
Member 

Joined: Sep 28, 2016 Posts: 2089
Preferred Pronoun Set: he/him/his/his/himself

Fair enough, but, you know, converging infinite sums are a thing that exists.
THE PROOF, shortened and improved:
1) The probability of never having a positive win rate is equal to 1 minus the probabilities of every possible way of getting to a positive win rate for the first time. 2) To get a positive win rate for the first time, you need to get exactly one more win than loss, and to never have more wins than losses before. 3) This means that you can only get a positive win rate after an odd game, because after even games you'll have either 2+ more wins, 2+ more losses, or equal number of wins and losses. 4) For your first time of getting a positive win rate, you can get a 1/0, 2/1, 3/2, 4/3, ... win/loss ratios. Let's examine them one by one. 5) The probability of getting a 1/0 as your first positive win ratio is, obviously, exactly N, as you need to win your first game. (And there is only one way of doing that. This remark may seem nonsense now, but more on that later.) 6) The probability of getting a 2/1 as your first positive win rate is a bit more complicated. You need to lose your first game, win your second game, then win your first game. Let's write that sequence as LWW. The probability of this is (N1)*N*N, or (N1) * N^2, where N1 is a probability of losing. (Still, only one way of doing this.) 7) The probability of getting a 3/2 as your first positive win rate is different, because there is more than one way of getting to it. Two, to be exact: LWLWW, and LLWWW. The number of losses and wins is same for both ways, so their combined probability is equal to (1N)*(1N)*N*N*N*2 or (1N)^2 * N^3 * 2 8) The probability of getting a 4/3 as your first positive win rate is more complicated again, because there are five ways to get there  LLLWWWW, LWLWLWW, LLWLWWW, LWLLWWW, LLWWLWW. Again, their combined probabilities are (N1)^3 * N^4 * 5. 9) Everything to the power of 0 is 1, and everything to the power of 1 is itself. This means that the pattern of powers is clear: for n, where n is the number of wins in your first positive win rate, the probability of any one of the paths to it is N^n * (1N)^(n1). 10) If you continue calculating numbers of paths for possible first positive win rates, they will be 1, 1, 2, 5, 14, 42, 132, ... Those are known as Catalan numbers, and play an important role in combinatorics. Their usual, simplest formula is (2n)! / (n+1)! * n! , but it gives 1, 2, 5, 14... for n= 1, 2, 3, 4  the first 1 in the sequence is the 0th term (for n=0). This means we'll have to use (2n2)! / n! * (n1)!, which is the same, but with 1 subtracted from each n. 11) This means, the probability to get n/n1 as your first positive win rate is N^n * (1N)^(n1) * (2n2)! / n! * (n1)! 12) This means, the probability to get a positive win rate at any point is SUM(n = 1, 2, ..., +infinity)(N^n * (1N)^(n1) * (2n  2)! / (n1)! * n!). And yeah, this series converges for 0<N<1, and not just to "1 or 0". 13) This means, the probability to NEVER get a positive win rate at any point is 1  SUM(n = 1, 2, ..., +infinity)(N^n * (1N)^(n1) * (2n  2)! / (n1)! * n!)
PROOF OF SIMPLIFICATION:
Let S = probability of ever getting a positive win rate (our infinite sum). Obviously, S is not equal to 0. M = 1  S.
Also, let 0<N<1/2.
If N would be 1/2 <N<1, by the law of large numbers, S would be 1, because win rate will tend to be positive, and so it definitely will be positive for infinite number of games. Yeah, I don't like this either, but I don' t have anything better.
S = N + N^2 * (1N) + N^3 * (1N)^2 * 2 + N^4 * (1N)^3 * 5 + N^5 * (1N)^4 * 14 + ...
S * (1N) = N * (1N) + N^2 * (1N)^2 + N^3 * (1N)^3 * 2 + N^4 * (1N)^4 * 5 + N^5 * (1N)^5 * 14 + ...
S * (1N) / N = (1N) + N * (1N)^2 + N^2 * (1N)^3 * 2 + N^3 * (1N)^4 * 5 + N^4 * (1N)^5 * 14 + ...
(1N) + N * (1N)^2 + N^2 * (1N)^3 * 2 + N^3 * (1N)^4 * 5 + N^4 * (1N)^5 * 14 + ... is equal to S, but with N replaced by 1N (and vice versa).
Since 0<N<1/2, 1/2<(1N)<1
By the law of large numbers (ugh) this means that (1N) + N * (1N)^2 + N^2 * (1N)^3 * 2 + N^3 * (1N)^4 * 5 + N^4 * (1N)^5 * 14 + ... = 1
Then, for 0<N<1/2 ,
S * (1N) / N = 1
S = N/(1N)
M = 1  S
M = 1  N/(1N)
EDIT: typo fix.
_________________
Last edited by UselessCommon on Mon Mar 19, 2018 12:38 am, edited 3 times in total.

