What is the cumulative probability? - page 4

 
You areconfusing the concepts of probability and frequency of occurrence of an event. They are only equal in the limit of infinity and only for tests under constant conditions
 
Avals писал (а) >>
You are confusing the concepts of probability and frequency of occurrence of an event. They are equal in the limit of infinity and only for tests under the same conditions.

That's why I'm asking for help from mathematicians. Can you solve the boxers problem?

And what I mean is probability. I couldn't think of a good example.

 
Mischek писал (а) >>

I mean, you have to be very careful with the logic that outputs the odds as a percentage. In this case, if you estimate your odds at 0, how do you

estimate the chances of a man with one arm and one leg against the same semi-loser?

If you got 100 somewhere in the output and are so sure of them then why dilute and compare.100 in principle cannot be.

I agree that it practically cannot be 100%. But theoretically, if P(A)=1, then for any P(B) other than zero the final probability of event X will be 1.

But if P(A)=0.99, the final probability is already I don't know what.

 
coaster писал (а) >>

That's why I'm asking for help from mathematicians. Can you solve the boxers problem?

Of course not, it is not solvable :) If we take purely the invariability of conditions (the boxers are always in the same shape and do not develop or degrade), then the statistics of the meetings between the two fighters will be decisive. I.e. the frequency of this event in the past. There is no precise formula for calculating this probability by overall boxer statistics. We can of course come up with expert estimates on the basis of these probabilities, but they are all approximate, and the quality of the result when the conditions change will be very poor.

 

Another variation on the solution.

A and B used to fight obviously not with each other, but with a certain boxer D.

And a virtual fight can be arranged through this G:


event probability
A beat G && B lost to G
0
.95*0.15=0.1425 (A beat B)
A beat G && B beat G0.95*0.85 (A & B - draw)
A lost to G && B lost to G
0.05*0.15 (A & B - draw)
A lost to G && B won G
0
.05*0.85=0.0425 (A lost to B)

In case there is a draw between A and B they get extra time, resulting in

as a result the ratio of m.y. probability of A's victory over B will not change,

0.1425 / 0.0425 There's no point in specifying further, the probability of A winning

A over B = 0.77.

P.S. I wanted to draw the table in monospace font, got something wrong.

 
Avals писал (а) >>

No, of course, it's not solvable :) If we take it purely in the context of the invariability of conditions (boxers always in the same shape and never develop or degenerate), then the statistics of the meetings between these two fighters will be decisive. I.e. the frequency of this event in the past. There is no precise formula for calculating this probability by overall boxer statistics. We can of course come up with expert evaluations on the basis of these probabilities, but this is only an approximation, and the quality of the result when the conditions change will be very low.

Let's not mention frequency at all for the moment. The task is theoretical. I couldn't think of a good example. Let the total number of sparring for both boxers be equal and tend to infinity. Boxer A wins all 95% of his fights. Boxer B has won all 85% of his fights. What is the probability that boxer A will retain his title in a single sparring.

Strengths are equal. >> Weight, age and even size XXL are equal. What other data do you need?

 

coaster, you need to know the reliability of the bulls' and bears' forecasts.

If the bears' forecast is 100% accurate, then you will choose their forecast (although they might only be 51% accurate).

In the simplest case if both experts give binary answers (yes/no) with probabilities A and B, in case of disagreement you choose the opinion of the best expert (max(A,B)).

If the answers are not binary, but probabilistic, and there are more than two experts, things are more complicated.

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Look for these answers in expert committees. It should be.

 
coaster wrote (a) >>

You are out of your depth. >> So far, Integer is making a point.

coaster, yes I seeyou've moved on to boxers.Whatare the axioms of probability theory when we're not talking about independent events? They're going to punch each other, aren't they? What kind of event space is that? How about winning with B at the same time (if the events are independent, then yes)? The example somehow does not fit your topic :)

 
coaster писал (а) >>

Let's not mention frequency at all just yet. The problem is theoretical. I couldn't think of a good example. Let the total number of fights of both boxers be equal and tend to infinity. Boxer A wins all 95% of his fights. Boxer B has won all 85% of his fights. What is the probability that boxer A will retain his title in a single sparring.

The power is equal. Weight, age and even size XXL are equal. What other data do you need?

Answer. From this data, no exact (mathematical) formula can be derived that boxer A will retain his title in the single sparring.

You must use data from their bouts with each other, i.e. calculate the probability from their bout statistics. All other stats are expert assessments, which can differ completely, and are of course inaccurate. Expert assessments are an attempt to find formulas that in practice give reasonably accurate results for a given area of application. I.e. specifically for boxing, for example.

 
Erics писал (а) >>

coaster, you need to know the reliability of the bulls' and bears' forecasts.

If the bears' forecast is 100% accurate, then you will choose their forecast (although they might only be 51% accurate).

In the simplest case if both experts give binary answers (yes/no) with probabilities A and B, in case of disagreement you choose the opinion of the best expert (max(A,B)).

If the answers are not binary, but probabilistic, and there are more than two experts, things are more complicated.

----------

Look for these answers in expert committees. It should be.

I need to know a reliable prediction of the probability of occurrence of a certain price when determining this probability by an Up-trend indicator on one side and by a Down-trend indicator on the other side. What will be the final probability?

Simply: A bullish indicator tells you: the price will appear in the area of interest with probability P1. A bearish indicator tells you: price will appear in that zone with probability P2. How do you determine the final probability?