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

Is this correct : fair game

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

Then if A begins, p(A wins)=[tex] p+(1-p)(1-q)p+...\frac{p}{p+q-pq}[/tex]...which is A wins at first, A loses, B loses, A wins, aso....

where as :

p(B wins)=[tex] (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 (there are no open rounds)..then the golden ratio is obtained as fair for the whole game, but of course not the single round....

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.... ?

well, it was not question of A and B both wins, because by construction, if A wins, then B loses, hence P(A and B wins)=0..if....but there can be that neiter A nor B wins, hence [tex]P(A and B lose)=p_{open}[/tex] ..but this does not mean that p(a and b win)<>0 which would mean not independent......what you mean is I suppose if the previous round influences the next one (if it is boxing, it is obvious...): p(A=winner(n+1)¦A=winner(n))....but here it's taken as independent.... rachmaninoff 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).

But I was in fact not speaking about theory, but on the basics : you say ok we make a faire game with a coin..it's fair ?? Yes, but if we repeat the coin tossing, it's not anymore...so is it fair or not ?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Is this correct : fair game ?

**Physics Forums | Science Articles, Homework Help, Discussion**