5

Note: The die is fair, normal 1 - 6 die.

So I understand that the expected number of rolls until each face occurs is 14.7 by the following post: Expected time to roll all 1 through 6 on a die

but what is the expected number of rolls until each face has appeared at least 2 times?

Community
  • 1
VelariaLol
  • 53
  • isn't it 14.7*2=29.4 – RE60K Apr 08 '15 at 08:01
  • @ADG that's my intuition, because you can simply restart the problem again. But i'm not sure if that's correct or not, just a guess. – VelariaLol Apr 08 '15 at 08:16
  • 1
    No, it's not. @ADG, because while waiting for the first complete "set", you will usually roll other faces for the second time. See here [http://www.jstor.org/stable/2308930?origin=crossref&seq=1#page_scan_tab_contents] for a formula. – martini Apr 08 '15 at 08:19
  • 1
    The $o(1)$ is an asymptotic expression, it makes sense only for $n \to \infty$, the number you look for is $$ E_2(6) = 6 \int_0^\infty \left[ 1 - \left( 1 - (1+t)e^{-t} \right)^6\right] , dt. $$ – martini Apr 08 '15 at 08:35
  • ... which by WolframAlpha is about 24.134 – martini Apr 08 '15 at 08:38

1 Answers1

4

This problem can be numerically solved by using the Poissonization trick. This rather unknown but powerful trick (due to A.N. Kolmogorov) is discussed in Chapter 4 of the book Understanding Probability of Henk Tijms. This trick leads to the answer

$\int_{0}^{\infty}\big(1-(1-e^{-t/6}-(t/6)e^{-t/6})^6\big)dt\approx 24.134.$

user164118
  • 691
  • My reply crossed the reply of Martini. I guess he used the same trick – user164118 Apr 08 '15 at 08:42
  • With an $n$-faced die with uniform probability, if you wait for all dices to appear $m$ times, the expected number of tries if $$ E_m(n) = n \cdot \int_0^\infty \left[ 1 - \left( 1 - e^{-t}\sum_{k=0}^{m-1} \frac{t^k}{k!}\right)^n \right], dt. $$ – martini Apr 08 '15 at 08:48
  • Yes, this would be the correct answer for a symmetric $n$-sided die (coupon collector's problem) – user164118 Apr 08 '15 at 08:49