# Outcome of rolling a fair die 6 times

I'm failing to understand how to come to the answer to this question.

If you roll a fair die six times, what is the probability that the numbers recorded are $1$, $2$, $3$, $4$, $5$, and $6$ in any order?

The answer given is $6!(1/6)^6 = 3/324$

Can anyone explain to me how to get to that answer? I would really appreciate the help! :)

-
The answer's numerator should be 5. $\;$ –  Ricky Demer Jul 31 '14 at 2:44

On your first roll, you need to get any of the six possible outcomes (that is, anything will do). This has probability 6/6. On your second roll, you need to get something different than your previous result. This has probability 5/6. On you third roll you need to avoid the two previous values, which has probability 4/6. Carrying on like this, the total probability is $$\frac66\times\frac56\times\frac46\times\frac36\times\frac26\times\frac16 = \frac{6!}{6^6}.$$
There are $6!$ ways to permute 1,2,3,4,5,6. Each of the permutations has $(1/6)^6$ chance of occurring. Since they are all mutually exclusive, the probability of any of the outcomes occurring is $6!(1/6)^6$.
@NamelessOne No, mutually exclusive. If your roll sequence was 1,2,3,4,5,6 then it certainly isn't 6,5,4,3,2,1. And that shows that they're definitely not independent. Let X be the event "The sequence was 1,2,3,4,5,6" and let Y be the event "The sequence was 6,5,4,3,2,1". $\Pr(Y) = 6!/6^6\neq\Pr(Y\mid X) = 0$ so the events are not independent. –  David Richerby Jul 31 '14 at 7:59
Apologies, I misunderstood. I see that Narut meant that the permutations are mutually exclusive (allowing us to simply add their probabilities), whereas the rolls are independent (allowing us to say that the probability of a specific permutation is $1/6^6$). @David your example has a mistake in that $Pr(Y)$ should be $1/6^6$, but I get the point that you are trying to make. –  Nameless One Jul 31 '14 at 10:53