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Topic: game (Read 628 times) |
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Benny
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Suppose, for instance, that scissors scores one point against paper, paper scores two against rock, and rock scores three against scissors. In this situation, would you automatically form a rock and hope to score three, or would you expect your opponent to form a rock, which you could beat by forming paper?
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towr
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Re: game
« Reply #1 on: Feb 20th, 2008, 1:07am » |
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I'd play randomly until I can determine my opponents strategy. If your goal is simply to beat your opponent, then it's a symmetric zero-sum game, and there is no a priori strategy to win.
If the goal is instead to get the maximum number of points, rather than beat your opponent, then I'd look for a tit-for-tat strategy. You'd want to alternate between rock and scissors. If your opponent doesn't play along, you can pick paper to decrease his gains (from 1.5 average to 1).
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Benny
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Re: game
« Reply #2 on: May 13th, 2008, 2:57pm » |
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In a tennis tournament there are X players. Let's assume that the initial pairings are done randomly, what are the odds that a certain pair will play each other?
It's easy to find that the number of possible pairings is X*(X-1)/2.
Since the initial pairings are done randomly, then the answer is:
(number of Matches) / (number of Pairings) = (X - 1) / X*(X-1)/2 = 2/X
The odds are 2/X
If there are 6 players, X=6, then the odds are:
(number of Matches) / (number of Pairings) = 2/6 = 1/3
Is this correct?
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Grimbal
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Re: game
« Reply #3 on: May 14th, 2008, 8:40am » |
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No  .
If there are 6 players, one player of the pair has 1/5 chances to play with the other player.
For X players, X even and >2, the chances are 1/(X-1).
For odd X, X>2, (assuming one player doesn't play), it is
(the probability to play)·(the probability to play with the other guy)
= (X-1)/X · 1/(X-1) = 1/X.
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Benny
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Re: game
« Reply #4 on: May 14th, 2008, 9:48am » |
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on May 14th, 2008, 8:40am, Grimbal wrote:No .
If there are 6 players, one player of the pair has 1/5 chances to play with the other player.
For X players, X even and >2, the chances are 1/(X-1).
For odd X, X>2, (assuming one player doesn't play), it is
(the probability to play)·(the probability to play with the other guy)
= (X-1)/X · 1/(X-1) = 1/X.
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Sorry, I don't understand your solution.
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Benny
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Re: game
« Reply #5 on: May 14th, 2008, 11:18am » |
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Do we not need to consider the number of possible pairings is X*(X-1)/2?
And since the initial pairings are done randomly, aren't the odds (number of Matches) / (number of Pairings), giving 2/X ?
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Eigenray
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Re: game
« Reply #6 on: May 14th, 2008, 12:51pm » |
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But there are only X/2 (or (X-1)/2) matches.
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Benny
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Re: game
« Reply #7 on: May 14th, 2008, 2:07pm » |
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on May 14th, 2008, 12:51pm, Eigenray wrote:| But there are only X/2 (or (X-1)/2) matches. |
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Suppose you have 4 people: A,B,C,D, then the number of matches are:
A with B : suppose A wins
C with D: suppose C wins
A with C: suppose A wins
That's 3 matches or (4-1) matches.
So, if we have X number of players, X needs to be an even number, the number of matches is X-1.
right?
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Grimbal
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Re: game
« Reply #8 on: May 14th, 2008, 2:31pm » |
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I see. I was only considering the first round. Sorry.
But the probability that a given pair plays depends a lot on the strength of these 2. The 2 best players are sure to play each other, eventually.
But if the outcome is random, I would say your reasoning is correct.
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Benny
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Re: game
« Reply #9 on: May 14th, 2008, 4:16pm » |
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on May 13th, 2008, 2:57pm, BenVitale wrote:In a tennis tournament there are X players. Let's assume that the initial pairings are done randomly, what are the odds that a certain pair will play each other?
It's easy to find that the number of possible pairings is X*(X-1)/2.
Since the initial pairings are done randomly, then the answer is:
(number of Matches) / (number of Pairings) = (X - 1) / X*(X-1)/2 = 2/X
The odds are 2/X
If there are 6 players, X=6, then the odds are:
(number of Matches) / (number of Pairings) = 2/6 = 1/3
Is this correct? |
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Formulae:
odds = probability / (1 - probability)
probability = odds / (1 + odds)
number of Matches) / (number of Pairings) = (X - 1) / X*(X-1)/2 = 2/X
Correct me if i'm wrong, but i think now this is a probability.
So, if p = 2/X, then the odds are:
odds = 2/X / (1 - 2/X)
odds = 2/X-2
If there are 6 players, X=6, then
odds = 2/6-2 = 2/4 = 1/2
Do you, guys, agree?
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temporary
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Re: game
« Reply #10 on: May 18th, 2008, 6:07pm » |
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The rps-321 game is not much different than regular rps. Winning takes more priority than getting more points since only the winning throw can get any points anyway.
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My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
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Benny
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Re: game
« Reply #11 on: May 18th, 2008, 6:19pm » |
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I'm still not satisfied with my own answer.
it seems likely that the worst player in the group will only play one game and be eliminated. Therefore it only makes sense to ask what the probability is that any specific player A is paired for the first game with any other specific player B, and this is clearly 1/(x-1)
What do you, guys, think?
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rmsgrey
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Re: game
« Reply #12 on: May 19th, 2008, 10:48am » |
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on May 18th, 2008, 6:19pm, BenVitale wrote:I'm still not satisfied with my own answer.
it seems likely that the worst player in the group will only play one game and be eliminated. Therefore it only makes sense to ask what the probability is that any specific player A is paired for the first game with any other specific player B, and this is clearly 1/(x-1)
What do you, guys, think?
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If you know how the two players you're asking about compare to the rest of the field, then it makes sense to take account of the effect of the initial pairings on when they get knocked out - which gets messy quickly for general pairings - but the average chance over all pairs of a given pair meeting during the tournament is going to be 2/X.
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