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Dishonest Strategy (Posted on 2005-07-16) Difficulty: 3 of 5
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?

See The Solution Submitted by Tristan    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Response | Comment 7 of 18 |
(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.

 


  Posted by Bob Smith on 2005-07-19 23:54:22
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