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...(adsbygoogle = window.adsbygoogle || []).push({});

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.... ?

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# Is this correct : fair game

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