## Probability question

This is a discussion on Probability question within the Psychology, Risk & Money Management forums, part of the Methods category; You have to make a yes/no decision. To help you, you have two independent decision factors A and B with ...

 Apr 13, 2007, 12:41pm #1 Joined Apr 2007 Probability question 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?
 The following members like this post: TWI
 Apr 13, 2007, 12:55pm #2 Joined Mar 2004 p(A)*p(B) pA(0,1) and pB(0,1)
 Apr 13, 2007, 1:12pm #3 Joined Dec 2006 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. __________________ Ex nihilo, nihil fit.
 Apr 13, 2007, 1:30pm #4 Joined Jan 2004 At a guess 1 - [(1-P(A))*(1-P(B)] Good question. __________________ "I refuse to tiptoe quietly through life only to arrive safely at death"
 Apr 13, 2007, 1:33pm #5 Joined Apr 2007 "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?
 Apr 13, 2007, 1:41pm #6 Joined Jan 2004 1-[(1-0.7)*(1-0.6)] = 0.88 T4 would say it improves your chances to use both. __________________ "I refuse to tiptoe quietly through life only to arrive safely at death"
Apr 13, 2007, 1:52pm   #7

Joined Apr 2007
Quote:
 Originally Posted by TWI 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?

 Apr 13, 2007, 2:07pm #8 Joined Jan 2004 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. __________________ "I refuse to tiptoe quietly through life only to arrive safely at death"
 Apr 13, 2007, 2:15pm #9 Joined Apr 2007 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!
Apr 13, 2007, 2:20pm   #10
Joined Dec 2006
Quote:
 Originally Posted by 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?

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
__________________
Ex nihilo, nihil fit.

 Apr 13, 2007, 2:23pm #11 Joined Jan 2004 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. __________________ "I refuse to tiptoe quietly through life only to arrive safely at death"
Apr 14, 2007, 11:22am   #12
Joined Apr 2006
Quote:
 Originally Posted by TWI 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

 Apr 14, 2007, 1:14pm #13 Joined Dec 2006 A and B are NOT independent in this instance. __________________ Ex nihilo, nihil fit.
 Apr 14, 2007, 1:16pm #14 Joined Nov 2004 If I understand your question correctly then this link might be useful, http://www.mathpages.com/home/kmath267.htm regards zu
Apr 14, 2007, 3:00pm   #15
Joined Apr 2006
Quote:
 Originally Posted by NotQuiteRandom 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?