# Finding the probability of 7 Trials, All 6 success

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• Mar 13th 2012, 09:21 AM
xEnOn
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.
• Mar 13th 2012, 09:48 AM
Plato
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,
• Mar 13th 2012, 10:04 AM
xEnOn
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!! :)