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Math Help - combining 2 probabilities

  1. #1
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    combining 2 probabilities

    I have 2 probabilities

    p1 is the probability that team A will score 0 goals
    p2 is the probability that team B will concede 0 goals




    how do i combine the 2 and calculate the probability of team A scoring 0 goals into team B?

    should the probability of A scoring 0 goals into team B not be the same as the probability of team B conceding 0 goals from team A?

    Is it as simple as p1*p2? I thought i might need to use conditional probability to work out the intersection of p1 and p2, but it doesnt seem to look right when I do it

    thanks for any help
    neil
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  2. #2
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    Unknown since we know nothing about Event A (Team A score 0 goals) or Event B (Team B concedes 0 goals - no idea what that means).
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  3. #3
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    ok i have the average number of goals that team A scores
    using the poisson formula = ((lambda^k) *( e^-lambda))/k! you can work out the probability that team A scores 0 goals( where k = 0, lambda is the average number of goals scored)

    I have the average number of goals that team B concedes
    using the poisson formula I can work out the probability that team B concedes 0 goals.

    is there any way of combining the 2 probabilities to work out the probability that team A scores no goals into team B?

    not sure how else to explain it
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  4. #4
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    concede means, to let in, have scored against them
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  5. #5
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    I have no idea how sports probabilities work (what assumptions and what not goes into them), and I would imagine there would be some conditional dependence based on who Team A is playing - for example a crappy team playing a crappy team versus a good team playing a better team; so for a real world model - I do not believe you can infer that the two probabilities are independent of each other. If it was a simple model - then yea, just multiply - your probability would be a sort of "general" probability that team A scores 0 against an arbitrary team, and Team B concedes 0 goals against an arbitrary team.

    That's my interpretation at the least.
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  6. #6
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    Quote Originally Posted by neeeel View Post
    I have 2 probabilities

    p1 is the probability that team A will score 0 goals
    p2 is the probability that team B will concede 0 goals




    how do i combine the 2 and calculate the probability of team A scoring 0 goals into team B?

    should the probability of A scoring 0 goals into team B not be the same as the probability of team B conceding 0 goals from team A?

    Yes, the probability of A not scoring against B is the probability of B not conceding against A more or less.
    (B could conceiveably concede against themselves ... an "own goal")
    However the probability of A not scoring "in a game" is based on their fixture history,
    if they failed to score in 30 of 100 on average, the probability of not scoring is 0.3
    and the probability of B not conceding is based on their past games also.

    However, this does not mean that the probability of A not scoring against B is 0.3..
    Statistically speaking, we'd need to examine the previous games between the clubs
    and many other current factors.


    Is it as simple as p1*p2? I thought i might need to use conditional probability to work out the intersection of p1 and p2, but it doesnt seem to look right when I do it

    Not in this case, since they are playing each other.
    If they were playing different teams, then the probability of A not scoring and B not conceding is the
    product of the probabilities, not taking into account the details of the opponents.

    For example, if B have always failed to keep a clean sheet,
    and A score in half their games,
    would the probability of a 0-0 score be close to 0.5 ?


    thanks for any help
    neil
    There are 4 possibilities

    (1) A score and B score
    (2) A score and B do not
    (3) A do not score but B do
    (4) neither score

    You could sum the probabilities of (3) and (4), but since some of the data is not available

    you need to calculate the probability that A do not score against B
    "given that" the probability of B not conceding a goal is p_2
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    you need to calculate the probability that A do not score against B
    "given that" the probability of B not conceding a goal is p_2
    this seems to imply using conditional probability

    if p1 = the probability that A do not score
    p2 = the probability of B not conceding a goal

    P(p1 | p2 ) = P(p1 intersection p2) / p2

    in this case though, i just end up with p1 as the answer, since (p1 intersection p2)/p2 = (p1 *p2)/p2 = p1

    Im still confused as to how i get the probability that team A do not score into team B
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  8. #8
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    Quote Originally Posted by neeeel View Post
    this seems to imply using conditional probability

    if p1 = the probability that A do not score
    p2 = the probability of B not conceding a goal

    P(p1 | p2 ) = P(p1 intersection p2) / p2

    in this case though, i just end up with p1 as the answer, since (p1 intersection p2)/p2 = (p1 *p2)/p2 = p1

    Im still confused as to how i get the probability that team A do not score into team B
    Do you see how you lost the information about team B ?
    You need to dig deeper.
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  9. #9
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    I havent lost it, I never had it in the first place :S

    I know u want me to work it out myself, I do too, but i need a push in the right direction
    neil
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  10. #10
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    Quote Originally Posted by neeeel View Post
    I havent lost it, I never had it in the first place :S

    I know u want me to work it out myself, I do too, but i need a push in the right direction
    neil
    What I mean is that the probability p_1 is just the ongoing average chance
    of Team A not scoring.

    That is not the probability of Team A not scoring against Team B.

    The solution will involve p_2

    so the solution tried does not work.
    It surely requires p_2 to be part of the solution, not cancelled.

    The probability of Team A not scoring if Team B do not concede is 1.
    The probability of Team B not conceding if Team A does not score is also 1
    (own goals ignored).

    P(AB) cannot be the multiplication of the probabilities because in this game
    Team A not scoring and Team B not conceding are not 2 events.
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  11. #11
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    Only thing I can come up with is taking the mean, ie (p_1 + p_2)/2 but i dont think thats a very good solution
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  12. #12
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    Quote Originally Posted by neeeel View Post
    Only thing I can come up with is taking the mean, ie (p_1 + p_2)/2 but i dont think thats a very good solution
    The reason the conditional probability solution you tried earlier doesn't work
    is.... it applies to A and B playing in different games.
    The probability of A not scoring against an arbitrary team
    given that B did not concede against another arbitrary team
    will end up giving you the probability that A do not score on average
    if you use that conditional probability formula.
    The events are then independent and multiplying the probabilities is feasible
    to give the probability of both of 2 events happening.

    In this case there is a complete overlap between A not scoring and B not conceding.
    Also, if the probability of B not conceding is very small,
    then the probability of A not scoring against B is much lower than
    if the probability of B not conceding is very high.
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  13. #13
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    ok im totally confused now

    In an earlier post, you said that multiplying the probabilities was not the solution

    now you are saying that it is the solution?

    so

    probability that A doesnt score into B = probability that A doesnt score * probability that B doesnt concede???
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  14. #14
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    No, I'm saying that multiplying the probabilities applies if A and B play different teams.
    You need to become clear on this.
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  15. #15
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    ok , the events are overlapping , so i can use p1 + p2 - (p1*p2) ?
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