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At what length of test series? Yura, you have to understand that on a long enough Bernoulli test series, this probability will be as small as one :)
P.S. On a series length of 14 trials, you're probably right (0.3^14). But 14 deals isn't serious.At what length of test series? Yura, you should understand that on a long enough Bernoulli test series this probability will be as small as unity :)
P.S. At a series length of 14 trials, you're probably right. But 14 deals is not serious.
And besides - you would need unreal deposit to cover 14 failures... And the 15th successful deal should have a sufficiently large rollback to cover the previous ones.
In my considerable experience of Martingale trading - more than 7 trades is a very big risk. Ideally 3-5 trades with a multiplier of 2.
Continuing the topic of calculating price consolidation levels...
Below is a picture for the pair Funtjen on TF15... Time frame 1000 bars...
Maximum at 200.10... Minimum - 192.61 and 202.29...
Now a time frame of 10,000 bars...
When we start talking about determining the maximum number of losing trades, we can draw an analogy to the martingale based on averaging. If we use averaging (i.e. if we catch the bounce), we can find the maximum length of a non-bounce period on the history and use it as a calculation one. And on its basis, determine the risks by calculating the lot sizes of each following deal. That is why this martingale system is no more risky than the usual MM system.
Here a "trial" is a trade that has two values - profitable/losses. The return time is such a number of Bernoulli trials (trades), at which the event "series of successive successful outcomes for the first time reached length r" occurs.
As we see, if we consider a losing trade as a success, its probability is 0.6 and the needed series is at least 15, we need 5400 trials (trades) on average to meet such a nasty series there. And if the parameters are different, then we apply formulas (7.7).
Martingale, relax early: for example, with a more realistic estimate of probability of a loss for the classical martingale (with doubling of a bet), equal to 0.75 (3 losses per 1 profit), applying the first formula of (7.7), we obtain (p=0.75, q=0.25, r=14) an average number of deals equal to about 0.982/0.25*0.75^14 ~ 220 deals.
2 Yura Reshetov: I did not understand your martingale exactly. Maybe it's not so aggressive.
2 Yuraz: Maybe I overestimated the probability of a loss (0.75)? What is the value of this figure in the system you made?
P.S. Analysis of a modified MoneyRain system:
p=0.6244, q=0.3756, r=11. Average number of trades according to Bernoulli's scheme, giving the same series of 11, by the same formula turns out to be 473. You have met it a bit earlier, Yura. So far it turns out that the tests are sort of like independent...
Remarkable picture... It's a pity no one has commented... Maybe, I'm in my wave and I don't notice something, which is obvious for everybody...
At the time interval of 10000 bars we have 4 zones of price consolidation, and we can see their borders ...
Example of a strategy that is not difficult to implement as MTS ...
Averaging towards the nearest boundary ... Exit by Takei (standard value) or Stop Loss on the border of moving from one zone to another ...
Waiting for comments...
13.12. Probability of making a loss in a series of successive trades .
of successive trades .
How does it apply to the subject we are considering?
P.S. The archive has been opened.
Thanks, lovova, but the archive doesn't open (if it's Bulashev, I have it). I looked it up. 13.12 is actually a statement of the binomial distribution and calculations for a small number of trades, while 13.13 seems to be numerical modelling of the loss series. Yet Feller gives it all in analytical form, while Bulashev only says that there are analytical difficulties involved and suggests a numerical procedure.