Good, Thanks. Is there an optimal Nstop? I guess that could be where loss level% ~ 25% because at loss level=50%, it takes 100% recovery to bring back C to Cstart. i.e. From loss level > 25% onwards, it becomes increasingly hard to recover. Maybe Uoptimal is also found likewise?
Sorry, like I said, I don't have the drawdown formula. Apparently at that website you posted, they do it by MC simulation?
They claim its a combination of MC + kaufmann improved ROR formula at bottom of that webpage...cannot reconstruct/arrange in this case:eek:
http://www.automated-trading-system.com/wp-content/uploads/2010/05/RiskRuinFormula1.png However, if we solely consider Kaufmann improved ROR, it does include 'ruin level' in the formula as shown above. If we equate this ROR formula to the one you gave, we will have all N, U & ruin level%. From which, we can fix any 2 variables constant & determine the third optimal one.
Best estimate of drawdown is in the attached paper. You'll need the Q function, which is available on Magdon-Ismail's weibsite.
Before posting links to such naive websites, I propose you first check them out. I used 0.3 for win ratio and 2 for win/loss ratio. I get the incredible number of 209% for the risk of ruin !!! Now try plugging in 0.6 for the win ratio and 0.5 for the win/loss ratio. I get 1294% for Risk of Ruin. !!! If you want to make progress in your life do your own homework. You threw this link in the thread 4 or 5 times without even checking the results. hmm...
It is just a non-exact MC simulation so if it is > 100%, it means total ruin is assured(for the figures you stated). Before even talking about progress, improvise some flexible common sense.
You cannot have ruin probability > 100%. Probability cannot be > 1. But IMO he wasn't talking about the MC result, I believe. He pointed to the formula result.
I know but I am not picky on exact figures as long as I can roughly gauge which combos of inputs leads to total ruin... total ruin is still total ruin. Anyway, that is not the main objective which is to find something that describes the effects of consecutive/string of non-stop losses with bet size changes on the prob. of ruin...Is it better to totally stop, pause but maybe restart a system later or adjust bet size constantly after N nonstop losses? If any 70% accurate system makes 100 trades, what is the probability of 30 losses clustered almost together in the first 50 trades?
OK I found the formula for risk of drawdown and I've put it in the form of the formula in the OP : The risk of drawdown (RoD) is the probability of a trader's account falling by a specified fraction (d). RoD = the lesser of 1 and ((1 - edge)/(1 + edge))^(d*C/U) where C = current trading account d = drawdown, a positive number no greater than 1 U = (p*W² + q*L²)^½ p = win rate of the trading system q = loss rate = 1 - p W = average winning trade L = average losing trade edge = (p*W - q*|L|)/U If d = 1, then RoD = RoR (risk of ruin)