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Homework Help: Combinatorics and Probability

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data
    in a poker game we want to calculate the Probability to get differnet combinations. in a card deck there are 52 cards from 4 different series. each series has 13 cards. we assume that each card get 5 random cards.

    What is the number of combinations to get the cards? (The order isn't important)
    my answer: 52*51*50*49*48

    what is the number of combinations if we don't distinguish?
    My answer: 52C5-13*12*4 (the number of combinations to choose 5 cards - the number of combinations which repeat themselves)

    What is the probability to get K,J,Q,A,Flush and all the cards are from the same series?
    My answer: (52/52)(12/52)(11/52)(10/52)(9/52)

    what is the probability to get a one pair (two card of the same kind)?
    My answer: 3!*(52/52)(3/52)(48/52)(47/52)(46/52)

    We play N times. What's the probabilty that until the Nth game we won't get one pair and in the last game we will get one pair?
    My answer:[(1-3!*(52/52)(3/52)(48/52)(47/52)(46/52))^(n-1)]*(3!*(52/52)(3/52)(48/52)(47/52)(46/52))

    Is it correct?
    thanks
     
  2. jcsd
  3. Mar 11, 2010 #2

    LCKurtz

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    The "series" are usually called "suits". Do you mean to ask how many possible different 5 card hands can be dealt? If so, your answer is wrong. You have given the number of permutations of 52 things 5 at a time, P(52,5), which counts different orders as different hands. You want combinations, C(52,5), which doesn't distinguish different orders of the same 5 card hands.
    Don't distinguish what? This is a confused paragraph; see above.

    No. How many royal flushes are there? How many possible 5 card hands are there? Use those figures.

    It's more complicated than that. Normally, when you say you have one pair it means you don't have a better hand. So you can't have a full house, which includes a pair or 3 or 4 of a kind.
     
  4. Mar 12, 2010 #3
    If we don't distinguish between the suits...


    I don't understand what you meant. I need to find the probability to get a shape/color and all the cards are from the same suit.


    So what do I do?

    The other answers are correct?
     
  5. Mar 12, 2010 #4

    LCKurtz

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    What in the world does "shape/color and all the cards from the same suit" mean?

    I doubt your answers are correct, but it's hard to tell given that I can't figure out what you are really intending to ask. If these questions are from a text, please copy them verbatim so I know what they really say.
     
  6. Mar 12, 2010 #5
    O.k , Forget about it, I need help only with the following qeustions:
    what is the number of combinations if we don't distinguish between the suits?
    And:
    We play N times. What's the probabilty that until the Nth game we won't get only one pair and in the last game we will get one pair?
    Trying to answer: Is like (1-get one pair)^(n-1) * (get one pair)?
     
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