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Is this correct : fair game

  1. Nov 24, 2005 #1
    Let suppose there are 2 opponents : A,B...for simplicity, Let suppose at each round A wins with prob. p, B with q..p,q, constant such that 0<p+q<1...

    Then if A begins, [tex] p(A=win)=p+(1-p)(1-q)p+...=\frac{p}{p+q-pq}[/tex] where as :

    [tex] p(B=win)=(1-p)q+(1-p)(1-q)(1-p)q+...=\frac{(1-p)q}{p+q-pq}[/tex]..

    Hence,p(A=win)+p(B=win)=1 even if p+q<1...However if A,B have at each round the same prob. of winning, then on summing rounds, the ones who begins has more chance to win ??? (but still we remember, those are juste probabilities...)

    However p(A=win)=p(B=win) (fair) => q=p/(1-p)...let suppose the game is a every round winner type one with p+q=1...this means, at each round, either A or B wins..then [tex] q=(1-q)/q=>q^2+q-1=0=>q=\frac{-1\pm\sqrt{5}}{2}[/tex] which is the golden ratio....

    if the game is a each round winner type, then it should be like counting the numbers of time who is the winner, since one of them surely wins.

    But if p+q<1 what is the context of the probability computation : based on : the first who wins, wins all...or even if he wins once, we continue, and count only determined issues (not open ones..)..p+q<1 could be determined by factors like : each round takes such time, if over, nobody won.... ?
    Last edited: Nov 24, 2005
  2. jcsd
  3. Nov 24, 2005 #2
    This is in the wrong subforum - it belongs in "Set Theory... Probability & Statistics".

    Anyways, your model is not good. Are the events "A wins" and "B wins" independent? They shouldn't be. What does it mean to say "A and B both win" (with probability p*q)? It's not meaningful.

    The general answer to your question is: Markov chains. It's too lengthy to explain here, so I suggest you look it up somewhere (assuming you're already good with linear algebra).
  4. Nov 25, 2005 #3
    Unfortunately I'm not very good with math so I can't help you with that.
    I can say though that I think every game gives an advantage to the player that goes first. The only exceptions to this that I know of are Go where the first player needs to win by a certain number of points which is supposed to be equivilent to the point advantage gained by being the first player. And Diplomacy where there isn't a first player because all players take their turns simultaneously.
    I'm probably not helping so I'll just shut up then. :smile:
  5. Nov 25, 2005 #4


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    I can think of many games where it pays to be last...even something as un-game-theory-like as OT in a College Football game.
  6. Nov 25, 2005 #5


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    kleinwolf : I think your calculation corresponds to an unrealistic game. Can you name a game where the opening player may win without the opponent even having gotten a chance ?
  7. Nov 25, 2005 #6
    Hmmm... I wasn't really considering sports.
    What sorts of games give an advantage to the second player?
  8. Nov 25, 2005 #7


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    (Most of the several variants of) knots and crosses, zertz, any game where you can learn from you opponent's move...ever played the really old computer game (I think it was written in BASIC) Gorilla ?
  9. Nov 25, 2005 #8
    Never played gorilla.
    I've never really played the others either but I know what sort of games they are more or less. I had always assumed that as long as player one knew what (s)he was doing they could negate any advantage the second player might find.
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