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    Help solving probability scenario




    I play in a pool league and 4 of my teammates and I were busy practicing when we happened across an interesting scenario:

    Between 5 of us we played 3 frames against each player, scoring 1 point for a win and 0 for a loss. At the end everybody ended with 6 points each. We thought it was quite remarkable. Assuming all of us have equal ability, what would be the probability that this would occur?

    In summary: 5 players. Each plays 3 frames against his opponents (so each plays 12 frames total). Everyone wins 6 and loses 6 of their frames. What's the probability this would occur?

    I have tried to calculate this in a number of ways (primarily using factorials) but ultimately I don't think I'm using the correct formula.

    Can someone help. I'm happy to detail the methods I have used if anybody wants to know.

    Thanks!!

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    Re: Help solving probability scenario

    Have not think of a good way to enumerate the cases.

    For 2-man case it is easy as a Binomial random variable will count.

    For 3-man case the things got more complicated, but still not hard to count.

    For more people case, still thinking of a systematic way to count. Would you mind sharing your method as a brain storm?

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    Re: Help solving probability scenario

    I took each frame as having 2 possible outcomes: win/lose (kinda like a coin toss).

    Then I worked out the total number of paths to the total frames played (effectively 30 frames) =30!
    Divided this by 15! * 15! (Overall 15 frames won and 15 not won)

    Ie Total number of paths of the desired outcome= 30!/(15!)*(15!)

    Then I divided this by the total number of combinations in 30 games which is 2^30

    This gives a ratio of 0.144 which is 14.4%

    Seems too high to be correct.

    So I then tried working out the same ratio for 12 frames (what each player plays) :

    Ie 12!/(6!) * (6!) divided by 2^12

    This gives a ratio of 0.23

    Then 0.23^5 = 0.00064 or 0.064% which seems more plausible.

    But somehow this method doesnt seem correct?

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    Re: Help solving probability scenario

    I didn't read too much into what you wrote and it's still early for me but it seems your methods aren't including the information that when one person wins that forces a loss on their opponent. Since you're interested in more than one person it seems you need to incorporate this info.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Help solving probability scenario

    I threw together a quick simulation and it seems the probability (assuming equal skill levels) is quite low: 0.00347

    That was based on 100000 trials of your process.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Help solving probability scenario

    Let X_{i,j} be the number games won by player i when facing player j. Then

    X_{i,j} \sim \text{Binomial}(3, 0.5)

    and

    X_{j, i} = 3 - X_{i,j}

    So with this property we are effectively handling 10 i.i.d. Binomial random variables, and we are required to calculate

    \Pr\left\{\bigcap_{i=1}^5 \sum_{\substack{j=1 \\ j \neq i}}^5 X_{i,j} = 6\right\}

    And this can be break down by law of total probability, with conditioning one by one, starting from X_{1, 2}, X_{1, 3}, \ldots. The list will end before the 10th r.v. due to the constraint given.

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    Re: Help solving probability scenario

    Thanks for the replies. They are very helpful, but I am an absolute layperson when it comes to stats and I have absolutely no idea how to use your formula in application to the problem. Does that cover the entire 5 people? I'm assuming this because of the "5" in the superscripts. I thought I knew some mathematics but what you've written is completely Greek to me!!! Can you try elaborate a little?

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    Re: Help solving probability scenario

    If you're just looking for an answer I've given you basically the answer. Your approaches weren't correct. Do you want to understand the math behind it or are you just looking for this particular solution?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Help solving probability scenario

    If you're just looking for an answer I've given you basically the answer. Your approaches weren't correct. Do you want to understand the math behind it or are you just looking for this particular solution?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Help solving probability scenario

    Thanks Dason. I was looking for an answer but I'm also interested to understand the math behind it - if it's not too complicated!!

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    Re: Help solving probability scenario

    The math amounts to "Look at all possible outcomes, count up how many meet your criteria (everybody ends up with 6) and divide by the number of possible outcomes". I didn't feel like enumerating all possible outcomes so I ran simulations which will converge to the true probability as the number of trials in my simulation gets large - 100000 is fairly large so we expect that the answer I get via simulation should be pretty close to the truth.

    BGM gave some math that would allow us to reduce the computation a tiny bit but it still isn't something that is fun to do.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Help solving probability scenario

    Thanks very much for your time. It's much appreciated!!!

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    Re: Help solving probability scenario


    From the problem description, it should be clear that there are ten distinct pairings of players and each pair plays three games, yielding 30 games played in the round. Each game has two equiprobable outcomes (i.e., player 1 wins/player 2 loses, or player 1 loses/player 2 wins).

    Therefore, for a given order in which the 30 games are played, there are 2^30 (=1073741824) different possible win/lose outcome sequences. This is the size of the sample space.

    We want to know the size of the subset of those outcome sequences in which each of the five players ends up with six wins and six losses. A computer program to enumerate the sequences of interest counts 3774624 of them, giving the desired probability as 117957/33554432, or near enough 0.0035154.

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