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Probability card questions

  1. Oct 9, 2007 #1
    So, there is a deck of 10 card and a player picks 3 cards one at a time without replacement. What is the probability that the three cards are selected in sorted (increasing) order?

    I am not fully understand this question, for me there are two possibilities:
    1. The final three cards are sorted in order. (e.g. 1st = 3, 2nd = 1, 3rd = 2)
    2. Or, it is strictly increasing, 1st card < 2nd card < 3rd card.

    I do favor the 2nd possibility but i can not convince my self, perhaps anyone can help me.

    So, my approach is (for the 2nd possibility) to group those 10 cards to a group of 3 cards (in increasing order). Lets call it A, so

    A = {(1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9),(8,9,10)}
    and S is 10x9x8.

    So P(A) = 8/(10x9x8) = 1/90

    Is this the right thing to do??
  2. jcsd
  3. Oct 9, 2007 #2


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    Science Advisor

    No, that would be "consectutive numbers", not "in sorted order".

    Yes, that is what they mean.

    No, you are missing such "orders" as (1, 5, 8), (3, 6, 9), etc.
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