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Math Help - discrete distributions

  1. #1
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    discrete distributions

    Suppose you pay $5 to the dealer, who then shuffles a deck of cards and
    turns over the top six of them. You receive $3 for every diamond.

    (a) Find the probability of winning (net) some money.
    (b) Compute the expected value and standard deviation of your (total)
    net win in one round of this game,
    (c) and in 15 independent rounds of this game.

    Answers:

    (a) I calculated the probability of getting 0 diamonds and 1 diamond, and subtracted that from 1

    Pr(0D) = (39C6)/(52C6) = 0.1603
    Pr(1D) = (13C1)(39C5)/(52C6) = 0.3677

    1 - Pr(0D) - Pr(1D) = 0.4720

    (b) For this I calculated the probability of 0, 1, 2, 3, 4, 5 6 diamond scenerios (similar to above), but then I multiplied each probability the the net loss or winnings that would occur.

    I got E(X) = -0.8155
    which makes sense since the probability of winning is less than 1/2

    However, I am having a hard time finding out the process to get the standard deviation in this case. Any help is appreciated.

    and for part (c), would the answer be the same as in part (b)? although there are 15 games, they are still independent of each other.
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  2. #2
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    Quote Originally Posted by chrisc View Post
    Suppose you pay $5 to the dealer, who then shuffles a deck of cards and
    turns over the top six of them. You receive $3 for every diamond.

    (a) Find the probability of winning (net) some money.
    (b) Compute the expected value and standard deviation of your (total)
    net win in one round of this game,
    (c) and in 15 independent rounds of this game.

    Answers:

    (a) I calculated the probability of getting 0 diamonds and 1 diamond, and subtracted that from 1

    Pr(0D) = (39C6)/(52C6) = 0.1603
    Pr(1D) = (13C1)(39C5)/(52C6) = 0.3677

    1 - Pr(0D) - Pr(1D) = 0.4720

    (b) For this I calculated the probability of 0, 1, 2, 3, 4, 5 6 diamond scenerios (similar to above), but then I multiplied each probability the the net loss or winnings that would occur.

    I got E(X) = -0.8155
    which makes sense since the probability of winning is less than 1/2

    However, I am having a hard time finding out the process to get the standard deviation in this case. Any help is appreciated.

    and for part (c), would the answer be the same as in part (b)? although there are 15 games, they are still independent of each other.
    (b) Let X be the random variable winnings in a single game.

    You've already calculated E(X).

    Now calculate E(X^2). Then Var(X) = E(X^2) - [E(X)]^2 .....


    (c) If you expect to lose $0.8155 in one game, then wouldn't you expect to lose (15)($0.8155) if you played 15 games ....?

    Let Y = 15X. Then E(Y) = 15 E(X) and Var(Y) = 15^2 Var(X) (since the games are independent) ....
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  3. #3
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    thank you
    also, my value for b was off. i forget to add a certain number. but its fixed
    Last edited by chrisc; November 19th 2008 at 07:53 AM.
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