Question: given that we have already seen 4 distinct faces of the dice so far, what is the expected number of rolls of the dice required until we see all 6? (The dice is fair.)
EDIT: The question is meant to indicate that we have seen EXACTLY 4 distinct faces so far.
I tried to model my solution in terms of the coupon collector problem/approach. Let $T$ be the number of rolls required until we see all $6$ faces. Let $F$ be the number of faces seen so far. Let $T_i$ be the number of rolls required to get the next new face given that we have already seen $4 + (i – 1) = 3 + i$ faces so far. Then $T_i – Geom((6 – (3 + i))/6) = Geom((3 – i) / 6)$ for $1 \leq i \leq 2$. Then $E(T | F = 4) = E(T_1) + E(T_2)$ via linearity, so it would be (because the expected value of a variable that takes distribution $Geom(p)$ is $1 / p$), $6/2 + 6/1 = 9$ rolls.
Is this reasoning/answer correct? I don't have too much intution for this kind of problem. If not please let me know where I went wrong!
Best Answer
If you already observed 4 faces, since the probability to observe one of the two faces remaining is $p = 2/6 = 1/3$, the expected number dice throw before we observe one of the two faces remaining is $3$ and with the same reasoning. When we will have observed the fifth face, the probability to observe the last face at the next dice throw is $p = 1/6$ so the expected number of dice throw before we observe the sixth given we observed the five first faces is $6$. In conclusion the number of dice throw before we observe the two last faces is $6 + 3 = 9$. Was writing it on my way home, so if it doesn'make sense please tell me and I will rewrite it in a clearer manner.