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?
(In reply to
Third time's the charm? by Bob Smith)
The next couple of terms are:
{208012, 742900, 2674440, 9694845, 35357670, 129644790}
That's up to 18 of 35. The terms are still not negligible and the total percentage is up to 86.79%. The terms are about linear on a log graph and extrapolating out to 42 of 83 gives a total percentage of 90.45%, just so you get an idea of total growth. The final increment is about 5E21 possible wins out of 10E24 total.
I guess Gamer made the conservative assumption that you lose the first n games and showed that, in the limit, you still win 100% of the time. Not making that assumption should certainly not mean that you win less, and it still appears to be heading in that direction.
re: Third time's the charm?
