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Not the Grail, just a regular one - Bablokos!!! - page 114

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Well, it's problematic to talk about anything without knowing the algorithm, I'll just say that I tried to make a robot on the rekill, implemented it on mt5. In the tester for months +, until I failed. Aggressiveness allowed me to double the depo several times, but the results are unstable and highly dependent on the moment of launch.
yosuf:
I suggest we move away from the eagle-reckoning topic as being malicious and focus on BP forex.
I wouldn't mind finding the topic malicious, but there is no evidence of it being unprofitable other than speculative.
Lastrer:

No eagles are not in a row, but total, a=3, b=4 (this is for example) then:

ororrro, rorrroorrro, oooh, rorrroorr, etc. eagle winnings

tails wins, orororrrr, ororrrrr, ororrrrr, etc.

need probability of tails winnings

So, for given values of a and b, eagles win if they manage to fall a total of 4 times BEFORE 3 consecutive tails fall?
Yes, a and b may be different
Lastrer:
Well, it's problematic to talk about anything without knowing the algorithm, I'll just say that I tried to make a robot on the rekill, implemented it on mt5. In the tester for months +, until the moment of failure.
Why, the algorithm is well known: either buy with the trend or sell against it. Half of the robots work according to the first scheme and the other half according to the second one. Let's see which strategy is more sustainable.
Lastrer:
Yes, a and b may be different
Got it...
Lastrer:
I wouldn't mind finding the topic harmful, but there is no proof of it being unprofitable other than speculative.
Is its "profitability" proven?
With a=3, b=7 we get a martin, with the number of knees equal to A. Only the series is continuous.
prikolnyjkent:
Got it...
Understand another thing - forex is not exactly a random process. There is a pattern here that does not always manifest itself, but at the most inopportune moment, from our point of view.

Here is Avals' solution, but for martin we get non-zero expectation, i.e. probabilities of getting series of three tails and seven eagles are not equal, that is why we need to find error

the problem is quite difficult to calculate. We have to consider different series lengths and for each series calculate the probability of getting A of tails and 4 of eagles in a row. The minimum length of the series is 3 (no event will occur at smaller lengths). The maximum series length is 12, because after the series rororororr with any outcome there is no point in counting further.

for series length=3. Probability of 3 tails in a row p(ppp)=0.125, probability of 4 eagles p(4o)=0. Hence, the probability of going to series 4 without getting any of these events = (1-0.125)*(1-0)=0.875

for series length=4. p(ppp)=0.125, p(4o)=C(4,4)/2^4=1/2^4=0.0625, where C is the number of combinations. Probability of going to series length 5 =0.875*(1-0.125)*(1-0.0625)=0.7177....

For series length=5. p(ppp)=0.125, p(4o)=C(4,5)/2^5=0.15625. Probability of going to series length 5 =0.7177*(1-0.125)*(1-0.15625)=0.53

etc.

and then multiply series probabilities by probability p(ppp) and add up.

0.125*1 + 0.125*0.875 + 0.125*0.7177 + 0.125*0.53 +...

1...107108109110111112113114115116117118119120121...650
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