If six fair dice are rolled what is probability that each of the six numbers will appear exactly once?
[Math] Probability of All Distinct Faces When Six Dice Are Rolled
diceprobability
Related Solutions
You've forgotten the ordering of the dice. It could be the first four dice that come up as threes. It could be the first two and the last two. And so on.
Each one of those orderings has the same probability of happening, and that probability is the one you've calculated. That means you need to multiply what you have by the number of different orderings, which is $\binom{10}4=210$.
Oh, and now that I've gotten a closer look at your formulas (since they've become well formatted), I can see that you've mixed up two different (correct) ways of thinking: either you take the probability of each die being a success and multiply them to get $(1/2)^6(1/6)^4$, or you take the number of possible good dice throws, and divide by the possible number of throws to get $\frac{3^6\cdot 1^4}{6^{10}}$. Remember to keep straight in your head which one of these you're using. They give the same answer, but do not go together well.
You have multinomial distribution with $n$ trials, $k = 6$ outcomes and all $p_i = \frac{1}{6}$. So probability of getting two equal rolls is $\sum\limits_{x_1 + \ldots + x_6 = n} \left(\frac{n!}{x_1! \ldots x_6!}\cdot \frac{1}{6^n}\right)^2$. It's unlikely there is good closed form for it. For asymptotic see, for example, this answer on mathoverflow.
For example, if we have $n = 2$, we need to sum over all variants to partitioning $2$ into $6$ non-negative terms. There are $15$ of them with $2$ ones and $4$ zeroes: $1+1+0+0+0+0$, $1+0+1+0+0+0$, ..., $0+0+0+0+1+1$ and $6$ with $1$ two and $5$ zeroes: $2+0+0+0+0+0$, $0+2+0+0+0+0$, ...
First variant will give in our sum $15$ terms equal to $\left(\frac{2!}{1!1!0!0!0!0!} \cdot\frac{1}{36}\right)^2 = \frac{1}{18^2}$. Second will give $6$ terms equal to $\left(\frac{2!}{2!0!0!0!0!0!} \cdot \frac{1}{36^2}\right)^2 = \frac{1}{36^2}$. So the answer is $\frac{1}{18^2}\cdot 15 + \frac{1}{36^2}\cdot 6 \approx 0.05$.
Best Answer
Imagine you throw one after the other. You consider a throw as a success if the number is different from all previous numbers. You start with one. This is always a succes so $P(\text{first}) = 1 = \frac{6}{6}$. Your second throw is a success if one of the remaining $5$ numbers shows, so $P(\text{second}) = \frac{5}{6}$. And so on. Since all the throws are independent, the total probability is the product of all separate probabilities:
$P(\text{all numbers are different}) = \frac{6}{6} \cdot \frac{5}{6} \cdot \frac{4}{6} \cdot \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6}$