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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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 re: less than 100% | Comment 3 of 18 |
(In reply to less than 100% by Larry)

" But if the opponent wins both, then you are 0 for 3"

But after losing the first (the circumstance of this) when 2 out of 3 was proposed, at the point of the second game it was 0 for 2, and it's at this point that the opponent claims his victory but the narrator proposes 3 out of 5. No skipping is needed for this to continue forever despite the opponent's victories.

 Posted by Charlie on 2005-07-16 18:04:41

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