We roll a $6$-sided die $n$ times. What is the probability that all faces have appeared?

Question

How do I solve these questions, there are so many cases I know the total outcomes is$6^n$. However, I am lost after that.

Answer 1

There are ${{6}\choose{1}}$ ways to a pick a sequence where only a single face appears in 10 throws.

There are ${{6}\choose{2}}$ ways to pick two faces and then $2^{10}$ unique sequences of throws consisting of those faces only.

Similarly, ${{6}\choose{3}}(3^{10})$ for three faces, ${{6}\choose{4}}(4^{10})$ for four faces, and ${{6}\choose{5}}(5^{10})$ for five.

By inclusion-exclusion we therefore have

$${{6}\choose{1}} - {{6}\choose{2}}(2^{10}) + {{6}\choose{3}}(3^{10}) - {{6}\choose{4}}(4^{10}) + {{6}\choose{5}}(5^{10}) = 44,030,736 $$ ways of rolling a die 10 times and not getting all six faces at least once.

The are $6^{10} = 60,466,176$ possible sequences.

Therefore, the probability that all faces appear at least once is $$\frac{60,466,176 - 44,030,736}{60,466,176} = \frac{16,435,440}{60,466,176} \approx 27.18\%$$