3 Die Probability

Question

The number of times you have to roll a set of 3 die to see a 6 on each die appears to be around ~10.555 according to a implementation of this in c++ and c#. How would this be statistically proven as instinct would suggest it should be 6 ?

Each die has 6 sides that are evenly distributed. All three die are thrown at the same time counting as one roll. Once all three die have displayed 6 at least once you can stop.

What I am looking for is the statistical proof that you are required on average to roll the set 10.555 times to have seen 6 on each of the 3 die at least once.

Many Thanks

Answer 1

$$\mathbb{E}(T)=\sum_{n=0}^\infty \mathbb{P}(T>n) = \sum_{n=0}^\infty 1-\left(1-(5/6)^n\right)^3={10566\over 1001}=10.55544456. $$

Answer 2

I can give you a hint why the expected value is not 6.

If you toss a single coin, the expected number of tosses $E_1$ before it shows a head is 2. Let's calculate the expected value $E_2$ that tossing two coins both of them have shown a head at least once. After the first toss, there are four cases, each with probability $\frac{1}{4}$: $HH, HT, TH, TT$. In the first case you are done, in the second and third case you need other $E_1$ tosses before the other coin shows a head, in the fourth case you are back to square 1. Therefore,

$$E_2 = \frac{1}{4}(1 + (1+E_1) + (1+E_1) + (1+E_2)) = \frac{1}{4}(8+E_2)$$

and therefore $E_2 = \frac{8}{3}$.