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Math Help - probability of 52 card deck

  1. #1
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    probability of 52 card deck

    I have a few simple probability questions regarding draws from a deck of cards.

    1) If two cards are drawn face down, what is the probability that the second card is an ace?
    2) If it is known the first card draw is an ace, how would that change the answer to (1)?
    3) What is the probability that 2 randomly drawn cards are both aces?
    4) If two cards are drawn from a deck, how many different combinations of the two cards are possible if the order is not considered and if the order is considered?

    For (1), I think the answer is 4/51, but do I need to include the probability that the first card is not an ace?

    For (2), I think it's 3/51.

    For (3), I think it's: 4/52 * 3/51 = 1/221

    For (4) - no idea
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  2. #2
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    1. You didn't learn anything about the 1st card since it's face down. So this is like choosing 1 card from 52, your probability is 4/52.

    2. Correct
    3. Correct
    4. Order doesn't count - 52 choose 2.
    Order does count - 52 x 51
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  3. #3
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    Hello, chemekydie!

    1) If two cards are drawn from a standard deck,
    what is the probability that the second card is an ace?
    I'll do this the Long Way.


    There are 4 Aces and 48 Others.


    We must consider both possibilities:

    [1] The first card is an Ace: . p(\text{1st Ace}) \:=\:\frac{4}{52}

    . . The second card is an Ace: . P(\text{2nd Ace}) \:=\:\frac{3}{51}

    . . Hence: . P(\text{Ace, then Ace}) \:=\:\frac{4}{52}\cdot\frac{3}{51} \:=\:\frac{12}{2652}


    [2] The first card is not an Ace: . P(\text{1st not-Ace}) \:=\:\frac{48}{52}

    . . The second card is an Ace: . P(\text{2nd Ace}) \:=\:\frac{4}{51}

    . . Hence: . P(\text{not-Ace, then Ace}) \:=\:\frac{48}{52}\cdot\frac{4}{51} \:=\:\frac{192}{2652}


    Therefore: . P(\text{2nd Ace}) \;=\;\frac{12}{2652} + \frac{192}{2652} \:=\:\frac{204}{2652} \;=\;\frac{1}{13}\;\;{\color{red}**}




    2) If it is known the first card draw is an ace,
    how would that change the answer to (1)?

    I think it's: . \frac{3}{51} . . . . Right!

    Just reduce it to \frac{1}{17}




    3) What is the probability that 2 randomly drawn cards are both aces?

    I think it's: . \frac{4}{52}\cdot\frac{3}{51} \:=\:\frac{1}{221} . . . . Yes!


    4) If two cards are drawn from a deck,
    how many different combinations of the two cards are possible if:

    (a) the order is not considered?
    (b) the order is considered?

    (a) If the order is not important, it is a Combination problem.

    . . _{52}C_2 \:=\:\frac{52!}{2!\,50!} \;=\;1326


    (b) If the order is important, it is a Permutation problem.

    . . _{52}P_2 \:=\:\frac{52!}{2!} \:=\:2652


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    The answer to #1 comes as a surprise ... until we think about it.


    Suppose they asked about the 37^{th} card.

    Do we have to consider what happened in the first 36 draws?
    . . No, those events do not matter.

    Spead the deck face down on the table.
    Point to any card and ask "What is the probability that this is an Ace?"

    Since there are 52 outcomes and 4 of them are Aces,
    . . the probability is: . P(\text{Ace}) \:=\:\frac{4}{52} \:=\:\frac{1}{13}

    So, you see, we don't care about the other cards.


    Bottom line: qmech is absolutely correct!

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