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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.)
More - part 2 | Comment 17 of 18 |

My solution converges more quickly for the P_win = 1/3 case, but only to about 38.2%, not the 50% that Gamer suggests.


  Posted by Bob Smith on 2005-07-25 13:48:28
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