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expectation problem cont.

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  • expectation problem cont.

    Let X be a random variable representing the number of times you need to roll (including
    the last roll) a fair six-sided dice until you get 4 consecutive 6's. Find E(X).
    A. 125 B. 1024 C. 1554 C. 2048 D. 15447
    Solution.
    Note that if X = n, then the last four rolls are 6’s, the one just before was not a 6, and there
    were no four consecutive 6’s on the n - 5 rolls before that.


    The solution is the p(x)=n= (1/6)^4 * 5/6* P (x> n -5)





    I do not understand the P x > n-5 part in this equation. Can someone explain this with an example by choosing a value for n.
    pete2323
    Actuary.com - Level II Poster
    Last edited by pete2323; December 26 2012, 02:29 PM.

  • #2
    I just need a verbal explanation

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    • #3
      Originally posted by pete2323 View Post
      I just need a verbal explanation
      Search here and on AO. This problem gets discussed every few months it seems.

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      • #4
        I only see solutions posted on AO. I understand that the last 4 rolls have to be 6 so (1/6)^4 and the one before must not be a six so 5/6.



        Lets assume x = 8, then one solution could be 6, 6, 6, no 6, 6,6,6,6 and another could be 6, no 6,6, no 6, 6,6,6,6

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        • #5
          Ohh I think I get it, the beginning rolls could be 1 through 6, the one before the last four cannot be six, and the last four have to be six??

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          • #6
            so if x =4 then its (1/6)^4, x = 5 its 5/6* (1/6)^4, x=6 its 1*(5/6) *(1/6)^4, x=7, its 1*1*5/6*1/6^4........ is this right??

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            • #7
              Then you just do Ex is = 1* p1 +2*p2+...... to infinity ??

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              • #8
                Did you read daaave's explanation? http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=111709&highlight=consecutive

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                • #9
                  I am going to read his solution again, it is very involved though..... Does my solution make sense or no?

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                  • #10
                    No response to my question, just what I figured from NoMoreExams

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                    • #11
                      1*P[1] + 2*P[2] +3*P[3] + 4*P[4]+......


                      I am trying to find a solution to this problem using the basic expectation formula. The first three terms would be 0. The fourth would be 4*1/6^4, and then every other term would be multiplied by 5/6 * 1/6^4. Can anyone help me sum this arithmetic series?

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