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Taking a mean average of two probabilities to predict an outcome?

  1. Sep 2, 2012 #1
    I hope this the right place to ask this question. I appologize if this is in the wrong place.

    I am not from a probability background. I have tried to search for an answer to my question, but I have not found it in any of the literature which I can understand. I'm not even sure exactly how to classify the question, which has hampered my search efforts! Maybe someone can point me in the right direction.

    The question could be put something like this:
    "If A occurs then there is a 50% of C occuring next. If B occurs then there is a 60% chance of C occurring next. What is the chance of C occuring next if both A and B occur at the same time?"

    This is not a "homework" question or a question from a textbook! It is a simplified version of a question I am facing in a piece of Arduino code I am trying to write which predicts changes in galvanic skin response (currently as a hobby, but potentially moving towards medical research if it yields results).

    I won't enter on a tangent by using the specifics of the code in this forum, but a similar non-coding analogy might be if a teenager asks his/her parents if they can stay out late, and the mother (A) is 50% likely to say yes (C) if asked on her own and the father (B) is 60% likely to say yes (C) if asked on his own - so what is the likelihood they will say yes (C) if asked together when they are both in the same room?

    Obviously A and B are not exclusive - but nor are they completely independent. Sometimes they are influenced by the same causes, and sometimes they can be influenced by independent causes. When they do occur together then their effects can "stack", and if the probability of C following B is higher than the probability of C following A then the overall probability of C following both A and B occuring at the same time is at least greater than that of C following the smallest probability (of A on its own).

    My problem is that I'm unsure how to approach the combined probability. The literature I have read seems to suggest I either add the probabilities or multiply them, but neither would work in this case.

    If I add 50% and 60% I end up with 110% - which would make C occuring a certainty when A and B both occur together, which testing has proven is false.

    If I multiply 50% by 60% I end up with 30% - which is less than either, and so clearly wrong!

    I'm guessing I might need to take a mean average of 50% and 60%? Is that a standard process in these situations?

    I am unsure if A has an affect on B when they both occur together. At the moment I just want to work out the answer as if they have no affect on each other. Later I might go back and try to work out the influence of one on the other. I'm guessing that I should take a mean average of the two, and if it later turns out that either of A or B influences the other when they both occur together (eg, if one takes a more prominent role over the other) then I can give one of the values a greater weighting in the average?

    I have tried to work out the probability of C when both A and B occur through observational testing - but the reason I have to do it by looking at them both separately is that they both occur together so rarely that they do not produce a large enough sample size to be confident in the answer. There are hundreds of times when A occurs on its own, and hundres of times when B occurs on its own, but less than about 10 times when they both occur together (in each test period).

    I was hoping to take advantage of the large sample sizes for A and B separately to give a more confident answer to when A and B occur together than just relying on a < 10 sample size.

    I hope I have explained it clearly enough. Sorry if this is an obvious question, but I have very little experience in probability and I haven't managed to find the answer by searching so far. Perhaps someone could tell me the name of this problem at least so that I know where to look it up. Thanks.
    Last edited: Sep 2, 2012
  2. jcsd
  3. Sep 2, 2012 #2
    Perhaps I should also mention that the use of only two events (A and B) leading to C is a gross oversimplification. There are in fact at least 5 influencing causes which I have so far identified (and likely more yet to discover) - each with their own probability of causing C when they occur on their own. I would eventually like to be able to predict the likelihood of C occuring after any given combination (or indeed all) of these precursor events. The number of times when all these precursor events occur together is much less than 10 per test period, and so the need to combine the probabilities of them all separately to get a more accurate result than relying on a small sample size is even greater.
  4. Sep 2, 2012 #3


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    In your 'parents' example, there is clearly an interaction between the causes, so it probably does not represent your problem well. E.g., with the other parent present, each might not want to be the one to give permission.
    So assume there is no interaction. Let the two probabilities be p and q. C will happen if either A or B happens:
    A & B: p*q
    A & not B: p*(1-q)
    B & not A: (1-p)*q
    Total: p + q - p*q
  5. Sep 2, 2012 #4

    Stephen Tashi

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    You haven't stated enough information to compute the answer, but since this is real life problems there are no doubt other facts that you haven't stated yet.

    The statement "If A occurs, the probability of C is 50%" is symbolized by the notation
    Pr( C | A) = 0.50 and the event is read as "the probability of C given A".

    You are asking to find Pr(C | A and B)
    By Bayes formula Pr(C| A and B) = Pr( C and A and B)/ Pr( A and B)
    There are various ways to find the probabilities on the right hand side.
    For example Pr(A and B) = Pr(A | B) Pr(B) = Pr(B | A) Pr(A)

    If A and B are "independent events", Pr(A|B) = Pr(A) so Pr(A and B) = Pr(A) Pr(B)

    The numerator of the fraction can be found in various ways.

    Pr(C and A and B) = Pr(A| B and C) Pr(B and C) = Pr(A and B| C) Pr(C) = etc.

    You have only stated Pr(C|A) and Pr(C|B). You must know more values to be able to compute P(C | A and B).
  6. Sep 3, 2012 #5
    Thank you for your replies. Now I have a handle on it, and I know which direction to go in.

    So, in order to know P(C | A and B) then I must first know Pr(C and A and B) ...

    I guess I can go back and find Pr(C and A and B) experimentally without having to work it out from other information. Do I read Pr(C and A and B) as "the probability of C ever occuring (regardless of its relation to A or B), plus the probability of A ever occuring (regardless of its relation to B or C), plus the probability of B ever occuring (regardless of its relation to A and C)."?

    I shall now go and take a closer look at Bayes.

    If I am reading (and calculating) this correctly then that is the same as: 0.5+0.6-0.5*0.6 = 0.8. So the probability of both parents saying "yes" when asked together is 80%.
    But if we take the reverse, and look at the probability of them both saying "no" then we get: 0.5+0.4-0.5*0.4 = 0.7. So that would mean there is an 80% probability of "yes" and a 70% probability of "no". Assuming there are only ever two possible outcomes, and they can't sit on the fence, then how is that even possible? Either I'm being a dufus here, or this proves that A and B cannot be independent and that there must be an interaction between them ...
  7. Sep 3, 2012 #6

    Stephen Tashi

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    No. Think of a "Venn diagram" showing 3 circles, A,B,C. The probability Pr(A and B and C) is the represented by the area where all 3 circles overlap. If you added up the areas of the three circles that would (in general) be a larger area than the area where they all overlap. The probability Pr(C | A and B) is the ratio of the the area where all 3 circles overlap to the area where circles A and B overlap. You can't visualize it as a simple area, it is a ratio of areas.

    What you calculated is the probability of at least one of the parents saying "Yes", not the probability that both say "Yes". If they act independently, the probability that both say "Yes" is (0.5)(0.6).
  8. Sep 3, 2012 #7
    OK, I think I understand. Thank you so much for your help! That would have taken me so long to figure out on my own. Now I know where I'm going, thanks.
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