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Math Help - A problem about mean (expectation)

  1. #1
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    A problem about mean (expectation)

    In a particular lottery, 3 million tickets are sold each weak for 50cents apiece. Out of the 3 million tickets, 12 006 are drawn randomly and without replacement and awarded prizes: 12 000 $25 prizes, 4 $10 000 prizes, 1 $50 000 thousand prize and 1 $200 000 prize. If you purchased a single ticket each week, what is the expected value of this game to you.

    I really need help on this one, my question is: Do the "high" prizes go first or the "low" prizes? How can we solve this?

    NB the answer in the book indicates -30.33 cents.

    Any help is appreciated...
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  2. #2
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    Let's count the probability of receiving a winning ticket. For the 25$-ticket we have P = 12000/3000000=1/250,
    for the 10000$-ticket it is P=4/3000000=1/75000,
    for both 200000$ and 100000$ - tickets it is 1/3000000.
    But for any ticket you should pay a price of 0.5$. So, when buying one ticket, the expected value of your gain is 25*(1/250)+10000*(1/75000)+200000*(1/3000000)+100000*(1/3000000)-0.5*1 =1/10+1/15+1/60+1/75-0.5=-0.3033$.
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  3. #3
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    Okay I got the solution, but I have something that's bothering me, since it was stated that the drawing was without replacement, shouldnt we account for that? The probabilities you gave obviously were for sampling WITH replacement, so we must have to use the combination some place, otherwise... Isn't that right?
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  4. #4
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    It was actually for without replacement. There's two reasons I can see you may be confused. First, with respect to how to calculate expectation:
    If you bet 1 on a coin flip, you have an expected value of 0. To calculate this you do 1*0.5 + -1*0.5 = 0. If when you won you got to flip again your equation would look like: -1*0.5 + 1*0.5 + 1*0.25 + 1*0.125... = 0.5

    And second, what expectation is: If you thought that you needed to change the probability for each ticket that wins. Remember that the expectation (in this case) is how much profit you can expect to make from purchasing a single ticket.

    Hope this help.
    -Matthew
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  5. #5
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    Quote Originally Posted by Rooo View Post
    [/tex],
    for the 10000$-ticket it is P=4/3000000=1/75000,
    for both 200000$ and 100000$ - tickets it is 1/3000000.
    But for any ticket you should pay a price of 0.5$. So, when buying one ticket, the expected value of your gain is 25*(1/250)+10000*(1/75000)+200000*(1/3000000)+100000*(1/3000000)-0.5*1 =1/10+1/15+1/60+1/75-0.5=-0.3033$.
    that doesn't even add up right! even when you put a 50 000 (the correct value) instead of 100 000... Plus I didn't get why you just multiplied the 0.5cents by 1...
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  6. #6
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    I just punched the second half in to a calculator. It all added up ok to me. The fractions look like they simplify correctly too (so long as you use 50,000). He multiplies the 0.5$ by 1 because that's the probability of losing that money as you have to pay that for the ticket.

    -Matthew
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  7. #7
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    GOD i dont know what's up with my calculator, I just checked them out... They're correct, thanks Mr. Mathew and Rooo...
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