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Help of permutations

  1. Jun 16, 2007 #1
    1. The problem statement, all variables and given/known data


    A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the 2nd half –deck. The half is then shuffled and a card is drawn from it. Compute the probability that this drawn card is an ace.



    2. Relevant equations

    -

    3. The attempt at a solution


    Ans: 4/27 + 3/27 +2/27+ 1/27(because cos the other half deck may have 4,3,2,1 aces after the original ace was placed there?)
     
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  3. Jun 16, 2007 #2

    Dick

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    By that argument basically ANY stack of cards has that probability. What do you know about the 'half deck'? It has 1 ace and 26 cards that could be anything. If you pick a card, you will either i) pick the ace or ii) pick one of the other cards. What is the probability of each and given that what are the odds you are holding an ace in each of the two cases?
     
  4. Jun 16, 2007 #3
    So you have to use conditional probility?

    Assuming A is the situation where the 2nd half-deck has only 1 ace
    B is the situation where the 2nd half-deck has 2 aces
    C is the situation where the 2nd half-deck has 3 aces
    I is the situation where the ace that was picked was choosen
    D is the situation where any other aces was picked

    then P(I|A) + P(I|B) + P(I|C) + P(D|A) P(D|B) + P(D|C) ?
     
  5. Jun 16, 2007 #4

    Office_Shredder

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    Do you know how partitioning with probability works?
     
  6. Jun 17, 2007 #5

    Dick

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    You don't need to count the aces in the 2nd half deck. If you pick a card from it that isn't the replaced ace, what are the odds that it in turn is an ace?
     
  7. Jun 17, 2007 #6
    sadly I don't really quite know how , I'm pretty weak in this chapter

    3/52 X 1/27?
     
  8. Jun 18, 2007 #7

    Dick

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    I like 3/51 better. We know where one ace is, so the other 51 cards have 3 aces spread amoung them. So one of the 'other cards' has 3/51 probability of being an ace. Now if you draw from the half deck what the probability you will get one of the 'other cards'?
     
  9. Jun 18, 2007 #8
    probability of drawing 'other cards' would be 48/51 x 1/27? why would you still need this anyway, I thought the answer would be just 3/51 X 1/27 + 1/27(the ace that was drawn from the first deck)
     
  10. Jun 18, 2007 #9

    Dick

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    Almost, but why 3/51 X 1/27? There are 26 'other' cards in the 'half-deck'.
     
  11. Jun 18, 2007 #10

    NateTG

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    There are 27 cards once you've added the ace.

    kabab00:

    The chance that a different card than the transferred ace comes up is
    26/27, and not 1/27, right?
     
  12. Jun 18, 2007 #11

    Dick

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    'Other' cards are cards other than the transferred ace.
     
  13. Jun 18, 2007 #12
    3/51 x 1/27 is the chance that you would pick one of the other aces from the half deck
     
  14. Jun 19, 2007 #13

    Dick

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    It is not! 3/51 X 26/27 is. There are 27 cards in the deck. 26 are not the transferred ace. You have a 26/27 chance of drawing one of them and each of them has a 3/51 chance of being an ace. Now add on the odds of the other way to get an ace.
     
  15. Jun 19, 2007 #14
    the other ways to get an ace would be 1/27 ??(the one that was drawn from the first deck)
     
  16. Jun 19, 2007 #15

    Dick

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    Exactly correct.
     
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