Thanks for the link. I would say the positive expectancy system became negative after the inclusion of $7 commission per trade. Hence it is not an example of a positive expectancy system going to ruin, but of a negative expectancy system which we already know will go to ruin at some point. Leaving that aside, what is the practical answer? Since we know that at some point positive expectancy systems will lead to ruin, what to do? Trade small and have an overall cumulative profit target?
No.... let's put it this way, if the probability of ruin over 10Y for your system is 0.0001%, do you care? Should you do anything differently? I think retail system traders are far too complacent about the ideas of placing profit targets and stop losses - these tools materially change the returns profile of a strategy is an extremely complex way.
I think you did not understand what you read. The only difference in the link I gave you between ruin and a high return was the starting capital. The expectancy was the same. When are you going to get these facts straight people? Example: fair coin toss. You make $2 with heads and lose $1 with tails. Expectancy is $0.5 If you start with only $2 your probability of ruin is 0.25 right away and 0.9999 in the next 100 tosses or so. If you start with $10, your starting probalility of ruin is (0.5)^10, very low and in the next 100 tosses it is only 0.04 (don't ask me how I got that). Expectancy is the same. Expectancy assumes infinite capital as it is only valid at the limit of large numbers.
Visaria, the one thing you should realize foremost is that Bill is almost never right. I gave a derivation of the Gambler's Ruin formulae in this thread. So if Bill is right, those formulae must be wrong. Go over that derivation and see if you can spot any errors. You won't find any but it's best that you verify this for yourself. Then go over Bill's childish formula again and you'll spot the following error: he says let p be the probability of ruin. If so, that's it. He's done. Taking p^n or (1-p)^n makes NO SENSE because the risk of ruin is an end product, not an intermediate product of be further manipulated. To repeat: IF p is Bill's symbol for the risk of ruin, then p^n makes no sense because the risk of ruin already accounts for multiple trials. Using such simple-minded notation has confused the imbecile into thinking there's more manipulation to be done, but there isn't. Remember: if ever you find yourself saying "intradaybill is right", doublecheck your steps leading to that conclusion because the odds are good that there's an error along the way. More evidence: as you rightly pointed out to imbecilebill, the example he linked to was a negative-expectation system due to commissions. But notice how he wants to turn that back on you by insisting it is a positive-expectation system and that "expectancy" (the correct word is expectation) can only be measured in the infinite.
I think Bill is correct about you not understanding probability theory. "IF p is Bill's symbol for the risk of ruin, then p^n makes no sense because the risk of ruin already accounts for multiple trials. " No, you don't understand probability theory. Bill argues that the probability of ruin asymptotically approaches 1 as time goes to infinity. Given any bankroll B, the probability that a series of consecutive losers will occur that will completely wipe it out approaches 1 as time grows large. If you don't understand this, you ought to refrain from posting in these threads. You wrote: "the example he linked to was a negative-expectation system due to commissions. " But it turned to a positive expectation system by changing the starting capital to $100,000 So where is the starting capital in the expectancy formula? E = p x avg. win - (1-p) x abs(avg. loss) Where do you see the starting capital there? It is no there. Bill tried to explain to you that the expectancy is realized at the limit of many trials but you don't seem to understand that. Expectancy can be positive, yet the system gets ruined after limited trials because of lack of capital. Are you that stupid after all? It appears that you not only lack education in this area but you are stupid too. DontMissThe Bus was right you are a crank. I would add, "a major one".
Interesting...if you have 8 consecutive losses straightaway, your prob(ruin) changes from 0.04 to 0.9999 in the next 100 tosses? What if you have only 4 consecutive losses that brings you to $6(point of capital path), what is the change in prob(ruin) for the next 100 tosses?
No, timmy, I mean Bill, I mean timmy, YOU don't understand. Bill's argument is wrong because, as I patiently explained, p^n or (1-p)^n makes NO SENSE because p (the "risk of ruin" in Bill's notation) IS ALREADY BASED ON MULTIPLE TRIALS. So you and he (assuming you're not just his sockpuppet) are wrong. Insanity is often defined as doing the same thing over and over and expecting a different result. I would say this definition also applies to a certain form of stupidity, the kind exemplified by you and Bill. Simply repeating the same nonsense over and over doesn't eventually make the nonsense correct, but it does emphasize what uneducable fools the two of you are.
Right you are indeed, my friend, right you are. It's definitely insane for anyone to try to talk any sense with kut2k2; I advise everyone to just stop trying. It's pointless. There's no chance he's going to recognize he's doesn't understand probabilities, or basic math, or basic boolean logic.
If you set T = oo I think you get: R{oo|X|C}= ((q/p)^(C-X) - (q/p)^oo)/(1 - (q/p)^oo) If q > p then we get R{oo|X|C}}= ((q/p)^(C-X) - oo)/(1 - oo) which for (C-X) < oo is equivalent to: R{oo|X|C}= ((q/p)^(C-X) - oo)/(1 - oo) = oo/oo Now, the limit is not defined as oo/oo doesn't equal to 1. This may imply one or more of the following: 1. The limit can be any number for T = oo 2. The method followed doesn't apply at the limit of oo 3. The method followed doesn't apply at all
Well done. Add the fourth possibility with P =1 4. Kut2k2bunka is an idiot and a crank who divides inf by inf and gets 1.