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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.)
Some Thoughts Just a guess | Comment 1 of 18
In the first case, for each sequence that leads to your opponent's win, there is another (symmetric) that makes you win. Since whenever your opponent wins you make the game longer, it seems to me you must eventually win, so the answer in the first case would be 100%... how's my guess?
  Posted by e.g. on 2005-07-16 13:41:20
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