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    Joint Probability Distribution from Drawing 5 Cards




    Consider drawing 5 cards from a standard poker deck without replacement. I am attempting to construct a joint probability distribution table for 2 random variables: J = the number of Jacks drawn (0-4), and S = the number of Spades drawn (0-5). While I've completed the table, it doesn't add up correctly.

    I'm guessing the problem is centered around the Jack of Spades; a card that is both a Jack and a Spade introducing dependence. For example, I can compute the probability of drawing no Jacks as:

    P(J=0) = 48C5 / 52C5 = 0.6588

    as there are 48 cards that aren't Jacks. That should equal the sum of the first column of the table, or:

    P(J=0) = P(J=0,S=0) + P(J=0,S=1) + P(J=0,S=2) + P(J=0,S=3) + P(J=0,S=4) + P(J=0,S=5)

    where:

    P(J=0,S=s) = 12Cs x 35C(5-s) / 52C5

    as there are 12 cards that are Spades (and not Jacks). Unfortunately, this logic doesn't work as it yields P(J=0)=0.5902, which doesn't match the first result. Using 13Cs in the calculation oddly produces the original value.

    Not sure what help I'm asking for here; perhaps just some general guidance on solving problems like this. Thanks.

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    Re: Joint Probability Distribution from Drawing 5 Cards

    perhaps just some general guidance on solving problems like this
    Two methods springs to mind:

    If you can assume independence:
    f(e1,e2) = f(e1)f(e2)
    Hence find the marginals and then find the simultaneous....

    If you cannot assume independence use conditioning:
    f(e1,e2) = f(e1|e2)f(e2)

    both marginal distributions - for either spades or jacks - are easily found... drawing without replacement problem...

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    Re: Joint Probability Distribution from Drawing 5 Cards

    Quote Originally Posted by JesperHP View Post
    If you can assume independence:
    You can't \hspace{1cm}
    I don't have emotions and sometimes that makes me very sad.

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    Re: Joint Probability Distribution from Drawing 5 Cards

    By your first direct method,

    \Pr\{J = 0\} = \frac {\displaystyle \binom {4} {0} \binom {48} {5}} {\displaystyle \binom {52} {5}}


    By your second method with law of total probability,

    \Pr\{J = 0\} = \sum_{s=0}^5 \Pr\{J = 0, S = s\}

    = \sum_{s=0}^5 \frac {\displaystyle \binom {4} {0} \binom {12} {s} \binom {36} {5-s}} {\displaystyle \binom {52} {5}}

    = \frac {\displaystyle \binom {4} {0}} {\displaystyle \binom {52} {5}} \sum_{s=0}^5 \binom {12} {s} \binom {36} {5-s}

    In this case you are essentially partitioning the decks into three groups:

    1) Jacks, a total of 4 cards.
    2) Spades excluding Jack of spade, a total of 12 cards.
    3) The remaining decks, a total of 36 cards.

    We can see the latter one actually equal to the previous one, as we know that a hypergeometric pmf will sum up to 1, or known as the Vandermonde's identity.

    http://en.wikipedia.org/wiki/Vandermonde%27s_identity

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    Re: Joint Probability Distribution from Drawing 5 Cards

    To generalize it into joint probability function,

    \Pr\{J = j, S = s\}

    = \frac {\displaystyle \binom {1} {0} \binom {3} {j} \binom {12} {s} \binom {36} {5-j-s}
+ \binom {1} {1} \binom {3} {j-1} \binom {12} {s-1} \binom {36} {5-j-s+1}} {\displaystyle \binom {52} {5}}

    Now you partition the deck into 4 groups:

    1) Jack of Spade, a total of 1 card
    2) Jacks excluding Jack of Spade, a total of 3 cards.
    2) Spades excluding Jack of spade, a total of 12 cards.
    3) The remaining decks, a total of 36 cards.

    Conditioning on whether the Jack of Spade is included, we obtain the above formula. Note that the above formula cannot directly apply to all cases because you have to ensure those numbers inside factorials are within ranges. In fact for those extreme cases it is still applicable, but one of the summand will vanish because the presence/absence of Jack of Spade will directly restrict certain extreme outcomes.

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    Re: Joint Probability Distribution from Drawing 5 Cards

    Thanks for the advice. Once I realized there are 36 cards in the deck that are neither Jacks nor Spades, the numbers looked a lot better. Enclosed is what I came up with, if anyone's interested.

    But here's another rub. Using these figures, the Correlation Coefficient between J and S is basically. 0. Either I made another dumb arithmetic mistake, or this is one of those situations where the values are dependent, but uncorrelated. Is this a reasonable conclusion? Seems counter-intuitive to me.
    Attached Images  

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    Re: Joint Probability Distribution from Drawing 5 Cards

    I have come up with the same table and get the covariance

    2.05998 \times 10^{-17}

    When you try to multiply the two marginal pmfs, you will find that it is very close to the joint pmf. Therefore although they are dependent, they are almost independent and the covariance/correlation is so close to zero.

    It is not too counter-intuitive, and the reason behind can be explained a little bit: The set of Jacks and set of Spades only has one card overlapping, and you are only drawing 5 cards. Most of the decks does not belong to these two sets so the joint pmf is concentrated around (0, 0). So number of jacks yields very little information for number of spades and they are almost independent.

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    Re: Joint Probability Distribution from Drawing 5 Cards

    OK, I think I buy that. I guess I was expecting a less subtle result.

    If anyone's interested, the post I made on all this is available at http://proloquor.net/prolog/?p=138.

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    Re: Joint Probability Distribution from Drawing 5 Cards


    Enough threads get posted that this will get buried at some point. It might not hurt to post a link to this specific thread in your acknowledgments.
    I don't have emotions and sometimes that makes me very sad.

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