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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: ( You must be logged in to post comments.)
  Subject Author Date
re: More - part 2Tristan2005-08-10 06:22:16
More - part 2Bob Smith2005-07-25 13:48:28
More responseTristan2005-07-22 21:14:22
Some Thoughtsre: Third time's the charm?Bob Smith2005-07-22 18:59:43
Hmm...Jesse2005-07-22 15:34:52
Some ThoughtsThird time's the charm?Bob Smith2005-07-22 14:54:43
New ideaGamer2005-07-21 20:57:12
My mistake in reasoningGamer2005-07-21 15:22:26
Gambler's RuinDevin Mahnke2005-07-21 06:04:46
re(3): ResponseBob Smith2005-07-21 01:51:20
re(2): ResponseTristan2005-07-20 05:17:46
re: ResponseBob Smith2005-07-19 23:54:22
QuestionResponseTristan2005-07-19 23:27:35
SolutionSolution (not 100%)Bob Smith2005-07-18 13:58:17
To win overall:Gamer2005-07-17 05:39:56
re: less than 100%Charlie2005-07-16 18:04:41
less than 100%Larry2005-07-16 14:53:36
Some ThoughtsJust a guesse.g.2005-07-16 13:41:20
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