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?
(In reply to
Response by Tristan)
Without elaborating every possibility
Game results - Overall result
LLWWW - Win
LL(LWW) - any one loss will go to 4 of 7 - 3 options
LL(LLW) - any two losses drive to 5 of 9 - 3 options
LLLLL - drive to 6 of 11 - 1 option
So given two losses ( 1/4 chance ), only 1/8 chance to win 3 of 5. Overall, 1/32 chance to win 3 of 5.
re: Response
