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Math Help - expected value

  1. #1
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    expected value

    you bet $5 in a game and there is a 251/ 495 probability that you wil lose that $5 and 244/495 probabilty taht you will win that you will win $5. given that your expected value is -7.07 cents, how much would you lose in the long run?

    this is my working:

    1st round: win -7.07 cents
    2nd rounnd: win -7.07 cents


    nth round: win -7.07 cents

    thus the answer should be a GP of 7.07 /( 1- 7.07) ...

    but my answer in my notes states that it is 1.4 cents.

    may i know where i have gone wrong? thanks!
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  2. #2
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    If you have an expected loss of 7.07 cents on each play, the total is NOT a geometric progression. It is just -7.07-7.07-7.07...= -7.07n cents where n is the number of times you played.
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  3. #3
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    then why would the amt lose in the long run be 1.4cents?
    how do i get that value?
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  4. #4
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    Quote Originally Posted by alexandrabel90 View Post
    you bet $5 in a game and there is a 251/ 495 probability that you wil lose that $5 and 244/495 probabilty taht you will win that you will win $5. given that your expected value is -7.07 cents, how much would you lose in the long run?

    this is my working:

    1st round: win -7.07 cents
    2nd rounnd: win -7.07 cents


    nth round: win -7.07 cents

    thus the answer should be a GP of 7.07 /( 1- 7.07) ...

    but my answer in my notes states that it is 1.4 cents.

    may i know where i have gone wrong? thanks!
    I think the book is giving an intermediate answer.
    The problem seems unrealistic, though
    (in terms of either losing the entire 5 dollars or not).

    Round1 E(loss) = \frac{7}{495}5\ dollars

    Round2 E(loss) = \frac{7}{495}\left(1-\frac{7}{495}\right)5=\left(\frac{7}{495}-\left(\frac{7}{495}\right)^2\right)5\ dollars

    Round3 E(loss) = \frac{7}{495}\left(1-\frac{7}{495}+\left(\frac{7}{495}\right)^2\right)5  =\left(\frac{7}{495}-\left(\frac{7}{495}\right)^2+\left(\frac{7}{495}\r  ight)^3\right)5\ dollars

    E(loss) long term = \left(\frac{7}{495}-\left(\frac{7}{495}\right)^2+\left(\frac{7}{495}\r  ight)^3-\left(\frac{7}{495}\right)^4+...\right)5\ dollars

    The two geometric sequences are

    \frac{7}{495}+\left(\frac{7}{495}\right)^3+\left(\  frac{7}{495}\right)^5+...

    -\left(\frac{7}{495}\right)^2-\left(\frac{7}{495}\right)^4-\left(\frac{7}{495}\right)^6-...

    The sum to infinity of the first is \frac{\frac{7}{495}}{1-\left(\frac{7}{495}\right)^2}=\frac{0.0141414}{0.9  998}

    The sum to infinity of the second is -\frac{\left(\frac{7}{495}\right)^2}{1-\left(\frac{7}{495}\right)^2}=\frac{0.00019998}{0.  9998}

    Hence the expected loss is 5 dollars multiplied by \frac{0.01414-0.00019998}{0.9998}=(0.01394)5\ dollars\ =\ 1.4c\ per\ dollar
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  5. #5
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    may i know why you how you get the equation for round 2 and hence forth?
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  6. #6
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    i think the last working doesnt give 1.4cents. 0.01395 (5) = 0.0697 dollars = 6.97 cents
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  7. #7
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    Quote Originally Posted by alexandrabel90 View Post
    may i know why you how you get the equation for round 2 and hence forth?
    The initial E(loss) is the "round 1" loss from $5.
    This value shrinks from round to round, while the probability of loss remains constant.

    \frac{244}{495}5-\frac{251}{495}5=-\frac{7}{495}5=-0.0707\ dollars

    Then in round 2 (being purely mathematical, considering that 7.07 cents had been lost),
    there is 5 dollars minus 7.07 cents to bet with, to put it that way!
    The expected loss is not 7.07 cents now, it's

    \frac{7}{495} of the "balance".

    Continuing on in this way develops the pair of geometric series.
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  8. #8
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    Quote Originally Posted by alexandrabel90 View Post
    i think the last working doesnt give 1.4cents. 0.01395 (5) = 0.0697 dollars = 6.97 cents
    That's what I meant by 1.4c per dollar.
    I think the book answer is pointing to this rather than a final answer of 1.4c.
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