A probability theory problem - page 10
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The figures are taken out of my head ... made up. You have to start somewhere, don't you?
Yes, let's assume that without conditions A, B and C the probability of the shooter hitting is 0.5, which is obtained with 100,000 trials and 50,000 hits.
And indeed:
purely intuitively - the result will improve by 33% (1.05 * 1.1 * 1.15 = 1.328) that is the final probability will be 0.5*33%=0.66% which in principle seems to be true. And a little better than the sample for the strongest factor C.
I'm not sure that's the right decision. Why? Because factors A and B, which favour event D, contribute almost nothing to the final probability. Individually factor C improves the odds from 0.5 to 0.65 and factors A and B additionally from 0.65 to 0.66 i.e. by 0.01 which is negligible. At intuition level, the result should be around 0.7-0.75
I agree. That's why I wrote that 0.5*0.5*0.5 is a finger in the sky.
Do you have an alternative solution to the problem or at least a hint?
There is no solution, of course, because there is no problem set. In general, in the probabilistic approach, setting a problem is not half the job, but why the whole thing. I can give you a hint from my side. We should not evaluate such an event as "growth" (it's very difficult to determine it), but the value of the expectation shift in an hour after the event A. Or in one day, or in one second - it depends on what event.
Why complicate things? The simplified term "growth" just implies a positive increment over some fixed (let it be an hour - in this case it does not matter) period of time.
Already reformulated the condition of the problem in relation to the arrow, which is harder to confuse. Let's try to solve it.
Why complicate it? The simplified term "growth" just implies a positive increment over some fixed (let it be an hour - doesn't matter in this case) period of time.
Already reformulated the condition of the problem in relation to the arrow, which is harder to confuse. Let's try to solve it.
Your formula is originally written correctly. To clarify, the formula is true for probabilities, not conditional probabilities. For conditional probabilities it is:
P(D) = P(A) * P(D|A) + P(B) * P(D|B) + P(C)*P(D|C)
For this formula we need to introduce the a priori probabilities of A, B, C, as I said before.
You have originally written the correct formula. Let's make it clear, that the formula is correct for probabilities and not for conditional probabilities. For conditional probabilities it is:
P(D) = P(A) * P(D|A) + P(B) * P(D|B) + P(C)*P(D|C)
For this formula you need to enter the a priori probabilities of A, B, C indicators, as I've already mentioned before.
Thank you.
P(D) = P(A) * P(D|A) + P(B) * P(D|B) + P(C)*P(D|C)
This is for the full group and not for independent events.
It seems that the condition with indicators and signals is misunderstood, immediately associating it with blinking, frequency of occurrence/occurrence, etc. Let's forget it as a bad dream and rephrase the same problem.
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
We have a shooter in position who can either hit or miss the target (event D).
The probability of hitting the target depends on some conditions/events:
Let's assume that the shooter took the firing line when conditions/events ABC coincided i.e. he is in good health, the wind does not blow the bullet away and the shooter's weapon is good.
Question: what is the probability of the shooter hitting the target P(D/ABC) when these conditions coincide?
Something is not right here. Events A,B,C may be independent (good gun was given, wind dropped, feeling better) - but they are not events of the shooting process itself. I don't know where to get the probability in the case of feeling good when there is no wind. There were no tests, the sampling frequency was not determined. The events themselves are independent, but the mechanism of their corporate influence on the result is unknown.
Seems like the same thing as trying to predict a patient's response to taking two different medications. Yes, independent (when we want, then we take each pill), yes, separately the reaction is known and described in the instructions of each of them. But the effect of their simultaneous use has not been assessed in any way. These drugs can interact in an unknown way. They can enhance each other's effects, or, conversely, weaken each other's. And not at all in the ways in which they act directly on the disease.
What if feeling good in the doldrums and sunny weather on the joy of being given a new weapon will cause the shooter's increased self-esteem and he will start firing with gusto almost without looking at the target?
Let's go through it again in order.
The formula proposed above (I will deliberately write it differently - through X, A, B, C):
P(X) = 1 - (1 - P(A)) *(1 - P(B)) *(1 - P(C))
will give the probability of a signal from at least one indicator. This is why the result is so high - three indicators signal more often. But this is essentially not what the problem statement is looking for.
By Bayes:
P(D|ABC) = P(ABC|D) * P(D) / P(ABC)
Here P(ABC) = P(A) * P(B) * P(C)
where a priori indicator probabilities are calculated as the number of signals of each indicator among the total sum of all indicators.
P(D) = 0.5 by default, when there is no super-trend, i.e. the probability of buying and selling signals are equal.
But I have doubts how to calculate P(ABC|D). The simplest way (due to independence):
P(ABC|D) = P(A|D) * P(B|D) * P(C|D)
and each such conditional probability must be calculated as the number of signals of each indicator on the set of all bars where the buy was correct.
But all this is not the final truth. ;-/