I have the perfect strategy to win any sort of game of chance. Just when I lose, I say , "2 out of 3?" and my opponent always accepts due to my infinite persuasiveness. If I lose again, I propose 3 out of 5, then 4 out of 7, etc.

Essentially, the effect of this strategy is that if the number of games I have won ever exceeds the number of games won by my opponent, then I win overall.

If I have a 50% chance to win any one game, what is the probability that I will eventually win overall (or rather, what does the probability approach)?

What if we play a game that involves a little strategy, and I only win 1/3 of the games?

If you are attempting to win w games out of 2w-1 games overall, the probability of you winning a particular set of games is 1/2^w (assuming you have lost the first w-1 games) for w= 1, 2, 3, ... follow from this formula. The terms for this series are 1/2, 1/4, 1/8, 1/16, ... and their sum converges to 1.

If instead there is only 1/3 chance of winning, the formula simply becomes 1/3^w which converges at 1/2. This means you will never win half of the time! Based on when your opponent will get suspicious around (or another infinitely persuasive clone shows up and... but we will leave infinity - infinity for another problem) or when you want to stop, it will be less than that.

Overall one could use the geometric series formula to find a sequence for a probability of winning p. Since the probability of winning a particular game with probabilility p is p^w, the sequence (starting at p) is p/1-p chance at winning the game. This also means that if you were better at the game than 1/2 (like 2/3) you would win even if you continued. (This also makes sense from the opponen's perspective, he won't win sometimes just because of having a less than half chance of winning.)

Posted by Gamer on 2005-07-17 05:39:56