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Probability involving a three card hand.

  1. Jan 14, 2014 #1
    Ha! First post! Hope you guys can help me. It's not really a homework problem, but it's close enough.

    1. The problem statement, all variables and given/known data

    What is the probability of drawing a king, a red card, and a jack from a fair deck of 52 cards, in that order, without replacement?

    3. The attempt at a solution

    If you draw a red king vs. a black king it can affect the probability of drawing a red card next. Similarly drawing a red jack vs. a non-jack from the red suit for the second card will affect the probability of drawing a jack for the third card.

    So I divided it into four possible cases, and added together their probabilities:

    Red King, Red Suit (non-Jack), Jack:

    [itex]\frac{2}{52}[/itex]x[itex]\frac{23}{51}[/itex]x[itex]\frac{4}{50}[/itex]

    Red King, Red Jack, Jack:

    [itex]\frac{2}{52}[/itex]x[itex]\frac{2}{51}[/itex]x[itex]\frac{3}{50}[/itex]

    Black King, Red Suit (non-Jack), Jack:

    [itex]\frac{2}{52}[/itex]x[itex]\frac{24}{51}[/itex]x[itex]\frac{4}{50}[/itex]

    Black King, Red Jack, Jack:

    [itex]\frac{2}{52}[/itex]x[itex]\frac{2}{51}[/itex]x[itex]\frac{3}{50}[/itex]

    ∴ summing the probabilities yields a solution of [itex]\frac{2}{663}[/itex] I believe.

    This makes sense to me, but I have a friend who is arguing with me and telling me this is wrong. I'd like some confirmation either way, and if it is wrong some help with the correct approach.

    I tried using WolframAlpha to check, but it wouldn't recognize my question whenever I tried to ask. If anyone knows the correct syntax for asking WolframAlpha (or Mathematica or Matlab), I'd really appreciate it.

    Thanks in advance.
     
  2. jcsd
  3. Jan 15, 2014 #2

    Dick

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    It looks correct to me.
     
  4. Jan 15, 2014 #3
    Thanks for the reply man, I appreciate it.
     
  5. Jan 16, 2014 #4

    haruspex

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    There is a more direct approach. Clearly it doesn't change the answer if you specify the order as K, J, red.
    The probability of drawing a K then a J is 16/(52*51).
    Since the K and J are equally likely red or black, on average you will still have equal number of red and black cards remaining, so the probability of now drawing a red card is 1/2.
    If your friend still disagrees, you could post your friend's reasoning.
     
  6. Jan 16, 2014 #5
    Looks good to me as well.
     
  7. Jan 17, 2014 #6
    Thanks for the replies, they managed to convince my friend. I think he was trying to do it a different way with the tools he knew (using 52 choose 3), but couldn't work it out to his satisfaction.

    I wish I knew how to represent it in Mathematica, since there is a sense in the back of my mind that it knows commands for it. I liked the more direct approach, haruspex, thanks for showing me there's almost always a better, simpler method. I suspect it'll come in handy again for me.
     
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