Throwing dice 10 times 
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.
 A: I'm afraid it's back to the drawing board.
Here is a simple simulation in R
based on a million performances of the experiment. Values approximated
by this simulation should be accurate to around 3 places--good enough to see that the result is not consistent with $1 - (5/6)^4$
m = 10^6;  n = 10;  u = s = numeric(m)
for (i in 1:m) {
  dice = sample(1:6, n, rep=T)
  u[i] = length(unique(dice))
  s[i] = sum(dice == 6) }
a = mean(u==6);  b = mean(s >= 2);  bs = mean((u==6)&(s>=2))
a; b; bs; bs/a
##  0.27155     # Aprx prob all six faces seen
##  0.515156    # Aprx prob at lest two 6's seen
##  0.133668    # Aprx prob both of above
##  0.4922408   # Aprx P(at least two 6's | all six faces seen)
mean((s >= 2)[u==6])
## 0.4922408    # Another way to get cond'l prob
1 - (5/6)^(n-6) # Incorrect formula
## 0.5177469    # Incorrect answer

I tried this with various $n$'s,  with the following
results for 8 through 11:
 n    Correct    Incorrect
 8    .299       .366
 9    .407       .427
10    .492       .518
11    .565       .598

The discrepancy between the simulated result and the proposed incorrect solution grows with $n$. I believe this is because
the number of 6's grows with $n$, and there are corresponding increasing ambiguities.
Essentially the proposed solution, for $n = 10$ is to take it as
given that we already have seen faces 1 through 6 (when ranked), and now as
a separate independent experiment we only need to get at least
one additional 6 in four additional rolls. In fact, we have already
done 10 rolls, and if there is already more than one 6, it is
unclear which to ignore.

It may help to consider a simpler experiment. Toss a fair coin
  four times. Given that we have at least one head and one tail,
  what is the probability that we have at least two heads.
  The correct answer based on a sample space of 16 equally likely
  outcomes has 14 outcomes with at least one head and one tail.
  Of these there are 10 with at least two heads. So the correct
  conditional probability is $10/14 = 5/7.$
However, a bogus argument corresponding to the incorrect answer
  to the main Problem would say, we already have at one head and one tail, now we just need to make sure at least one of the other two tosses
  is a head, for an obviously incorrect probability of $3/4$.

I'm not an expert in combinatorics (especially at this time of day), so I won't try to give an
argument for a correct analytic solution to the main problem.
But I have reasonable confidence in my simulation and my reasons
for believing we do not yet have a correct analytic solution.
A: cause all the numbers appeared at least once you have 1 certain 6 among your 6 throws and you should calculate the probability of appearing at least one 6 in 4 throws, so you should calculate the probability of no 6 appearing which is equal to $\left ( \frac{5}{6} \right )^{4}$ and then subtract it from 1. So I think your answer is correct.
A: Hint: You know there was at least one roll each of 1, 2, 3, 4, and 5. Therefore, 5 of the rolls can be counted definitely out. You also know that a 6 was already rolled once. Therefore, all you need to calculate is the probability that in 4 more dice rolls, you'll get a 6 at least once again.
Does that simplify the problem enough for you to solve it?
