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Math Help - Probability problem involving combinations

  1. #1
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    Probability problem involving combinations

    Here's the problem statement:

    A deck of 52 cards contains 4 aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 13 cards, what is the probability that all 4 aces will be received by the same player?

    The answer is \frac{4 \binom{13}{4}}{\binom{52}{4}} but I'm really having trouble interpreting that answer and seeing how to get it. I understand that it's written as (# of favorable outcomes) / (total # of outcomes) but I would think that the total number of outcomes would be \frac{52!}{(13!)^4}, the number of ways to deal 52 cards to 4 people in groups of 13. I don't get where \binom{52}{4} comes from. If anyone has any advice or a hint or something I'd love to hear it.
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  2. #2
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    Your result isn't correct.Take a look here :Hypergeometric distribution - Wikipedia, the free encyclopedia
    (from wikipedia)
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  3. #3
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    Hello, tonyc4l!

    A deck of 52 cards contains 4 Aces.
    If the cards are delat randomly to four players so that each player receives 13 cards,
    what is the probability that all 4 Aces will be received by the same player?

    The answer is \frac{4 \binom{13}{4}}{\binom{52}{4}}
    but I'm really having trouble interpreting that answer and seeing how to get it.

    I understand that it's written as (# of favorable outcomes) / (total # of outcomes)
    but I would think that the total number of outcomes would be \dfrac{52!}{(13!)^4}
    the number of ways to deal 52 cards to 4 people in groups of 13.

    This is correct, but unnecessary.

    I don't get where \binom{52}{4} comes from. .
    Neither do I.
    If anyone has any advice or a hint or something, I'd love to hear it.

    Consider one of the four players, say, \text{Mr.}\,X.

    What is the probability that he gets all four Aces?
    (Note that we don't care what the other players get.)

    There are: . {52\choose13} possible hands he could get.


    He must get the four Aces: . {4\choose4} = 1 way.

    And he gets 9 of the other 48 cards: . {48\choose9} ways.

    Hence, there are: . 1\!\cdot\!{48\choose9} hands which contain the four Aces.


    \displaystyle \text{So: }\; P(\text{X gets 4 Aces}) \;=\;\frac{{48\choose9}}{{52\choose13}} \;=\;\frac{48!}{9!\,39!}\cdot\frac{13!\,39!}{52!} \;=\;\frac{13!\,48!}{9!\,52!} .[1]

    . . . . . . =\;\dfrac{13\cdot12\cdot11\cdot10}{52\cdot51\cdot5  0\cdot49} \;=\;\dfrac{11}{4165}



    Since this could happen to any of the four players:

    . . P(\text{someone gets 4 Aces}) \;=\;4\cdot\dfrac{11}{4165} \;=\;\boxed{\dfrac{44}{4165}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The given answer has: . \displaystyle \frac{{13\choose4}}{{52\choose4}} \;=\;\frac{13!}{4!\,9!}\cdot\frac{4!\,48!}{52!} \;=\;\frac{13!\,48!}{9!\,52!}

    . . which happens to equal [1].


    But I too can't see the reasoning behind that answer.
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  4. #4
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    Nevermind! I wrote something dumb.
    Last edited by undefined; October 4th 2010 at 01:52 PM.
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  5. #5
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    I too wish someone could explain the reasoning behind the given answer.

    \dfrac{\binom{49}{9}\frac{(39!)}{(13!)^3(3!)}}{\fr  ac{52!}{(13!)^4(4!)}} also gives the correct answer.
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  6. #6
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    Quote Originally Posted by tonyc4l View Post
    Here's the problem statement:

    A deck of 52 cards contains 4 aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 13 cards, what is the probability that all 4 aces will be received by the same player?

    The answer is \frac{4 \binom{13}{4}}{\binom{52}{4}} but I'm really having trouble interpreting that answer and seeing how to get it. I understand that it's written as (# of favorable outcomes) / (total # of outcomes) but I would think that the total number of outcomes would be \frac{52!}{(13!)^4}, the number of ways to deal 52 cards to 4 people in groups of 13. I don't get where \binom{52}{4} comes from. If anyone has any advice or a hint or something I'd love to hear it.
    I had difficulty seeing the "book solution", but here is, I think, a plausible explanation.

    Suppose one of the players is Fred, and let's see if we can find the probability that he gets all 4 aces. For convenience of analysis, let's say the first 13 cards in the deck are dealt to Fred. If we simply distinguish between aces and non-aces, and we consider the aces indistinguishable, then there are \binom{52}{4} possible locations of the aces in the deck, all of which we assume are equally likely. Of these, \binom{13}{4} have all the aces among the first 13 cards. So Fred's probability of getting all the aces is
    \frac{\binom{13}{4}}{\binom{52}{4}}.

    Multiply by 4 to obtain the probability that some player, not necessarily Fred, gets all the aces.
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  7. #7
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    Quote Originally Posted by awkward View Post
    I had difficulty seeing the "book solution", but here is, I think, a plausible explanation.

    Suppose one of the players is Fred, and let's see if we can find the probability that he gets all 4 aces. For convenience of analysis, let's say the first 13 cards in the deck are dealt to Fred. If we simply distinguish between aces and non-aces, and we consider the aces indistinguishable, then there are \binom{52}{4} possible locations of the aces in the deck, all of which we assume are equally likely. Of these, \binom{13}{4} have all the aces among the first 13 cards. So Fred's probability of getting all the aces is
    \frac{\binom{13}{4}}{\binom{52}{4}}.

    Multiply by 4 to obtain the probability that some player, not necessarily Fred, gets all the aces.
    Thanks! This problem had been bugging me.
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  8. #8
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    Can someone explain to me how you guys worked the problem? I used the hypergeometric distribtion and it was really simple.

    4* \frac{{{4}\choose{4}}{{48}\choose{9}}}{{{52}\choos  e{13}}}

    When you work it out into decimal form, gives the same solution as posted in the original post; and I think it's much easier to read and understand.

    I'd love to understand how you guys came to your answer though.

    Nevermimd. Awkward's post makes sense. I still like the hypergeometric method more though. Thanks!
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  9. #9
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    Quote Originally Posted by downthesun01 View Post
    Can someone explain to me how you guys worked the problem? I used the hypergeometric distribtion and it was really simple.
    4* \frac{{{4}\choose{4}}{{48}\choose{9}}}{{{52}\choos  e{13}}}
    Did you read reply #3?
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  10. #10
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    Haha... Sorry. I missed that post
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