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    Tricky problem solution




    Hi,

    A soccer team plays matches.

    -- If in a previous match it won, it has a 0.6 chance to win in the next match (if it failed, the chances to win in the next match are 0.51).

    -- If in a match before the previous one it had won, the chances of winning next time are 0.55, but if it had failed, the chances of winning are 0.49.

    So the outcome of the next match is statistically dependant on two past events, and the outcome of a previous match depends on the match before too, respectively.

    If the team have won both previous matches, what are the chances it will win the next time?

    Could you please indicate possible way of solving this problem?

    Alexey

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    Re: Tricky problem solution

    Hello,

    So, I am interested in multivariate conditional probability. I have never read any literature on it. May I aks you to advise me on the authors to read or even specific articles?

    Alexey

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    Re: Tricky problem solution

    Quote Originally Posted by alexeymosc View Post
    Hello,

    So, I am interested in multivariate conditional probability. I have never read any literature on it. May I aks you to advise me on the authors to read or even specific articles?

    Alexey
    It just seems to me to be an ill-defined problem.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Tricky problem solution

    Quote Originally Posted by Dason View Post
    It just seems to me to be an ill-defined problem.
    I am not a pro in probability theory (I had a basic course in the university only). However, I am interested in analysis of time series with hidden dependancies and what some calls the long memory. It is just a hobby. I will try to explain what my diea is.

    I have recently came across a time series of high frequency currency quotes which made me think. I transformed the series into returns and if the return has a positive sign, let's denote it as a 1, and if the sign is negative, then 0.

    below are probabilities of occuring 1 given 1 or 0 at previous step(s):

    lag 1
    0---0,56
    1---0,44

    lag 2
    0---0,45
    1---0,55

    lags 1 and 2
    0---0---0,51
    0---1---0,41
    1---0 ---0,59
    1---1---0,49

    So the conditional probability when analyzing 2 events is different from that based on 1 previous event, which means according to my understanding that the uncertainty about the latest event becomes less.

    So, given only lag 1 and lag 2 pair-wise probabilities, is it possible to come at the 3-event conditional probability wihout having additional knowledge about the process?

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    Re: Tricky problem solution


    I have already discussed this problem at a probability forum in Russia (where I am from ), and I was told by a probability associate professor that the problem that I formulated cannot be solved using only pair-wise probabilities. I also experimented with frequencies on the real data and came to that conclusion as well. That means if I know conditional probabilities between A & B, and between B & C, I still cannot figure out a conditional probability of A given C. It was intriguing to me, but seems unresolvable. Kolmogorov's axiomatization of conditional probability involves knowing the frequency of a joint event A & C (in the following equation they are A&B): . There is no other way to get the figure. However, I think that a thought giant in probability - who I am not - can have a ground to dig this problem and look for possible ways to estimate this figure.
    Last edited by alexeymosc; 09-04-2013 at 01:07 AM.

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