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?
You will win at least 50% of the games on the first try
If you lose the first toss/roll/etc., then you go to "2 out of 3", at which point you must win twice to win overall. You only have a 25% chance of doing that which means you will win 1/8 (overall) of the games on 3 tosses/rolls/etc. (in addition to the 50% won on the first roll)
So the sum is not 1/2 + 1/4 + 1/8 ... It's 1/2 + 1/8 + 1/32 which converges to 2/3. The remaining games will continue on as you infinitely expand your opportunities to win.
Likewise, with a 1/3 chance of winning, the series converges to 3/8.
Edited on July 18, 2005, 9:35 pm
Solution (not 100%)
