Another idiot heard from. Where did I claim the proof was my own? If you or any of the other google-challenged slackwits reading this had any clue, you'd have found this proof yourself pages ago. For the record, the theorem about the gambler's ruin was proven by Christiaan Huygens. As I already stated, this is old stuff. That was a huge hint to all but the google-challenged slackwits that infest ET.
I have met many cranks but never one like you. This was the problem statement: "Define: 1. A trading system that generates entries and exits with some frequency F > 0 per unit time and starts with bankroll B. 2. A Cummulative profit objective P 3. A time to quit for good objective T 4. The probability of ruin R such that bankroll B reaches 0. The questions (problems) are: (1) Given P = Pquit and/or T = Tquit are there conditions that may lead to R =1? In other words, even if there is a quit time or quit profit, under which conditions ruin R = 1 is certain? Please exlude the case where expectancy is negative. (2) Alternative, if there are no P and T constraints is ruin certain (R = 1)? I think (2) has been answered already. (1) remains. " You don't read the whole thread and you jump like an autistic child to sh*t all ovver the place. The above was the problem the other guys tried to answer you fuc*head. Where do you see the similiarity with your answer? Where is the universal qualifier you idiot? You keep on answering (1) when everybody has admitetd form start that the naswer was known.
Obviously kut2k2's problem formulation didn't account for total profit stop or an exit time so it is an answer to (2). If q < p and win = loss then the probability of ruin is >0. However, even that doesn't answer (2) completely. in (2) it is asked whether P of ruin < 1, not if P of ruin > 0. Obviously, 1 > 0. As the starting capital grows large at every step of the game, P of ruin goes to 1. In case (2) ruin is certain as T goes to infinity, something that has been answered already by DontMissTheBus I think. Thus it seems that kut2k2 did not provide an answer to the first question but stated the well-known fact that if a trading startegy has an edge p> q then it has positive expectancy if win = loss. That is what you proved kut2k2. I don't think everyone else is an idiot and maybe you should stop insulting all ET members here because it is impossible everyone else is wrong and you are correct.
Didn't insult ALL ETers, dude, just the google-challenged slackwits who have trouble thinking for themselves. I gave you a formula: P{T|S} = (1 - (q/p)^S)/(1 - (q/p)^T) This formula answers all of your questions, except the special case where p = q, which is certain ruin for trading/gambling without time or profit limits (T = infinity). There is a serious abundance of pantywaists in ET who need to be hand-held through every fucking trivial step of an idea before the lightbulb turns on, if it ever does. This does not describe everybody in ET, but it sure as hell describes far too many of you. If you're going to be a trader, man up and learn to think for yourself. Otherwise find a different pastime. All this fucking time and bandwidth spent on Gambler's Ruin. I wouldn't have believed it if I hadn't seen it for myself but I should have believed. After all, this is ET.
Can you tell me how does this answer the case of trend-follwing systems with win rate < 50% and avg. win >> avg. loss? The formula you gave assumes an edge in terms of win rate and also assumes win = loss. I think you underestimate the problem. This is not a problem about gambler's ruin. It is a problem about trading system risk of ruin. Why are you equating the two? I'm starting having questions about your comprehension ability.
It's hard for him to do that: he doesn't actually understand the math he googled and posted, nor the subject that was being discussed when he interjected. I'm done discussing this topic with that fella; He's clearly unhinged and I'm just wasting my time.