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 Dishonest Strategy (Posted on 2005-07-16)
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)

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 less than 100% | Comment 2 of 18 |
Consider the options immediately after losing the first game and saying "2 out of 3".  You play 2 more games.  If you win both, then you win overall.  If you split then the opponent wins 2 out of 3 overall, and you say "3 out of 5".  But if the opponent wins both, then you are 0 for 3, so saying "3 out of 5" doesn't help. You've already lost that.  So it depends on how dishonest your strategy can be.  If you can skip "3 out of 5" and go immediately to "4 out of 7", then your chances of winning approach 100% as the game approaches infinity.  And my guess is the same is true for the second part where you only win 1/3 of the games.

But if you can't skip; if once you're down 0 to 3 the game is over, then the answer is less than 100%.  But I haven't worked it out yet.
 Posted by Larry on 2005-07-16 14:53:36

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