- | Joined Mar 2013 | Status: Resurrection | 225 Posts
Discussion is an exchange of knowledge;argument is an exchange of ignorance
- Joined Oct 2013 | Status: Forex Shaman | 1,468 Posts
"There's a sucker born every minute" - P.T. Barnum
Martingale, Anti-martingale, and Compounding 40 replies
Martingale, Reverse Martingale, Modified Martingale, Maths 1 reply
How to change this Martingale to Reverse Martingale ? 3 replies
Martingale vs. Non Martingale (Simplified RoR vs Profit) 1 reply
Does this simple system by Richard Donchian really work? 134 replies
DislikedIn addition to the links I posted in post #155 (examples of martingales that have blown accounts), here are some definitive answers about martingale — and other staking systems — by math experts: Mike Shackleford (The Wizard of Odds); also here Michael Bluejay (VegasClick.com)...Ignored
He says that 1-8 levels is sufficient. It's my view that, given enough time, every price pattern that can possibly occur, will eventually do so.
Will he ever encounter 8 consecutive losses during his trading lifetime?
[EDIT]
He seems typical of what I would call the "martingale mentality". Despite his comment about price behavior being predictable, he apparently can't find an edge that allows him to overcome his losses. Hence he assumes that (1) nobody else can, either; and (2) that if he can eliminate the losses, he has solved his problem. Hence he turns to martingale, and justifies his use of it by saying that 8 consecutive losses is an impossibility.
Disliked@ybfjax: With respect, the guy you're quoting seems to be contradicting himself. On the one hand, he's saying that it's possible to gain an edge from price behavior ("The market never goes in a single direction forever; it fluctuates up and down like the tides of a beach. This in itself already gives you a high positive expectancy of the price movement itself"). Sure, FWIW I agree. But then he suggests that professional traders offer no workable (profitable?) alternative to martingale ("....I have not seen them ever offer any transparent alternative....Ignored
QuoteDislikedIf so, that's clearly false. Reasons: 1. There are profitable traders who don't use martingale.
QuoteDisliked2. It's a mathematical fact that staking systems can't alter expectancy (PipMeUp has posted a neat mathematical proof here). Progressive staking systems increase the average bet size, and consequently increase risk of ruin exponentially. In other words: no long term benefit, but significantly increased risk.
QuoteDisliked3. An 'edge' is not necessarily the result of trying to avoid losses. It's simply an outcome where profits exceed losses, over time, and irrespective of staking. Bottom line: his entries and exits either give him an edge, or they don't: If he does have an edge, there's no need to employ progressive staking (that's analogous to the winning formula used by casinos). If he doesn't have an edge, progressive staking is merely trading off guaranteed recovery in the short term, for possible disaster in the long term.
QuoteDislikedHe says that 1-8 levels is sufficient. It's my view that, given enough time, every price pattern that can possibly occur, will eventually do so. Will he ever encounter 8 consecutive losses during his trading lifetime? If not, then martingale will "work" for him, exactly as he says. If so, then how easily can the (consequential) loss be shrugged off? That's a personal question that I believe every trader must consider.
QuoteDisliked[EDIT] He seems typical of what I would call the "martingale mentality". Despite his comment about price behavior being predictable, he apparently can't find an edge that allows him to overcome his losses. Hence he assumes that (1) nobody else can, either; and (2) that if he can eliminate the losses, he has solved his problem. Hence he turns to martingale, and justifies his use of it by saying that 8 consecutive losses is an impossibility.
DislikedIf you use a system with RR=10 you need a win rate greater than 9.1% If you use a system with RR=5 you need a win rate greater than 17% If you use a system with RR=2 you need a win rate greater than 33% If you use a system with RR=1 you need a win rate greater than 50% If you use a system with RR=0.5 you need a win rate greater than 67% If you use a system with RR=0.1 you need a win rate greater than 91% What win/loss do you need for a system with RR=0??? Martingale tries to get 1 unit of profit risking 2^n units of risk, the RR decreases exponentially...Ignored
DislikedBut everything is not necessarily equal; we must take the probabilities into account. If upon reaching level B, the probability of outcome "1" is significantly greater than outcome "2", than it was at A, then "Z", who has weighted his position more heavily at "B" will fare best, and "X" worst. No matter what the outcome, Y's performance will always lie somewhere between X's and Z's, because he has hedged his bets. Scaling in is a form of diversification.Ignored
DislikedHowever, if price takes path "3", then it depends on the exit point. If the exit point is higher than A, then X will outperform Y, and both will outperform ZIgnored
Disliked"Suppose every trade has a common TP that is 50 pips higher than point B, and a SL of 10 pips below its entry. And suppose that, having reached point A, the probability of price reaching the TP is 60%; and if price reaches point B, the probability of its continuing to the TP is now 90%."Ignored