Finding the probability of 7 Trials, All 6 success

Suppose I roll a fair dice for 7 times. What is the probability that I will roll all the 6 sides of the dice?

What I have done is this:

$\displaystyle \left ( \frac{1}{6} \right )^6\times 7! = \frac{35}{324}$

But is this correct?

I am not sure if the above is correct because I thought it could be a Negative Binomial distribution too. The 7 throws with 6 sides show up is somewhat daunting and makes the question confusing.

Re: Finding the probability of 7 Trials, All 6 success

Quote:

Originally Posted by

**xEnOn** Suppose I roll a fair dice for 7 times. What is the probability that I will roll all the 6 sides of the dice?

What I have done is this:

$\displaystyle \left ( \frac{1}{6} \right )^6\times 7! = \frac{35}{324}$

But is this correct? I am not sure if the above is correct because I thought it could be a Negative Binomial distribution too. The 7 throws with 6 sides show up is somewhat daunting and makes the question confusing.

I get one-half your answer.

The string $\displaystyle 1123456$ can be arranged in $\displaystyle \frac{7!}{2!}$ ways.

We can have six times that number to account for the fact each of the six sides can be doubled.

There are a total of $\displaystyle 6^7$ possible strings,

Re: Finding the probability of 7 Trials, All 6 success

Thanks, Plato! You're right! The division of 2! to remove the repeated number during the 7th throw. Thanks!! :)