What is the cumulative probability?
It is good that there is no guarantee, because I don't agree with this result.
In this case with P(A) equal to 1 the result will be 1 independent of P(B) (or vice versa with P(B)=1, P(A && B)=1 independent of A).
But in that case, if P(A)=0, the result should be (similarly to the previous 100% guarantee) equal to zero, regardless of P(B). Which does not happen according to this formula.
I.e. probability equal to zero means 100% probability that the event will not happen.
I have a variant answer: 2*P(A)*P(B). But this is still at the level of a hypothesis. I would like to know the real formula.
If one of the probabilities is 1 (let's say A), then the event will happen anyway, we don't need to look at probability B. This is the reasoning behind it: throw two coins and you need at least 1 eagle. Or throw 2 dice and you need at least 1 six.
I doubt that fifti/fifti can make any difference to 100%. Let alone knock it down to the 75% level.
If one of the probabilities is 1 (let's say A), then the event will happen anyway, you don't need to look at probability B. That's the reasoning behind it: throw two coins and you need at least 1 eagle. Or throw 2 dice and you need at least 1 six.
And if one of probabilities is 0 (let A), so the event won't happen anyway, you don't need to look at probability B.
I would add to all this that the combination P(A)=1 with P(B)=0 is impossible (and vice versa). Why? I don't think it's possible to comment on that.
2*P(A)*P(B) is the wrong formula at all, because it can result in 2, which probability cannot have. Simply, multiplication is the probability of two orals falling out at the same time when two coins are tossed - two events coinciding at the same time.
Really wrong, I agree. I'm wrong :)
I doubt that fifti/fifti can make any difference to 100%. Let alone knock it down to 75%.
And if one of the probabilities is 0 (let's say A), then the event will not happen anyway, you don't need to look at probability B.
I would add to all that the combination P(A)=1 with P(B)=0 is impossible (and vice versa). Why? I think it does not need to be commented.
It means that the task is not set precisely.
If you can't describe the task formally, explain it on your fingers: throw coins, dice, take balls out of the bag, divide apples among schoolchildren, etc.
Then the task is not set accurately.
If you can't formally describe the task, explain it on your fingers: throwing coins, dice, pulling balls out of a bag, dividing apples between pupils, etc.
Why not exactly:
The bull says: -Event X will happen with a 35% probability.
Bear says: -No. Event X will happen with 51% probability.
Of course I'm going to believe the Bull. But just how much should I believe him? After all, the witchdoctors don't have definitively vague predictions. (Vague is 50/50).
I have a question for the mathematicians. Although it looks like an offtopic, it is applicable to MTS.
Problem:
Let there be an event X whose probability of occurrence is equally dependent separately on two events A and B independent of each other.
If the probability of A-dependent event X is P(A)=0.4,
and the probability of occurrence of event X, dependent on B, is defined as P(B)=0.2,
then the question is:
What is the resulting probability of occurrence of event X: P(A && B) ???
There is not enough data to decide.
For example, the conditions are:
-if a man has a ring on the ring finger of his right hand, he is married p=0.5 (women are married)
-any man is married with p=0.5 (there are singles, children, widowers)
but if both conditions are met - a man has a ring on his right ring finger, he is married. The probability of such an event is close to 1. That is, the probability p(X/A) and p(X/B) cannot be calculated from the probabilities p(X/AB)
The formula p(x) = 1 - (1-p(A))*(1-p(B)) for two consecutive independent events, and the result is the probability that at least one of events A or B will occur. For example, the probability of hitting an enemy missile with the first line of defense =0.7, with the second line of defense 0.5. What is the probability of hitting one of the lines? p=1-(1-0.7)*(1-0.5)=0.85
In the case of dependent events, we need conditional probabilities in the formula, but that's still not it. It's all about calculating the probability of at least one event occurring in successive outcomes.
Also, in the case of the market there is such a thing as robustness, which results in the problem having a different solution.
For example, from The New Market Magician" (Erckhardt):
"...Are there other practical implications of robust methods that would differ from the results of studies assuming a normal probability distribution?
- An important application concerns the situation where you have multiple indicators for a particular market. The question arises: how to combine several indicators in the most efficient way? Based on certain precise statistical measurements, it is possible to assign weights to different indicators. However, the choice of weights assigned to each indicator is often subjective.
You will find in the robust statistics literature that in most cases the best strategy is not to weight, but to assign a value of 1 or 0 to each indicator. In other words, to accept or reject an indicator. If an indicator is good enough to be used in principle, it is also good enough to be assigned a weight equal to the others. And if it does not meet this standard, it is not worth bothering with.
The same principle applies to the selection of trades. How do you best allocate your assets to different trades? Again, I will argue that the allocation should be even. Either the trade idea is good enough to be executed - in which case it should be executed in full - or it is not worthy of attention at all."
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I have a question for the mathematicians. Although it looks like an off-topic, it is applicable to MTS.
Problem:
Let there be an event X whose probability of occurrence is equally dependent separately on two events A and B independent of each other.
If the probability of A-dependent event X is P(A)=0.4,
and the probability of an event X depending on B is P(B)=0.2,
then the question is:
What is the resulting probability of occurrence of event X: P(A && B) ???