Let p be the probability that the event {Ruin} will occur (according to some stupid formulas you wrote down). 0 < p <1. The probability that the event does not occur in n trials is equal to p^n Now, if you made it past elementary school, you should now that as n --> inf p^n --> 0 with p < 1. Thus, Prob{ruin does not happen} -> 0 as n --> inf Therefore, Prob{ruin happens} --> 1 What you don't understand from the above? That is why you have to enforce a quit time or profit stop. That was the main question I posed which you still don't understand. You will never understand. Gambler's ruin formula is exactly that you stupid. It says that the longer the gambler stays in the game the risk of ruin goes to 1. Take your winnings and run you moron...
I'll post this once more for posterity, but not for the mathematically challenged like you, imbecilebill. You've already proven that you are logic-immune. This is a modification (my own) of the Gambler's Ruin theorem; the original was published by Christiaan Huygens in 1657. This is not a copy-and-paste job as several idiots have claimed. We want to calculate the risk of ruin for a gambler. We start with the simple situation of a gambler placing a series of even-payoff bets against a richer opponent, e.g., a casino. Let p be the probability of winning one betting unit. Let q be the probability of losing one betting unit. p + q = 1 Let R{T|X|C} be the probability of failure (aka risk of ruin) for a gambler with a current bankroll of C betting units who is attempting to reach a target bankroll of T betting units before falling to a surrender bankroll of X betting units. Let P{T|X|C} be the probability of successfully reaching target bankroll T before falling to surrender bankroll X. P{T|X|C} = 1 - R{T|X|C} P{T|X|X} = 0 P{T|X|T} = 1 P{T|X|C} = p*P{T|X|C+1} + q*P{T|X|C-1} (p + q)*P{T|X|C} = p*P{T|X|C+1} + q*P{T|X|C-1} P{T|X|C+1} - P{T|X|C} = (q/p)*(P{T|X|C} - P{T|X|C-1}) P{T|X|X+2} - P{T|X|X+1} = (q/p)*(P{T|X|X+1} - P{T|X|X}) P{T|X|X+2} - P{T|X|X+1} = (q/p)*P{T|X|X+1} P{T|X|X+3} - P{T|X|X+2} = (q/p)*(P{T|X|X+2}-P{T|X|X+1}) P{T|X|X+3} - P{T|X|X+2} = (q/p)²*P{T|X|X+1} P{T|X|X+n+1} - P{T|X|X+n} = (q/p)ⁿ*P{T|X|X+1} P{T|X|C} = SUM[ P{T|X|X+n+1} - P{T|X|X+n} ]_n=0...C-X-1 P{T|X|C} = SUM[ (q/p)ⁿ*P{T|X|X+1} ]_n=0...C-X-1 P{T|X|C} = P{T|X|X+1}*SUM[ (q/p)ⁿ ]_n=0...C-X-1 Case 1 : p = q = ½ P{T|X|C} = P{T|X|X+1}*(C-X) P{T|X|T} = P{T|X|X+1}*(T-X) 1 = P{T|X|X+1}*(T-X) P{T|X|C} = (C-X)/(T-X) R{T|X|C} = 1 - P{T|X|C} R{T|X|C} = (T-C)/(T-X) R{T|X|C} < 1 if C > X and finite T R{infinity|X|C} = 1 for all values of C Case 2 : p <> q P{T|X|C} = P{T|X|X+1}*(1 - (q/p)^(C-X))/(1-(q/p)) P{T|X|T} = P{T|X|X+1}*(1 - (q/p)^(T-X))/(1-(q/p)) 1 = P{T|X|X+1}*(1 - (q/p)^(T-X))/(1-(q/p)) P{T|X|C} = (1 - (q/p)^(C-X))/(1 - (q/p)^(T-X)) R{T|X|C} = 1 - P{T|X|C} R{T|X|C} = ((q/p)^(C-X) - (q/p)^(T-X))/(1 - (q/p)^(T-X)) R{infinity|X|C} = 1 if q > p R{infinity|X|C} = (q/p)^(C-X) if q < p R{infinity|X|C} < 1 if q < p AND C > X Bill, your insistence that the risk of ruin with no target limits ( R{infinity|X|C} ) is a certainty regardless of expectation has been thoroughly disproven. Everybody but you realizes this. If you have any sense of shame, you'll abandon your foolish position.
If you have no shame at all and you want to distinguish yourself from a stubborn 3-year old with asperger syndrome you should stop confusing infinite time with infinite target capital. You have shown with several mistakes in parroting some other proof you found that for q < p the probability of reaching infinite capital before going broke is essentially R{infinity/C} = 1 - (q/p)^C Does this say what will happen if the gambler plays forever? This equation is of no practical significance because there is no infinite bankroll and all markets have liquidity constraints. It is an academic exercise. Try to understand the problem you moron. It has nothing to do with accumulating infinite capital before getting broke. It has to do with what happens if you keep on playing forever in real life where infinities do not exist.
Infinite target capital takes infinite time to accumulate, slow learner. Even a child can figure that out, but not you evidently. Your entire thread was an academic exercise. You just aren't smart enough to recognize that. You can't even parrot my nomenclature without doing a retarded screw-up of it. Just like a shortbus-riding paste eater. Back on ignore for you, uneducable fool. You are a proven waste of time, space and good oxygen.
Where have you seen infinite capital you autistic child? You still cannot comprehend that these two are different problems you moron. When you are cornered you just insult. You are now plonked for good. Please don't come to any threads I start in the future because you will prove you are autistic. Plonk
intradaybill - give it up; This guy is clearly unbalanced and sits at the front of the short bus; Getting into such an argument with such people only serve to make you regret how terribly silly you are for arguing with such people.
Reaching infinite capital is not the same as playing for an infinite time. You can reach infinite capital in time t = 1 units if the profit function has the form: N = b+ at/(t - 1), a > 0, b > 0 Then you can get ruined at t = 2 if the profit function assumes the form: N = [at/(t -1 ) ] -2a , a > 0, t > 2 In other words, you do not know the path of the profit function. which is time dependent, or trial dependent, based on the distributions of market returns. The more you stay in the market, the more you risk riding a profit function that will invert your unlimited profit potential and turn it into a devastating loss. I think you cannot think in terms of clumps in coin toss experiments. In that case applies not the simple Fib numbers but the k-step Fib number. The problem is too hard to model and solve. You didn't even scratch the surface. But, listen to this, the probability of a clump that will ruin you is always 1 as the number of trials goes to infinity. This is trivial to prove and it is left to the reader as an excercise Kut2k2, I think you have underestimated your opponents, the difficulty of the problem and overestimated your own understanding of the issues involved.
================ vl-money; Highest probabilities are the first 50 trades are going to have much more losses than the last 50 trades in 7 years.Not a prediction. Same reason applies,starting out, to have the best risk of ruin/wipe out a target; load 5 or 6 rounds in a six shooter.{Russian roulette is 5 of 6 rounds loaded, do not try that @ home}.
You can try that experiment yourself. Point was, if the prob(m consecutive H or T) increases with sample size, just like prob(ruin)->1 as trials approaches infinity, what is optimal Nth trial to halt? Basic cases of prob(for m=2 & 3) changes as no. of trials increases, i.e.1-prob(formula given for Not). http://www.bjmath.com/bjmath/probable/flips.htm