The "Maybe we'll get lucky" counsellor - page 6
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Then the probability is very high, practically 1.
95% of variants will be with frequency of profitable between 350-2*27 and 350 + 2*27, i.e., roughly speaking, between 300 and 400.
99.7% of variants will be with profitable frequency between 270 and 430.
I don't think it's likely that there won't be a single one that fits or is better (out of 10,000). Or am I wrong?
That is, we will surely choose the best option, what is the probability that in 10,000 there won't be the right one?
Here's an optimised version of just 3 losing trades out of 81 for 2011:
Once again:
There is a system that produces roughly equal frequencies of profit and loss on 700 trades - no matter how its parameters change. Right?
You want to know what are the chances that, having checked all possible variants of parameters (10,000 variants in total), we won't encounter a single case of "640 times 60 and worse"?
I can estimate the probability, but why do you need it?
2 yosuf: you are very persistent, to put it mildly. What is the SL/TP ratio in the last figure?
Here's an optimised version of just 3 losing trades out of 81 for 2011:
Yusuf, screw such tests and such optimization. These are insignificant pictures and figures :)
Here's an optimised version of just 3 losing trades out of 81 for 2011:
Again:
There is a system that produces roughly equal frequencies of profit and loss on 700 trades - no matter how its parameters change. Right?
You want to know what are the chances that, having checked all possible variations of the parameters (10,000 variations in total), we won't encounter a single case of "640 on 60 and worse"?
I can estimate the probability, except why would you need it?
That's right. (I mean we don't meet any cases of 640 over 60, 641 over 59 etc.)
We could talk about the probability of not finding a good ("in the mathematical sense") system that is only a fit, given enough parameters.
Here's an optimised version of just 3 losing trades out of 81 for 2011:
Да правильно.(в смысле не встретим ни одного случая 640 на 60, 641 на 59 и тд)
Well, formulas are easier to write than to calculate.
The probability of having any combination excluding anything better than 640 by 60 in a single test is very close to 1.
The variance of the binomial distribution is 700*0.5*0.5, i.e. the s.c.o. is about 13.23. The number 640 is about (640-350)/13.23 ~ 21.92 sigmas away from 350.
The required probability in a single test is approximately 1 - 3.34*10^(-107).
Accordingly, the probability that any combination, excluding anything better than 640 by 60, in each of 10000 tests is equal to ( 1 - 3.34*10^(-107) )^10000. This number is still extremely close to 1.
You can sleep now.
It's a pity I don't have anything to calculate it on
Although I'm curious how many sigmas any EA can be fitted with at 10 000 passes with more than 50% probability.
At 460 passes we already obtain 3 sigmas. (1 - (1 - 0.003/2)^460 = 0.5) (correct me if I'm wrong).
I think the millionth pass cannot be calculated in this formula, although perhaps mathematicians intuitively understand how much should come out.
If sl falls 10 times less often than tp (like Yusuf's), what would the solution to the original problem look like?
If I'm not mistaken, the sigmas will be much less than 21.