Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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! :)

share|improve this question
    
The answer's numerator should be 5. $\;$ –  Ricky Demer Jul 31 at 2:44

2 Answers 2

up vote 12 down vote accepted

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 $$ \frac66\times\frac56\times\frac46\times\frac36\times\frac26\times\frac16 = \frac{6!}{6^6}. $$

share|improve this answer

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 $6!(1/6)^6$.

share|improve this answer
    
I am sure that you mean to say 'independent', not mutually exclusive. –  Nameless One Jul 31 at 7:16
2  
@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 at 7:59
1  
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 at 10:53
    
@NamelessOne Thanks for the correction! If only one could edit comments... –  David Richerby Jul 31 at 12:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.