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Math Help - Probabilities of a Poker Hand

  1. #1
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    Smile Probabilities of a Poker Hand

    Hi,

    Here is my question

    I randomly draw 5 cards from a pack. (Ace is the highest card) What would be the probability of drawing a king, Queen, Jack etc as your highest card?


    I think below I have worked out the probability of an ace being your highest card:

    The probability of drawing an ace = 1- the probability of drawing 0
    aces
    The probability of drawing 0 aces is: C(48.5)/C(52,5)
    =1712304/2598960
    =0.65

    David

    So The probability I think of drawing an ace = 1-0.65 = 0.35

    Any help with this question would be much appreciated,

    Thank you
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  2. #2
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    Suppose we assign numerical values to the different cards.
    An ace=1 for the highest; king=2 for second highest, etc until 3=12 for lowest possible.
    Then P(X = k) = \frac{{\sum\limits_{j = 1}^4 {\binom {4}{j}\binom {52-4k}{5-j}} }}{{\binom {52}{5}}}

    So the probability that the hand has the a jack as the highest card is P(X=4).
    Attached Thumbnails Attached Thumbnails Probabilities of a Poker Hand-cards.gif  
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  3. #3
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    Smile An additional 2 questions

    Great Help, Thank you so much!!

    I have just 2 more questions which follow on from this that I would be grateful you could answer:

    1) What is the highest card likely to be?

    2) Someone offers you a "all or nothing bet", that in a randomly selected set of five cards, the highest card will be either King or Ace. Would you place the bet?
    (An "all or nothing bet" just means that if you bet $1, you will get $2 back if you are right and nothing if you are wrong)
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by IronMan View Post
    Hi,

    Here is my question

    I randomly draw 5 cards from a pack. (Ace is the highest card) What would be the probability of drawing a king, Queen, Jack etc as your highest card?


    I think below I have worked out the probability of an ace being your highest card:

    The probability of drawing an ace = 1- the probability of drawing 0
    aces
    The probability of drawing 0 aces is: C(48.5)/C(52,5)
    =1712304/2598960
    =0.65

    David

    So The probability I think of drawing an ace = 1-0.65 = 0.35

    Any help with this question would be much appreciated,

    Thank you
    The probability of not drawing an ace is:

     \frac{48}{52}\times \frac{47}{51}\times \frac{46}{50}\times \frac{45}{49} \times \frac{44}{48} \approx 0.66

    Probability of not drawing an ace or king (which is the probabilty that the highest card is a Q or less) is:

     \frac{44}{52}\times \frac{43}{51}\times \frac{42}{50}\times \frac{41}{49}\times \frac{40}{48} \approx 0.42

    P(highest~card~is~K)=1-P(highest~card~is~less ~than~Q)-P(highest~card~is~A)

    and so on

    RonL
    Last edited by CaptainBlack; March 9th 2008 at 12:59 PM.
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    The probability of not drawing an ace is:
     \frac{48}{52}\times \frac{47}{51}\times \frac{46}{50}\times \frac{45}{49} \approx 0.72RonL
    But five cards are dealt not four.
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  6. #6
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    Cool Thanks

    Hi thanks so much for the help. I'm a little confused though, which answer is right lol? I'm finding it a little hard to comprehend the question.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Plato View Post
    But five cards are dealt not four.
    Of course just need to put in the extra terms

    RonL
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  8. #8
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    Smile Thanks

    Ok got it. I really can't thank you enough, forever indebted thanks
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