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Math Help - Expected Value

  1. #1
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    Expected Value

    Confused in this section.. help appreciated..

    If anyone can help with this... is there not a calculator function that will help??

    Two examples:

    In a lottery 3,000,000 tickets are sold at a dollar a piece. Out of these 12,006 are drawn at random without replacement and awarded prizes of 12,000 $25, 4 $10,000 prize, 1 $50,000 prize and 1 $200,0000 prize. What is the expected value of 1 per week.

    Next one,

    In a state lottery a three-digit integer is selected randomly. If a player bets $1, on a number, the payoff is $500 minus $1 paid for ticket. Let X equal the payoff of -1 - 499 and find E(X)


    Just a heads up I may need help on Bernoulli next.
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  2. #2
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    Quote Originally Posted by skyslimit View Post
    Confused in this section.. help appreciated..

    If anyone can help with this... is there not a calculator function that will help??

    Two examples:

    In a lottery 3,000,000 tickets are sold at a dollar a piece. Out of these 12,006 are drawn at random without replacement and awarded prizes of 12,000 $25, 4 $10,000 prize, 1 $50,000 prize and 1 $200,0000 prize. What is the expected value of 1 per week.
    Hi there

    You should consider each prize and the probability of recieving thast prize.

    For $25: probabilty = 12,000/3,000,000

    For $10,000: probabilty = 4/3,000,000

    For $50,000: probabilty = 1/3,000,000

    For $200,000: probabilty = 1/3,000,000

    and finally if you don't win the prize is $-1 so

    For $-1: probabilty = (3,000,000-12,006)/3,000,000 = 2,987,994/3,000,000

    Now multiply each return by its respective probabilty and sum the products together. This is the expected value.


    E(X) = 25\times\frac{12000}{3000000}+

     10000\times\frac{4}{3000000}+

    50000\times\frac{1}{3000000}

    -1\times\frac{2987994}{3000000}

    =-0.80
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  3. #3
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    Very Helpful, thank you.

    On the second, I understand I must simply multiply the odds of of winning and losing (+- one dollar) by the winnings and loss. However, I can't seem to remember the odds of a random 3 digit integer... could it be simply would. Obvously 3! is too small... would it be, however, 27 * 27 * 27??
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  4. #4
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    Quote Originally Posted by cjmac87 View Post
    Very Helpful, thank you.

    On the second, I understand I must simply multiply the odds of of winning and losing (+- one dollar) by the winnings and loss. However, I can't seem to remember the odds of a random 3 digit integer... could it be simply would. Obvously 3! is too small... would it be, however, 27 * 27 * 27??
    With the second question I read it as there are 900 three digit numbers i.e (100, 101, 102, ..... ,997,998,999) as i would not include numbers like 001 or 023, because these are really 1 or 2 digit numbers (maybe or maybe not for this exercise!?)

    So  E(X) = (500) \frac{1}{900} + (-1) \frac{899}{900} = -0.44

    If you are including the numbers I left out befroe then the queation would be:

    So  E(X) = (500) \frac{1}{999} + (-1) \frac{899}{999} = -0.40
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