coolnfundu
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Re: PLAYERS WINNERS GAME
« Reply #2 on: Nov 10th, 2004, 9:25pm » |
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say 100 players are there with one being out if he looses 2 matches so according to you we need a minimum of 198 matches ... to eliminate 99 people ...
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| « Last Edit: Nov 10th, 2004, 10:28pm by coolnfundu » |
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ThudnBlunder
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Re: PLAYERS WINNERS GAME
« Reply #3 on: Nov 10th, 2004, 10:04pm » |
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Quote:| the question is not how to eliminate but how to declare a winner ... |
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Listen carefully, coolnfundu.
You see that young lady over there, the only person who hasn't been eliminated?
Well, you just go up to her and say, "Honey, I declare you the winner!! And, would you believe it, the 1st prize happens to be a week's holiday in my hometown!" And don't worry if she appears somewhat crestfallen at this news. This will quickly vanish when she discovers that the 2nd prize is two weeks holiday in your hometown!
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| « Last Edit: Nov 11th, 2004, 4:54am by ThudnBlunder » |
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coolnfundu
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Re: PLAYERS WINNERS GAME
« Reply #4 on: Nov 10th, 2004, 10:14pm » |
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oye hoye ...
this is not what was meant ...
I meant how about doing the hard core insertions ..
Say 5 players with 1 losses and them being out of the tournament ... so we would need 1 vs 2, 2 vs 3, 3 vs 4, 4 vs 5 so putting original formula in here works rather well ...
Lets say two losses and you are out we would add a game for 1, 2, 3 and 4 each ... so lets see we get 8 matches I think ...
Ok the formula works I think ...
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ThudnBlunder
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Re: PLAYERS WINNERS GAME
« Reply #5 on: Nov 10th, 2004, 10:24pm » |
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Quote:how about doing the hard core insertions ..
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Nope, I'm afraid you will only be allowed to talk to her. 
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| « Last Edit: Nov 11th, 2004, 4:49am by ThudnBlunder » |
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coolnfundu
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Re: PLAYERS WINNERS GAME
« Reply #6 on: Nov 10th, 2004, 10:36pm » |
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okiii ... here is a variation of the question ...
There are two players with one dice each, each of them roll it in turn. If it comes same color, A wins, otherwise B wins. One dice has 5 red faces and 1 black face. If probability is conserved (Conservation of probability), then what should other dice hold on each face?
Damn peasy question ...
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ThudnBlunder
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Re: PLAYERS WINNERS GAME
« Reply #7 on: Nov 10th, 2004, 10:53pm » |
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:By the little-known Conservation of Probability Theorem, the die must have 3 red and 3 black faces.
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| « Last Edit: Nov 11th, 2004, 4:45am by ThudnBlunder » |
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Grimbal
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Re: PLAYERS WINNERS GAME
« Reply #8 on: Nov 11th, 2004, 1:55am » |
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on Nov 10th, 2004, 9:25pm, coolnfundu wrote:say 100 players are there with one being out if he looses 2 matches so according to you we need a minimum of 198 matches ... to eliminate 99 people ...
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Of course. If 99 people need to loose 2 times each, that is 198 games. I assume a game is one-to-one and there is one winner and one looser. Else, you should explain how a game is played, how many are playing and how many are loosing in a game.
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