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Probability and Expected value

  1. May 1, 2008 #1
    This is for a discrete math homework set.

    1. Suppose the cards in a deck are given the following values: Ace has value 1, two has value 2,..., ten has value 10, Jack has value 11, Queen has value 12, and King has value 13. A player selects a card. If it is a heart, the player receives half the value of the card. If it is a diamond, the player receives twice the value of the card. If the card is black the player loses 15 What is the player's expected value in the game?



    2. I'm having issue dealing with the black cards in my attempted solution.



    3. I am going about the solution in this manner
    (52choose1 )* .5(13choose1)*2(13choose1)*[the value obtained from the -15 points and 26choose1]


    Am I approaching this question correctly?
     
  2. jcsd
  3. May 2, 2008 #2

    HallsofIvy

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    What you are doing looks much too complicated and I see no "probability" in your formula at all. Here's how I would do it:

    First, assuming the card drawn is a diamond, what is the expected value? What is the probability of drawing a diamond?

    Second, assuming the card drawn is a heart, what is the expected value? What is the probability of drawing a heart?

    Finally, assuming the card drawn is black, the expected value is -15. What is the probability of drawing a black card?
     
  4. May 3, 2008 #3
    so the probability of drawing a heart or a diamond is 1/4
    the probability of drawing a black card is 1/2

    The expected value once a heart is drawn is [(13+12+11+...+2+1)/13)]*2
    The '*2' because of the problem definition.
    The expected value once a diamond is drawn is [(13+12+11+...+2+1)/13)]

    The expected value once a black card is drawn is [(13+12+11+...+2+1)/13)]-10
    The '-10' term is from the problem definition.

    would that mean then that the total expected value would then just be
    .25*[(13+12+11+...+2+1)/13)]*2 + [(13+12+11+...+2+1)/13)] + [[(13+12+11+...+2+1)/13)]-10]?
     
  5. May 3, 2008 #4

    exk

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    The expected value of a heart would not be multiplied by 2 since you are receiving half the value, divide by 2 instead. You forgot to multiply by .25 and by 2 the expected value of the diamonds and finally the expected value of the black cards has nothing to do with their actual value. Since the probability of getting a black card is 1/2 and the value you get for a black card is -15 your expected value is.....?
     
  6. May 3, 2008 #5

    HallsofIvy

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    Then you had better go back and read your problem again. You told us
    " If it is a heart, the player receives half the value of the card."

    "If it is a diamond, the player receives twice the value of the card. "

     
  7. May 4, 2008 #6
    ah, it looks like I switched those.

    .25*[(13+12+11+...+2+1)/13)]*2 + .25*[(13+12+11+...+2+1)/13)]*.5 + .5*[(13+12+11+...+2+1)/13)]-15]

    is this correct then?
     
  8. May 4, 2008 #7

    HallsofIvy

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    Yes. And it looks to me like you really want to know what 1+ 2+ ...+ 12+ 13 is!
     
  9. May 4, 2008 #8
    :P
    yes I could add that up, but I tend to leave things written out like that so that I can quickly see where I got it from (to double check my work)
    thank you!
     
  10. May 4, 2008 #9

    exk

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    If you lose 15 points with certainty for any black card then why do you still need the expected value of a black card in your calculation:
    5*[(13+12+11+...+2+1)/13)]-15]
     
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