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Math Help - Help me troubleshoot my answer

  1. #1
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    Help me troubleshoot my answer

    You are playing a poker game. You have 55 and the board reads 358 giving you 3 of a kind. There are two more cards to come. What's the probability you make a full house or better?

    Here's my answer. You can only make a fullhouse or 4 of a kind. So on the first of the two cards to come, you can hit 7 out of 47 cards (3,5,or8). If you miss on the first card, you can hit 10 out of 46 cards. It's 10 because you can not only hit a 3, 5, 8 but also make a fullhouse by getting a pair in the last two cards. So, 1 - (prob(missing both cards)) = 1-(40/47 * (36/46)

    But it should also add up to the probability of hitting on the 1st card + probability of hitting on the second card - the probability of hitting on both 1st and second card. So, 7/46 + 10/46 - 6/46. But my two answers don't tie. What am I doing wrong?
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  2. #2
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    I think your answer is definitely correct (I got the same answer doing it a different way than you).

    I think it's your checking that's gone wrong! I don't really understand where you're getting your numbers in the second part. The probabilities are

    A) Miss first and second (40/47)*(36/46)=1440/2162
    B) Miss first, hit second (40/47)*(10/46)=400/2162
    C) Hit first (no matter what happens with second) 7/47 = 322/2162
    These add up just fine.

    Edit because I think I was wrong:
    If you take 10/46+7/47-(7/47*10/46) = 722/2162 you will notice that this is the same as 1-P(A) (what you wanted).
    Last edited by MissTK; June 11th 2007 at 07:22 AM.
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  3. #3
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    Hello, CrazyAsian!

    You are playing a poker game.
    You have \{5,5\} and the board has \{3,5,8\} giving you 3-of-a-kind.
    There are two more cards to come.
    What's the probability you make a full house or better?
    You are complicating the problem by ordering the draws.

    There are: {47\choose2} = 1081 possible draws.

    Four-of-a-kind
    You must get the fourth 5 . . . There is 1 way.
    Your fifth card can be any of the remaining 46 cards.
    . . There are: . 1 \times 46 \:= 46 ways to get 4-of-a-kind.


    Full House . . . There are 3 ways.

    [1] Draw another 3 and some other card.
    There are: {3\choose1} = 3 ways to draw another 3
    . . and 44 choices for the fifth card.
    There are: 3 \times 44 \:=\:132 to get "5s over 3s".

    [2] Draw another 8 and some other card.
    There are: {3\choose1} = 3 ways to draw another 8.
    . . and 44 choices for the fifth card.
    There are: 3 \times 44\:=\:132 ways to get "5's over 8's".

    [3] Draw a different pair (other than 3s or 8s).
    There are: 10 choices of the value of the pair
    . . and {4\choose 2} = 6 ways to get the pair.
    There are: 10 \times 6\:=\:60 more ways to get a Full House.

    Hence, there are: . 132 + 132 + 60 \:= 324 ways to get a Full House.


    Therefore, there are 46 + 324\:= 370 ways to get a Full House or better.

    . . The probability is: . \frac{370}{1081}

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