A fair dice was thrown 10 times, and it's been registered that all numbers 1-6 have appeared at least once. If this is true, what's the probability that at least 2 sixes have appeared?
At first I thought I should calculate the amount of combinations of die throws that the 6 numbers that have definitely appeared as $\binom{10}{6}$, then dividing it by the amount of all possible outcomes, $6^{10}$, and then multiply it with the probability of throwing at least 1 six in the remaining 4 throws, since one has already been registered, $P(A)= 1-\left(\frac56\right)^4$.
Then I began to have serious doubts, as $6^{10}$ is an insanely large number, and whether I'd even need the possible outcomes of the 6 numbers that were thrown, and the only solution I could think of is that the all that matters would be the 4 throws we don't know about, regardless of in which throw we got one of the numbers 1-6, and that the answer would be the already mentioned: $$P(A)= 1-\left(\frac56\right)^4$$
Now I'm having minor doubts about this approach, and don't have a solution to this problem, so I would really appreciate if someone could tell me if this way of thinking is right or wrong.
Computing the Probabilities
Using Inclusion-Exclusion to compute the number of ways to get all five faces ($1$-$5$) with $d-n$ dice, we get that the number of ways to have $n$ sixes with $d$ dice is $$ \binom{d}{n}\left[\binom{5}{0}5^{d-n}-\binom{5}{1}4^{d-n}+\binom{5}{2}3^{d-n}-\binom{5}{3}2^{d-n}+\binom{5}{4}1^{d-n}-\binom{5}{5}0^{d-n}\right] $$ Therefore, the probability of getting $2$ or more sixes, given that we have at least $1$ six is $$ \frac{\displaystyle\sum_{n=2}^d\binom{d}{n}\left[\binom{5}{0}5^{d-n}-\binom{5}{1}4^{d-n}+\binom{5}{2}3^{d-n}-\binom{5}{3}2^{d-n}+\binom{5}{4}1^{d-n}-\binom{5}{5}0^{d-n}\right]}{\displaystyle\sum_{n=1}^d\binom{d}{n}\left[\binom{5}{0}5^{d-n}-\binom{5}{1}4^{d-n}+\binom{5}{2}3^{d-n}-\binom{5}{3}2^{d-n}+\binom{5}{4}1^{d-n}-\binom{5}{5}0^{d-n}\right] } $$ The results for different values of $d$: $$ \begin{array}{c|c} d&p&\text{approx}\\\hline 6&0&0.000000\\ 7&\frac16&0.166667\\ 8&\frac{17}{57}&0.298246\\ 9&\frac{17}{42}&0.404762\\ \color{#C00000}{10}&\color{#C00000}{\frac{1606}{3261}}&\color{#C00000}{0.492487}\\ 11&\frac{293}{518}&0.565637\\ 12&\frac{18868}{30083}&0.627198 \end{array} $$
Applying the Principle of Inclusion-Exclusion
Above, it is mentioned that we used inclusion-exclusion to compute the number of ways to get all five faces with $d-n$ dice.
To do this, we let $A_k$, for $k\in\{1,2,3,4,5\}$, be the arrangements without face $k$. Thus, the number of ways to arrange $d$ dice with some face missing, is $$ \begin{align} \left|\bigcup_k A_k\right| &=\overbrace{\ \sum_k\left|A_k\right|\ }^{\binom{5}{1}4^d}-\overbrace{\sum_{k_1\lt k_2}\left|A_{k_1}\cap A_{k_2}\right|}^{\binom{5}{2}3^d}+\color{#9000B0}{\overbrace{\sum_{k_1\lt k_2\lt k_3}\left|A_{k_1}\cap A_{k_2}\cap A_{k_3}\right|}^{\binom{5}{3}2^d}}\\ &-\underbrace{\sum_{k_1\lt k_2\lt k_3\lt k_4}\left|A_{k_1}\cap A_{k_2}\cap A_{k_3}\cap A_{k_4}\right|}_{\binom{5}{4}1^d}+\underbrace{\sum_{k_1\lt k_2\lt k_3\lt k_4\lt k_5}\left|A_{k_1}\cap A_{k_2}\cap A_{k_3}\cap A_{k_4}\cap A_{k_5}\right|}_{\binom{5}{5}0^d} \end{align} $$ Let's explain how to compute the purple term, the total of the intersections of three of the $A_k$. The others will be similar.
$\binom{5}{3}$ is the number of ways to choose the $3$ faces to be excluded. Once those have been excluded, there are $2$ faces onto which to map the $d$ dice; that is, $2^d$ ways. Thus, the purple sum is $$ \binom{5}{3}2^d $$ Thus, to count the number of ways to get all five faces with $d$ dice, we subtract from the number of ways to arrange all $d$ dice: $$ \binom{5}{0}5^d-\binom{5}{1}4^d+\binom{5}{2}3^d-\binom{5}{3}2^d+\binom{5}{4}1^d-\binom{5}{5}0^d $$ which is used above.