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?
Okay, I've been wrong twice, so let's go down swinging... (Thank you Tristan for an entertaining problem!) :)
Borrowing a page from Charlie, I fired up the computer
1 of 1
Winner : W
2 of 3
Winner : LWW
3 of 5
Winner : LWLWW
Winner : LLWWW
4 of 7
Winner : LWLWLWW
Winner : LWLLWWW
Winner : LLWWLWW
Winner : LLWLWWW
Winner : LLLWWWW
5 of 9
Winner : LWLWLWLWW
Winner : LWLWLLWWW
Winner : LWLLWWLWW
Winner : LWLLWLWWW
Winner : LWLLLWWWW
Winner : LLWWLWLWW
Winner : LLWWLLWWW
Winner : LLWLWWLWW
Winner : LLWLWLWWW
Winner : LLWLLWWWW
Winner : LLLWWWLWW
Winner : LLLWWLWWW
Winner : LLLWLWWWW
Winner : LLLLWWWWW
So at stage i, there are w_i possible winners, each with a chance of 1/2^(2i-1) of occurring.
w_i = {1,1,2,5,14,42,132,429,1430,4862,16796,58786,...}
This yields, using up to "stage" 12 of 23 so far, a probability of 83.88% of winning. Please note that the terms of the series are not insignificant even at this "stage", and this computer estimate should not be considered converged.
Third time's the charm?
