Probability question

[WinstonSmith]

You have to make a yes/no decision.
To help you, you have two independent decision factors A and B with probabilities p(A) and p(B) of being correct. The probability of making the correct choice is p(A) if you use decision factor A alone, and p(B) using B alone.
What is the probability of being correct using both A and B?

 

[pssonice]

p(A)*p(B)

pA(0,1) and pB(0,1)

 

[NotQuiteRandom]

Your question is not phrased as I would expect and would benefit from some clarification to be sure that I am reading it correctly. As I understand the question you are asking:

"If I have a single event and can choose between 2 decision factors (A and B) to determine the probability of the event, where A and B have probabilities of being correct of P(A) and P(B), what is the total probability of making the correct decision if I use both factors".

Assuming this to be the case and there is only 1 event but there are 2 unrelated decision factors it is additive: P(A)+P(B).

It would be P(A)*P(B) if we were considering the probability of 2 independent events occurring each with probabilities P(A) and P(B) but, judging by your question it is the former and not the latter case in which you are interested.

 

[TWI]

At a guess

1 - [(1-P(A))*(1-P(B)]

Good question.

 

[WinstonSmith]

"If I have a single event and can choose between 2 decision factors (A and B) to determine the probability of the event, where A and B have probabilities of being correct of P(A) and P(B), what is the total probability of making the correct decision if I use both factors".

Yes that's correct. Let's give an example: You have 2 decision factors A and B to decide whether to enter a trade. P(A)=0.7 and P(B)=0.6. You propose to use B as a filter to be used if A is satisfied. Are you better or worse off?

 

[TWI]

1-[(1-0.7)*(1-0.6)] = 0.88

T4 would say it improves your chances to use both.

 

[WinstonSmith]

1-[(1-0.7)*(1-0.6)] = 0.88

T4 would say it improves your chances to use both.

Thanks TWI. Could you explain your reasoning?

 

[TWI]

You have to look at th chance of being incorrect rather than chance of being correct.
so 1-P(A) and 1-P(B)
P(A) and P(B) are independent so the probability of both occurring is the multiple.
Now you have the chance of being wrong when they are combined.
Simply subtract this from 1 to get the chance of being correct.

 

[WinstonSmith]

Thanks TWI. I don't think it's totally correct though.

What if P(B) is less than 0.5? You'd expect a worsening of total probability in that case, but that formula always improves the total.

Not that I have any better ideas. I am completely stumped!

 

[NotQuiteRandom]

"If I have a single event and can choose between 2 decision factors (A and B) to determine the probability of the event, where A and B have probabilities of being correct of P(A) and P(B), what is the total probability of making the correct decision if I use both factors".

Yes that's correct. Let's give an example: You have 2 decision factors A and B to decide whether to enter a trade. P(A)=0.7 and P(B)=0.6. You propose to use B as a filter to be used if A is satisfied. Are you better or worse off?

Ah, this example changes everything The factors are not truly independent as you initially stated because they both rely on the same variable (market).

Your problem is therefore more complicated as it requires a look at the overlap. To solve the problem we would need to make certain assumptions about volatility (make it static) and correlation between A and B.

NQR

 

[TWI]

If either prob < 0.5 then it is a contrary indicator and should do opposite getting > 0.5
I think you will find the formula works for these smaller probs also run the numbers with 0.1 on each
get chance to be right of 0.19, about right.

 

[andread]

At a guess

1 - [(1-P(A))*(1-P(B)]

Good question.

= 1 - ( 1 - pa - pb + pa*pb ) = pa + pb - pa*pb

which is what the theory says

 

[NotQuiteRandom]

A and B are NOT independent in this instance.

 

[zupcon]

If I understand your question correctly then this link might be useful,

http://www.mathpages.com/home/kmath267.htm

regards
zu

 

[andread]

A and B are NOT independent in this instance.

right. Didn't read with enough attention the following posts. Sorry

So let me see if I get it.
You have one event A, with probability pA. If A occurs, you look at the probability of another event B, pB|A. This is the probability of B occurring after A occurred. In this case pB, your probability a priori, should be pA * pB|A
If you don't know the probability of B after A occurred, but you only know the probability of B in general, then you have a problem, because you don't have any information about the relationship between A and B.

Hope I got it right. Winston any comments?

 

[new_trader]

"If I have a single event and can choose between 2 decision factors (A and B) to determine the probability of the event, where A and B have probabilities of being correct of P(A) and P(B), what is the total probability of making the correct decision if I use both factors".

Yes that's correct. Let's give an example: You have 2 decision factors A and B to decide whether to enter a trade. P(A)=0.7 and P(B)=0.6. You propose to use B as a filter to be used if A is satisfied. Are you better or worse off?

I could be way off the mark but I don't think probability works that way. A decision factor is different to a probability.

eg/
You have 2 bags:

Bag 1. Has 7 white balls and 3 Black balls giving a probability to choose a white ball of 0.7
Bag 2. Has 6 white balls and 4 Black balls giving a probability to choose a white ball of 0.6

The decision you must make is how many balls you take and whether you take them from different bags or the same bags. You can only calculate the probability after you have made a decision.

Eg/ You want to choose 2 white balls. If you choose to take the 1st ball from bag 1 which gives a better probability and it is black then your filter is no longer valid. Decision 1 has not been satisfied. So, you will only take a Ball from Bag 2 if the 1st ball from Bag 1 is white. In which case, it is an AND function and someone already mentioned it is a multiplication factor.

I believe the odds of taking 1 white ball from bag 1 AND 1 white ball from bag 2 is 0.7 x 0.6 = 0.42.

What you want to do is calculate a way of improving your probability in which case probability B must be entirely dependent on the outcome of A and can only be calculated after result A, not before. In this case, you take Balls from only 1 bag.

Bag 1. Has 7 white balls and 3 Black balls giving a probability to choose a white ball of 0.7

You want to Choose 2 white balls but you will only do so if the 1st ball you choose is White. In which case:

P(A) is 0.7 on the 1st pick only. If it is White and the ball is NOT replaced, the probability of choosing another white is now 0.6666

The chance of picking 2 white balls if one is not replaced is 0.7 x 0.6666 = 0.46

The decision to choose a second white ball is yours. So you could say the chance of choosing a second ball is 0.7



I think