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Consider the game of Three.

  1. Feb 12, 2013 #1
    Consider the game of Three."

    1. The problem statement, all variables and given/known data
    You shuffle a deck of three cards: ace, 2, 3. With the ace worth 1 point, you draw cards at random without replacement until your total points are 3 or more. You win if your total points are exactly 3. What is the probability that you win?


    2. Relevant equationsn
    Permutation?

    3. The attempt at a solution
    I think the desirable outcomes are given by 3P1. The sample space is 3!. So the chance of winning would be (3P1)/3!.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 12, 2013 #2

    HallsofIvy

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    Re: Consider the game of Three."

    When you say "With the ace worth 1 point", do you mean the other cards are worth 0?
     
  4. Feb 12, 2013 #3

    Ibix

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    Re: Consider the game of Three."

    I don't think your result is correct. Perhaps you could explain a little about your reasoning for your figures?

    It's worth noting that this is a very small sample space. There are only three cards and you are drawing without replacement so enumerating every possible sequence ad simply counting the outcomes is a perfectly reasonable approach.
     
  5. Feb 12, 2013 #4

    Dick

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    Re: Consider the game of Three."

    If so, how would you get to more than 3?
     
  6. Feb 12, 2013 #5

    HallsofIvy

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    Re: Consider the game of Three."

    Ah- I missed the "without replacement". Thanks.
     
  7. Feb 12, 2013 #6

    rude man

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    Re: Consider the game of Three."

    Do we have a winner?
     
  8. Feb 12, 2013 #7
    Re: Consider the game of Three."

    The cards are worth 1, 2, and 3 respectively.
     
  9. Feb 12, 2013 #8

    LCKurtz

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    Re: Consider the game of Three."

    No, the number of points in the sample space (outcomes) is 6
    No. The sample space could be viewed as {12, 13, 21, 23, 3, 3} or {123, 132, 213, 231, 312, 321}, depending on how you want to view it.

    Why?
     
  10. Feb 12, 2013 #9

    rude man

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    Re: Consider the game of Three."

    So what is that number? So we can vote if it's right or not ...
     
  11. Feb 12, 2013 #10

    Dick

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    Re: Consider the game of Three."

    3P1 is 3!/(3-1)!=3 is fairly standard notation. 3/3!=1/2. You can vote if you want. But it's still wrong.
     
  12. Feb 12, 2013 #11

    rude man

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    Re: Consider the game of Three."

    We're not all probabilists.

    I agree, that answer is wrong.
     
  13. Feb 13, 2013 #12

    HallsofIvy

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    Re: Consider the game of Three."

    The probability of winning on the first card, that is, drawing a 3, is 1/3.

    If the first card is not a 3 it might be an ace, with probability 1/3, in which case, to win you must next draw the 2, not the three. What is the probability of that?

    If the first card is not a 3 it might be a 2, with probability 1/3, in which case, to win you must next draw the ace, not the three. What is the probability of that?
     
  14. Feb 13, 2013 #13

    rude man

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    Re: Consider the game of Three."

    Ah, I smell a winner from H of I!
     
  15. Feb 13, 2013 #14

    LCKurtz

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    Re: Consider the game of Three."

    I expect everyone here except countzander knows the correct answer. We're waiting for an argument from him...something more than just a wrong answer.
     
  16. Feb 13, 2013 #15

    rude man

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    Re: Consider the game of Three."

    I agree.
     
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