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?

Instead of calling the games 2 out of 3? 3 out of 5? I will call them stages. Stage -1 represents winning (the goal stage) when the game stops. Stage X represents a time in the game when you have X more losses than wins or X net losses.

In this situation, when you are on stage N, you would have to go back and forth between the stages (in some order) until you go forward to stage N+1 or are able to go back N+1 stages to -1.

From looking at the way the approximations fell (using small N and not including games that weren't on stage -1 or N+1 by round 30) it seems there is a 1/(N+2) chance of going to stage -1 instead of stage N+1, but I don't know the proof of that.

From the way this looks, it means there's a:

1/2 (1/2) chance of winning on stage 0
1/6 (1/2*1/3) chance of winning on stage 1
1/12 (1/2*2/3*1/4) chance of winning on stage 2
1/20 (1/2*2/3*3/4*1/5) chance of winning on stage 3
1/30 (1/2*2/3*3/4*4/5*1/6) chance of winning on stage 4
1/42 (1/2*2/3*3/4*4/5*5/6*1/7) chance of winning on stage 5
(continuing on in the same way...)

These approach 1 when added up, so the probability of you winning is 1.

Posted by Gamer on 2005-07-21 20:57:12