# Expected number of rolls on a die until each face has appeared at least twice

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?

• isn't it 14.7*2=29.4 Commented Apr 8, 2015 at 8:01
• @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. Commented Apr 8, 2015 at 8:16
• 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 [jstor.org/stable/… for a formula. Commented Apr 8, 2015 at 8:19
• The $o(1)$ is an asymptotic expression, it makes sense only for $n \to \infty$, the number you look for is $$E_2(6) = 6 \int_0^\infty \left[ 1 - \left( 1 - (1+t)e^{-t} \right)^6\right] \, dt.$$ Commented Apr 8, 2015 at 8:35
• ... which by WolframAlpha is about 24.134 Commented Apr 8, 2015 at 8:38

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

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

• My reply crossed the reply of Martini. I guess he used the same trick Commented Apr 8, 2015 at 8:42
• With an $n$-faced die with uniform probability, if you wait for all dices to appear $m$ times, the expected number of tries if $$E_m(n) = n \cdot \int_0^\infty \left[ 1 - \left( 1 - e^{-t}\sum_{k=0}^{m-1} \frac{t^k}{k!}\right)^n \right]\, dt.$$ Commented Apr 8, 2015 at 8:48
• Yes, this would be the correct answer for a symmetric $n$-sided die (coupon collector's problem) Commented Apr 8, 2015 at 8:49