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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)

 Subject Author Date re: More - part 2 Tristan 2005-08-10 06:22:16 More - part 2 Bob Smith 2005-07-25 13:48:28 More response Tristan 2005-07-22 21:14:22 re: Third time's the charm? Bob Smith 2005-07-22 18:59:43 Hmm... Jesse 2005-07-22 15:34:52 Third time's the charm? Bob Smith 2005-07-22 14:54:43 New idea Gamer 2005-07-21 20:57:12 My mistake in reasoning Gamer 2005-07-21 15:22:26 Gambler's Ruin Devin Mahnke 2005-07-21 06:04:46 re(3): Response Bob Smith 2005-07-21 01:51:20 re(2): Response Tristan 2005-07-20 05:17:46 re: Response Bob Smith 2005-07-19 23:54:22 Response Tristan 2005-07-19 23:27:35 Solution (not 100%) Bob Smith 2005-07-18 13:58:17 To win overall: Gamer 2005-07-17 05:39:56 re: less than 100% Charlie 2005-07-16 18:04:41 less than 100% Larry 2005-07-16 14:53:36 Just a guess e.g. 2005-07-16 13:41:20

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