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Card Probability

  1. Oct 13, 2003 #1
    Useing a common deck of cards, you are dealt a hand of 7 cards. Whis is the probability that all the different suits are present in your hand?
     
  2. jcsd
  3. Oct 14, 2003 #2

    jcsd

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    I can't think of an easy way to do this, I think you'd just have to calculate and add the probailties of the different permutuations of cards (though of course treat all cards of the same suit as the same card) that gives you 4 different suits. It would probably be easier to work it out the other way by calculating and adding the probailties of all the different permutuations of cards that don't give you four different suits and because of unity take this figure away from 1.
     
  4. Oct 15, 2003 #3

    jcsd

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    After some consideration, I have found a simplish way to do it:

    Firstly, the denominator of all terms in the sequence that will give us the probilty will be the same:

    (52*51*50*49*48*47*46)

    So with the denominator out of the way the numerator of this fraction can be found, which is:

    4(7!/2!2!2!)(134*123) + 12(7!/3!2!)(134*122*11) + 4(7!/4!)(134*12*11*10)

    Which gives the total probailty as about 0.5696
     
  5. Oct 15, 2003 #4
    [?]

    jcsd:

    Please recheck your result. It seems to work out to a probability of about 1.7557.
     
  6. Oct 16, 2003 #5

    jcsd

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    Nope, I've re-checked it, I still get the same answer and I'm pretty sure that the method is correct.
     
    Last edited: Oct 16, 2003
  7. Oct 16, 2003 #6

    selfAdjoint

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    Your denominator is based on first card may be one of 52, second card may be one of 51, and so on. But this assumes a particular ordering of the hand. The cards might be in any order so you have to multiply by the number of possible arrangements, which is 7!. See the derivation of the binomial coefficients.
     
  8. Oct 16, 2003 #7

    jcsd

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    No the denominator will always be the denominator given, different orderings are taken into account in the numerator (the 7!/2!2!2!, etc. terms and the first coefficent takes into account differemt combinations), no specfic ordering is assumed.
     
  9. Oct 16, 2003 #8

    jcsd

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    As this is causing some confusion, I'll explain the exact meaning and resoning behind every term:

    The denominator must be (52*51*50*49*48*47*46) as the probailty of getting any unique permutuation of any combination of seven cards is 1/(52*51*50*49*48*47*46), thefore any other probanilty relating to the premutuations of the combination of seven cards will be an additon (or integr mutiple) of this term and will always have the same denominator.

    Given that four of the cards must be different from each other there are 3 possible combinations relating to the numbers per suite of each card in the hand: 2 cards each of three suites and 1 card of one suite: there are four possible ways of rotating the suites through this combination; 3 cards from one suite, 2 cards from one suite and 1 card from two suites: there are 12 possible ways of rotaing the suites through this combanation; and finally 4 cards from one suite and 1 card each from three suites: there are 4 possible ways of rotaing the suites through this combination.

    For each seperate combination there will also be permutations and these are represented by the factorials in brackest, using the rule that when you have a number of objects the same the number of permutautions are equal to: n!/p!q!...
     
  10. Oct 16, 2003 #9

    Njorl

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    GAH!

    I just wrote up a nice neat solution to this and it vaporized!

    jscd's solution is correct. I duplicated it with a different method.


    Njorl
     
  11. Jul 21, 2004 #10
    I know this is an old thread but reading it has helped me refresh my understanding of probability calculations.

    However - I am having trouble getting the exact solution jcsd came up with. I approached the problem for the case where you have a 7 card hand with only three suits represented. The only way to have that seniario is for NO cards of a given suite to be in the dealt hand which means you only have 39 of the 52 cards in the deck from which to make up your hand. The total number of unique 7 card hands that can be made from only 39 cards (3 suites present) would be combine(39,7), or 15,380,937. Since any of the 4 suites could be absent you would multiply that number by 4 to get 61,523,748.

    Repeating that process for the case where 2 suits are missing (combine(26,7) * 6) and the case where 3 suits are missing (combine(13,7) * 4) you come up with a total number of hands that don't have at least 1 of each suite represented by adding these three, or 61,523,748 (3 suites) + 3,946,800 (2 suites) + 6,864 (all 1 suite) = 65,477,412.

    Dividing the number of possible hands without all four suits present by the total number of 7 card hands that could be made from 52 cards and subtracted from 1 should be the probability of having all four suites present. The solution I get from the above exercise is 0.5106, not 0.5696.
     
  12. Aug 3, 2004 #11
    You've made a mistake. In combine(39,7) you don't have, necessarily, 3 suites presents.

    The jcsd's solution is ok.
     
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