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?

Both Gamer and Bob Smith have posted solutions, but they disagree with one another!

Both of you got to your answers by summing an infinite series.  I am not surprised your series are different; the answer can probably be represented as many different infinite series and products.  However, it is not entirely clear how you each got your series.

So, question, for both of you and anyone else:

What is the probability that I will lose the first game, lose the 2 out of 3, and then win 3 out of 5?

If I read correctly, Bob Smith says 1/32, and Gamer says 1/8.

Edited on July 19, 2005, 11:38 pm

Posted by Tristan on 2005-07-19 23:27:35