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  • rolling a dice problem

    A fair dice is rolled 7 times. Let Y1 , Y2 ,..., Y7 be the numbers turned up. Let Y=min( Y1 , Y2 ,..., Y7 ).

    Compute E(Y)

    a. 1.327
    b. 1.359
    c. 1.346
    d. 1.358
    e. 1.332
    The answer should be c, but i don't know

  • #2
    To calculate E[Y] You'd need the pdf for Y

    Can you think of a way to come up with one?

    (hint: Utilize the relationship between CDF and pdf; namely that the derivative of the CDF gets you the PDF. )

    Edit:
    Whoops! The above is for a continuous distribution! Sorry!
    For this discrete case try...

    Calculate
    1 * prob[1 is smallest roll] + 2*prob[2 is smallest roll] +...6*prob[6 is smallest roll]

    I THINK this is how you'd solve.
    Last edited by trsspidey; January 13 2012, 08:21 PM.

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    • #3
      Is this the entire question?

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      • #4
        Originally posted by sophiehan View Post
        A fair dice is rolled 7 times. Let Y1 , Y2 ,..., Y7 be the numbers turned up. Let Y=min( Y1 , Y2 ,..., Y7 ).

        Compute E(Y)

        a. 1.327
        b. 1.359
        c. 1.346
        d. 1.358
        e. 1.332
        The answer should be c, but i don't know
        You just make a discrete distribution Y for each possibilty for min(Y1:Y7)

        These are 1, 2, 3, 4, 5, 6.

        For instance, you could find the probability that min(Y1:Y7) is 6; or put more simply, that all die rolls are a 6.

        You then find the probability that min(Y1:Y7) is 5; or put more simply, that one roll is 5, and the rest are at least 5.

        You continue until you've made the entire distribution, and then find E(Y) of the discrete distribution.

        At least this is what I would do.

        Maybe there is an easier way?

        On a side note, I got 1.34588 which is c.

        The discrete distribution I got was:

        Y | P(Y)

        1 | .7209
        2 | .2206
        3 | .0507
        4 | .0073
        5 | .0005
        6 | .0000

        I rounded, but it's close to the distribution I got for Y.
        Last edited by bsd058; January 15 2012, 03:51 AM. Reason: redundancies

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        • #5
          I forgot, the easiest way to calculate the probabilities for the distribution seems to be as follows:

          P(Y=1) = 1 - P(Y>1) = 1 - (5/6)^7 = .7209.
          P(Y=2) = 1 - P(Y>2) - P(Y=1) = 1 - (2/3)^7 - .7209 = .2206.
          P(Y=3) = 1 - P(Y>3) - P(Y=2) - P(Y=1) = 1 - (1/2)^7 - .7209 - .2206 = .0507.
          .
          .
          P(Y=6) = 1 - P(Y=1) - P(Y=2) - P(Y=3) - P(Y=4) - P(Y=5) = .0000 or P(All die roles are 6) = (1/6)^7 = .0000. (although this is not exactly 0, it is somewhat negligible)

          Just in case you were wondering how I came to the distribution's probabilities. If you need an explanation, please let me know.
          Last edited by bsd058; January 14 2012, 03:46 PM.

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          • #6
            Thank you so much! I get it! Sometimes I get really fuzzy about maximum/minium problems...but this totally makes sense

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