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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 response | Comment 16 of 18 |
I'm actually going on vacation quite soon, so regretfully, I must leave you all off with some last words on this puzzle, at least until I return.

You all seem to be reaching the correct solution, at least for the first question.  It should not be too hard to apply the same idea for the second question (which, by the way, has a different answer).

If you find it too difficult to reapply the same method to the second question, try to find easier methods.  There are many, many ways to solve this one.  I myself used a completely different series to get the answer.  And please, have fun!

And on that note, I'm off, or at least I will be tomorrow.

  Posted by Tristan on 2005-07-22 21:14:22
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