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  1. #1
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    Question from someone via email

    While playing cards what is the probability of being dealt a hand of 8 cards that is already sorted if we ignore suits and just sort numerically, either high to low, or low to high?
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by MathGuru View Post
    While playing cards what is the probability of being dealt a hand of 8 cards that is already sorted if we ignore suits and just sort numerically, either high to low, or low to high?
    Oh, a sorted question makes this one tough. Would this follow a hyper geometric distribution?
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  3. #3
    Eater of Worlds
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    I may be misinterpreting and way off base, but this is sort of like a straight only with 8 cards instead of 5. Are there restrictions?. Are we allowed, say, J,Q,K,A,2,3,4,5?.

    Or just numbered cards, 2,3,4,5,6,7,8,9, for example?.

    The number of consecutive groups of 8 cards is 3. If we use A,2,3,...,10.

    In each sequence of 8 we can choose each card in 4 ways, this gives 4^{8}=65,536 ways.

    So the total is 65,536*3=196,608.

    But this includes straight flushes as well.

    \frac{196,608}{C(52,8)}.

    It's, no doubt, more complicated than this, but perhaps it's a start.
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  4. #4
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    There are some real issues with this one.
    Suppose we allow multiple denominations.
    That is say 2246667Q is a sorted hand of eight.
    Then there are 120055 possible hands.
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  5. #5
    Eater of Worlds
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    There are some real issues with this one.
    That's exactly what I was thinking, but I gave my 2 cents anyway. Probably worthless, but...........
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  6. #6
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    Hello, MathGuru!

    What is the probability of being dealt a hand of 8 cards that is already sorted
    if we ignore suits and just sort numerically, either high to low, or low to high?

    Without stretching the interpretation of the problem,
    . . I assume we want an "eight-card straight " dealt in numerical order.

    There are: . {52\choose8} \:=\: 4,515,258,175 possible eight-card hands.


    Let's assume increasing order . . .

    There are 7 possible straights . . . from A-to-8 to 7-to-A.

    For each straight, there are: 4^8 possible selections.

    Hence, there are: . 7 \times 4^8 \:=\:458,752 ways.


    Therefore: . P(\text{8-card straight, increasing order}) \;=\;\frac{458,752}{4,515,228,900} \;=\;\frac{16,384}{161,258,175} \;\approx\;0.0001

    . . and: . P(\text{increasing or decreasing}) \;\approx\;0.0002

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