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?
This reminds me a lot of the gambler's ruin problem. It can be proven by counting probabilities
(not by me!) that, if you have a certain amount of money, and you play a game with some odds, you'll lose all your money if you play for long enough.
From the problem above, you can always play again (have unlimited money), but you can choose to quit as soon as you're ahead (so you opponent has 'finite' money). Eventually, you will pull ahead, adn then you can quit.
The same result is not produced if you don't have the option to quit; that game would go on infinitely because there are no boundary conditions.
Gambler's Ruin
