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        riddles >> general problem-solving / chatting / whatever >> game
        
(Message started by: BenVitale on Feb 19^th, 2008, 9:15pm)

Title: game
Post by BenVitale on Feb 19^th, 2008, 9:15pm

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?

Title: Re: game
Post by towr on Feb 20^th, 2008, 1:07am

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).

Title: Re: game
Post by BenVitale on May 13^th, 2008, 2:57pm

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?

Title: Re: game
Post by Grimbal on May 14^th, 2008, 8:40am

No :P.

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.

Title: Re: game
Post by BenVitale on May 14^th, 2008, 9:48am

on 05/14/08 at 08:40:53, Grimbal wrote:

Sorry, I don't understand your solution.

Title: Re: game
Post by BenVitale on May 14^th, 2008, 11:18am

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 ?

Title: Re: game
Post by Eigenray on May 14^th, 2008, 12:51pm

But there are only X/2 (or (X-1)/2) matches.

Title: Re: game
Post by BenVitale on May 14^th, 2008, 2:07pm

on 05/14/08 at 12:51:07, Eigenray wrote:

But there are only X/2 (or (X-1)/2) matches.

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?

Title: Re: game
Post by Grimbal on May 14^th, 2008, 2:31pm

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.

Title: Re: game
Post by BenVitale on May 14^th, 2008, 4:16pm

on 05/13/08 at 14:57:19, BenVitale wrote:

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?

Title: Re: game
Post by temporary on May 18^th, 2008, 6:07pm

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.

Title: Re: game
Post by BenVitale on May 18^th, 2008, 6:19pm

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?

Title: Re: game
Post by rmsgrey on May 19^th, 2008, 10:48am

on 05/18/08 at 18:19:13, BenVitale wrote:

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.