Search results

http://www.elitetrader.com/vb/showthread.php?s=&postid=3355540#post3355540 http://www.elitetrader.com/vb/showthread.php?threadid=231633

A trader is using a system where, after each trade, he wins W units with a probability of p (0 < p < 1) or he loses L units with a probability of q = 1 - p. The trader has a starting bankroll of B0 units and he has in mind a surrender bankroll of X units (B0 > X). If the trader's bankroll...

The probability of ruin is based on ALL possible paths to ruin, not just the single path of going directly from C to X. C might grow to 10,000C before dropping back to X. As long as C < T, the many paths to ruin are still ongoing. R{infinity|X|C} = p*R{infinity|X|C+1} + q*R{infinity|X|C-1}...

How often do brokers go out of business? Less than 1% of the time? Somebody forgot to tell that to Warren Buffet. The contention of the majority of the ruin-is-certain cult here in ET is that positive expectation makes no difference. According to them, a system can be consistently...

Seriously, dude, you've fallen for this certainty-of-ruin rubbish also? :( When the broker goes bust, hopefully you've already pocketed the profit you made from using a positive-expectation system.

Notice how you never got a response, my friend. That's because math cranks are real good at challenging mathematical claims by others, regardless of evidence, and really bad at defending their own mathematical claims.

You are obviously a mentally deficient person so I will explain this s-l-o-w-l-y one last time before putting you on ignore. There is a gambler who wants to increase his current bankroll C. He participates in an even-payoff game where he wins or loses one unit per trial. In other words, at...

Actually I calculated the probability of failing to reach the target T as R{T|X|C} = ((q/p)^(C-X) - (q/p)^(T-X))/(1 - (q/p)^(T-X)), (assuming p does not equal q). This derivation has nothing to do with assuming T approaches infinity. That comes later. If you didn't have serious...

Definitely crankspeak, isn't it? This person has no clue about basic math yet projects that onto others as their flaws. Sad.

It is amazing how much contention there is over basic mathematical concepts here in ET. I often why people with poor math skills assume they can make it as traders. It's not as easy as buying a lotto ticket.

Thanks for posting. You've given us a textbook example of crankism: somebody who thinks he understands limit math but clearly doesn't. "I have a question for you: how much you think one should lose to be left with X capital after reaching infinite capital T?" Who said you ever reached...

Of course it is. T is chosen by the gambler. It is in no way dependent on C, X, p or q. Can you? So give us the "dependent" relationship you talk about. Be specific. You have reading comprehension problems. The risk of ruin is the probability of FAILURE. Obviously you won't reach infinite...

I recently had a series of unpleasant exchanges with some ET members in which I was accused of being a crank, and I thought the same of them (and still think so). But it got me wondering: have any other members here experienced something similar? It turns out there is a whole history of...

R{T|X|C} = ((q/p)^(C-X) - (q/p)^(T-X))/(1 - (q/p)^(T-X)) Now divide the numerator and denominator by (q/p)^(T-X) : R{T|X|C} = (((q/p)^(C-X)/(q/p)^(T-X)) - 1)/(1/(q/p)^(T-X) - 1) So the limit of R{T|X|C} as T increases to infinity is R{infinity|X|C} = (0 - 1)/(0 - 1) = 1 IF q > p

+1 Exactly. Repubs wonder why Obama can't ride roughshod over the Democrats in Congress the way Bush, Bush and Reagan rode over the Congressional Repubs. Totally different worldviews.

I see you've been fantasying about circle jerks for a long time, Bonzo. Folks we have a real homo chimpanzee here. :D

Seriously, this is the worst thread of ET: http://www.elitetrader.com/vb/showthread.php?threadid=151802 The OP has been gone 2½ years, the thread didn't make sense when he started it and it has morphed into some undying monster about who knows what. If somebody knows how to stake this...

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)...

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...

And here we note that reg has finally mastered that "Monkey See, Monkey Do" skill. :D