# Probability question

#### WinstonSmith

##### Member
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?

TWI

#### NotQuiteRandom

##### Active member
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

##### Well-known member
At a guess

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

Good question.

#### WinstonSmith

##### Member
"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

##### Well-known member
1-[(1-0.7)*(1-0.6)] = 0.88

T4 would say it improves your chances to use both.

#### WinstonSmith

##### Member
TWI said:
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

##### Well-known member
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

##### Member
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

##### Active member
WinstonSmith said:
"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

##### Well-known member
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.

#### NotQuiteRandom

##### Active member
A and B are NOT independent in this instance.