Does Repetition Affect the Fairness of Probability Games?

[kleinwolf]

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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)=p+(1-p)(1-q)p+...\frac{p}{p+q-pq}...which is A wins at first, A loses, B loses, A wins, aso...

where as :

p(B wins)=(1-p)q+(1-p)(1-q)(1-p)q+...\frac{(1-p)q}{p+q-pq}...:
..

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... ?

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).

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 P(A and B lose)=p_{open} ..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...

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 ?

 

[EnumaElish]

When you write "A wins" you mean "A wins at least once," correct? Suppose A and B are rolling a die. A wins if 4 is rolled, B wins if 3 is rolled. But both lose if neither 3 nor 4 is rolled. So A can keep on winning round after round if there is a continuous run of 4's. I suggest you explain this using a concrete example.