Dishonest Strategy (Posted on 2005-07-16)

	Submitted by Tristan
Rating: 4.0000 (2 votes)
Solution:	(Hide)
First, there is 1/2 probability that I will win the first game, and therefore win overall. Otherwise I am one win behind. If I lose the first game, then consider every game afterwards in pairs. For each pair, I have a 1/4 chance of winning twice, 1/4 chance of losing twice, and 1/2 chance of breaking even. When we break even, it does not affect us at all, so we can ignore those pairs. Completely ignoring the times we break even, we each have a 1/2 chance of winning the next pair. If I win the next pair I win overall. If my opponent wins, I am behind by 3 wins. If I lose the pair discussed above, we can consider the rest of the games in groups of 4 (ignoring tied pairs). Again, we can completely ignore groups of 4 that are overall ties. There is a 1/2 chance I will win the next group of 4, winning overall. If I lose, we can consider the following games in groups of 8, then 16, and so on. If I have a 1/2 probability of winning at each steps described above, I will eventually win 100% of the time, assuming an infinite number of games. In the game involving strategy, the calculations are the same, but the results are not necessarily the same. In the first game, I have 2/3 chance of losing. In the second step where I consider the games in pairs, I have 4/5 chance of losing the next non-tying pair (the probability for each pair is 4/9 to lose and 1/9 to win). In the third step where games are considered in groups of 4, I have a 16/17 chance of losing the next non-tying group of 4 (the probability for each group of 4 is 16/25 to lose and 1/25 to win). In general, the probability for the nth step will be 1-1/[1+2^(2n-1)]. It can be proven by induction that the product of the first n steps is 1/2 + 1/[2*2^(2n)-2]. This approaches 1/2 as n approaches infinity. Generally speaking, it can be shown by generalizing the above proof that for any probability p of winning a single game, the probability of winning overall is p/(1-p). It should be noted that there are many ways to get to the solution, and this is not necessarily the simplest.

	Subject	Author	Date
	re: More - part 2	Tristan	2005-08-10 06:22:16
	More - part 2	Bob Smith	2005-07-25 13:48:28
	More response	Tristan	2005-07-22 21:14:22
	re: Third time's the charm?	Bob Smith	2005-07-22 18:59:43
	Hmm...	Jesse	2005-07-22 15:34:52
	Third time's the charm?	Bob Smith	2005-07-22 14:54:43
	New idea	Gamer	2005-07-21 20:57:12
	My mistake in reasoning	Gamer	2005-07-21 15:22:26
	Gambler's Ruin	Devin Mahnke	2005-07-21 06:04:46
	re(3): Response	Bob Smith	2005-07-21 01:51:20
	re(2): Response	Tristan	2005-07-20 05:17:46
	re: Response	Bob Smith	2005-07-19 23:54:22
	Response	Tristan	2005-07-19 23:27:35
	Solution (not 100%)	Bob Smith	2005-07-18 13:58:17
	To win overall:	Gamer	2005-07-17 05:39:56
	re: less than 100%	Charlie	2005-07-16 18:04:41
	less than 100%	Larry	2005-07-16 14:53:36
	Just a guess	e.g.	2005-07-16 13:41:20