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    There's this poker game that is popular among the Asian community.
    It is a 4-player game where each player is dealt with 13 cards.

    What is the probability of getting the following cards among the 13 cards in their hands:
    (a) 4 of a kind (i.e. any set of 4 cards with the same rank);

    edit: Sorry. That was a bit confusing. I just copied it from wikipedia and it says '4 of a kind'. Edited now.

    (b) A straight flush (i.e. straight and flush).
    For example,
    A 2 3 4 5 of the same suits,
    2 3 4 5 6 and so on ....
    ......
    ......
    J Q K A 2.


    I'm just curious to see which is harder to be dealt.
    edit: There is always an argument among the players on which is harder to be dealt. Hence, I would like to see the probabilities from a mathematical point of view.
    I can try to calculate it myself but I don't trust my statistics.
    I would rather do maths!!
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    (Original post by UglyDuckling)
    There's this poker game that is popular among the Asian community.
    It is a 4-player game where each player is dealt with 13 cards.

    What is the probability of getting the following cards among the 13 cards in their hands:
    (a) 4 of a kind (i.e. spades, hearts, clubs and diamonds);

    (b) A straight flush (i.e. straight and flush).
    For example,
    A 2 3 4 5 of the same suits,
    2 3 4 5 6 and so on ....
    ......
    ......
    J Q K A 2.


    I'm just curious to see which is harder to be dealt.
    4 of a kind defined as standard and not what you wrote?

    It's not a very simple question (which I suspect is why it hasn't been answered yet), I'll give it a go later, in a hurry now.

    edit: Also that's a non-standard definition of straight flush. It's not normally allowed to 'go around' an A.
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    (Original post by nota bene)
    4 of a kind defined as standard and not what you wrote?

    It's not a very simple question (which I suspect is why it hasn't been answered yet), I'll give it a go later, in a hurry now.

    edit: Also that's a non-standard definition of straight flush. It's not normally allowed to 'go around' an A.
    Oh I see. Nevermind. Just change the definition.
    Make sure you define the scope of a straight flush before doing the calculation. Thanks.

    edit: so I suppose 10 J Q K A should be the last of the straight flush.
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    (Original post by UglyDuckling)
    There's this poker game that is popular among the Asian community.
    It is a 4-player game where each player is dealt with 13 cards.

    What is the probability of getting the following cards among the 13 cards in their hands:
    (a) 4 of a kind (i.e. any set of 4 cards with the same rank);

    edit: Sorry. That was a bit confusing. I just copied it from wikipedia and it says '4 of a kind'. Edited now.

    (b) A straight flush (i.e. straight and flush).
    For example,
    A 2 3 4 5 of the same suits,
    2 3 4 5 6 and so on ....
    ......
    ......
    J Q K A 2.


    I'm just curious to see which is harder to be dealt.
    edit: There is always an argument among the players on which is harder to be dealt. Hence, I would like to see the probabilities from a mathematical point of view.
    I can try to calculate it myself but I don't trust my statistics.
    I would rather do maths!!
    Not that i play this poker game but sometimes you can't rely just maths alone to win every hand. Psychology
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    it's all luck and facial expressions
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    (Original post by fourdigit)
    Not that i play this poker game but sometimes you can't rely just maths alone to win every hand. Psychology
    thank you for your words of wisdom
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    I think they should do a poker statistics degree. I would definitely do that
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    (Original post by lifeadd1ct)
    I think they should do a poker statistics degree. I would definitely do that
    LOL!!
    Then the act of gambling will be lifted up to a whole new level!!
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    How about (13*4C4*48C9)/(52C13)=399/12495 for the first probability? (I may have made a mistake, so just wanting to check if someone else gets the same answer...) Note that I'm only worrying about getting at least one set of four of a kind here, I haven't discounted hands with e.g. 4 of A, 4 of 2 and 5 random cards.

    For the second part, I think this works (again I would like someone to check me);

    (4*10*5C5*47C8)/(52C13)=399/4998 (4 as there's symmetry wrt what suit and 10 as there are 10 cards a straight can start with (A through 10))
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    a) Assuming you mean at least one set of a 4-of-a-kind. This is how I worked it out.

    N choose R (nCr) gives the number of possible hands, with 52 cards and 13 in your hand:
    (52! / (39! * 13!)) = 635013559600

    There are 13 possible combinations of 4-of-a-kinds, one for each suit. 13

    Using nCr again to calculate the remaining number of possible combinations of the other 9 out of 48 cards to get dealt into your hand: (48! / (9! * 39!)) = 1677106640

    (13*1677106640) / 635013559600

    = 21802386320 / 635013559600

    = 0.0343337335

    ~3.4%

    I've probably gone wrong somewhere there, mind. Can anybody verify??
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    That looks like the same answer and method as I used.
 
 
 
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