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Probably weird

  1. Jun 29, 2005 #1
    The cards in a set of 36 are numbered 1 to 36. The cards are shuffled and four cards are dealt. What are the chances of them being dealt in descending order?
    Major help needed
    I have tried everything in this problem my answer was 1/8 which was wrong I hope you guys can do better.
     
  2. jcsd
  3. Jun 29, 2005 #2
    Well how many different sets of 4 cards can you get that are in descending order? (i.e. {36, 35, 34, 33} is one, {35, 34, 33, 32} is another... how many are there?)
     
    Last edited: Jun 29, 2005
  4. Jun 29, 2005 #3

    matt grime

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    either the second is higher than the first, or the third is higher than on of the previous two or the 4 is higher than one of the previous 3. hmm, A or B or C... inc exc principle then a bit of coditional prob. anyone?
     
  5. Jun 29, 2005 #4
    It's actually fairly simple. First you can easily determine how many different sets of 4 cards can you get that are in descending order, and you can probably figure out how many total hands there are. And well, yea... you know what to do from there.
     
  6. Jun 29, 2005 #5
    but there can 1000000000000000s of possibilties for eg 7,6,5,1 or 10,6,3,2
    like that
     
  7. Jun 29, 2005 #6
    Actually, your estimate is much larger than the actual number of possibilities. Try to think of it this way:

    How many possibilities are there for the first card? (Ans: Obviously 36)
    How many possibilities are there for the second card? (Hint: Not 36)
    How many possibilities are there for the third card?
    How many possibilities are there for the fourth card?

    Now you can multiply them up and it gives you the total number of possibilities for a 4-card deal.
     
  8. Jun 29, 2005 #7
    its not as easy as that.
     
  9. Jun 29, 2005 #8

    matt grime

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    given 4 distinct cards, a,b,c,d, each hand containing these 4 is equally likely to be dealt, exactly one of these is descending. how many ways of dealing a,b,c and d are there ie how many ways can you reciecve a hand with the cards 1,2,3,4 in? how many diffeent hands are there in total?
     
  10. Jun 29, 2005 #9
    I thought you meant consecutively descending. Ok, so the way I thought about it was this:

    You can add the number of ways you can get a descending order where your highest card (call it M) is 4 (since M can't be less than 4), plus the number of possibilities when M=5, M=6, all the way to M=36.

    The way of counting the number of ways to get a descending order when your highest card is M is (at least I think) easiest to see in the form of code:

    possible_descending_hands = 0;
    for (i=3, i < M, i = i + 1){
    for (j=2, j < i, j = j + 1){
    for (k=1, k < j, k = k + 1){
    possible_descending_hands = possible_descending_hands + 1;
    }
    }
    }

    Then of course, you loop again for M = [4,36] and that would be your number of possible descending hands. Note that I'm not suggesting you write a program for this. It's just that it's easiest to explain how to count your descending hands with the use of loops (you can yet convert that into a mathematical expression).

    [Edit: Hmm.. not allowed to put spaces at the beginning of lines. I guess that means no indentation for you :P]
     
  11. Jun 29, 2005 #10

    matt grime

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    well, you could do that but the answer is can be obtained just from thinking about it if it's any help the extreme cases are

    1. dealing 1 card in descending order: 1/1

    2. Dealing 36 cards in descending order 1/36!

    and i thought i gave sufficient hints last time!
     
  12. Jun 30, 2005 #11

    uart

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    For the case of consecetively decending it's also very easy.

    The first card must be 4 or more which has a probabilty of 33/36. After that there is no choice which card must come next so the probability is just 33/36 * 1/35 * 1/34 * 1/33 = 1/(34*35*36).
     
    Last edited: Jun 30, 2005
  13. Jun 30, 2005 #12

    matt grime

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    the question doesn't say they are in sequence.
     
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