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 re(2): Response by Tristan)

Yes, of course...wait a minute...

 

Roll             Result

L                 2 of 3

W                W

So far, 1/2

 

LL               3 of 5

LW              -

 

LLW            -

LLL             4 of 7

LWL            3 of 5

LWW           W

So far, 1/2 + 1/8

 

LLWL          4 of 7

LLWW         -

LLLW          -

LLLL           5 of 9

LWLL          4 of 7

LWLW         -

Hmmm, 2 of these will win 3 of 5 on the next turn.  This would make the series start 1/2 + 1/8 +1/8 or 1/2 + 1/4.   I suppose I need to go figure out what I missed the first time, huh?  :)

Posted by Bob Smith on 2005-07-21 01:51:20