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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.

Digitallis
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saop293
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    Are you familiar with the Inclusion-Exclusion Principle? – N. F. Taussig Jun 11 '23 at 20:01
  • yes, but how does that apply in this case – saop293 Jun 11 '23 at 20:42
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    @saop293, it applies because (for example) the set of sequences of rolls that include no 4s, 5s, or 6s will be included in (be a subset of) the set of sequences that include no 5s or 6s, which will in turn be included in the set of sequences that include no 6s. – Cameron Buie Jun 11 '23 at 20:50
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    Your goal is to find the probability that all the faces have been seen. This is the complement of the probability that 1 hasn't been seen or 2 hasn't been seen or 3 hasn't been seen or.... To find the probability of that disjunction, inclusion-exclusion is optimal. – Cameron Buie Jun 11 '23 at 20:54

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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 $$\frac{60,466,176 - 44,030,736}{60,466,176} = \frac{16,435,440}{60,466,176} \approx 27.18\%$$

RTF
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    +1

    More generally, the probability is $$\frac{6^n-6\cdot 5^n+15\cdot 4^n-20\cdot 3^n+15\cdot 2^n-6}{6^n}$$ for $n$ throws, or equivalently, $$1-6\cdot\left(\frac56\right)^n+15\cdot\left(\frac23\right)^n-20\left(\frac12\right)^n+15\left(\frac13\right)^n-6\left(\frac16\right)^n.$$

     – Cameron Buie Jun 11 '23 at 20:44