# Determine the probability the number $j$ will appear $n_j$ times when $n$ dice are rolled

Suppose that $n$ balanced dice are rolled. What is the probability that the number $j$ will appear exactly $n_j$ times $(j=1,\ldots,6)$, where $n_1 + n_2 + \cdots + n_6 = n$?

I have derived two solutions: $$\frac{1}{{6+n-1 \choose n}}$$ and $$\frac{{n \choose n_1,n_2,\ldots,n_6}}{6^n}$$

They are not equivalent. Which one is correct? What is the incorrect solution falsely assuming?

-
This is a Binomial distribution: en.wikipedia.org/wiki/Binomial_distribution –  julien Dec 27 '12 at 19:19

Probability that a dice shows $j$ is $\dfrac16$ and doesn't show $j$ is $\dfrac56$.

Hence, the probability that out of the $n$ dices, $n_j$ dices show $j$ is $$\dbinom{n}{n_j} \left(\dfrac1{6} \right)^{n_j} \left(\dfrac56 \right)^{n-n_j}$$

EDIT

If you want the probability that $n_1$ dices show $1$, $n_2$ dices show $2$, $n_3$ dices show $3$, and so on, then the desired probability is $$\underbrace{\dbinom{n}{n_1} \left(\dfrac16 \right)^{n_1}}_{\text{Probability 1 occurs n_1 times}}\\ \times \underbrace{\dbinom{n-n_1}{n_2} \left(\dfrac16 \right)^{n_2}}_{\text{Probability 2 occurs n_2 times given that 1 has occurred n_1 times}}\\ \times \underbrace{\dbinom{n-n_1-n_2}{n_3} \left(\dfrac16 \right)^{n_3}}_{\text{Probability 3 occurs n_3 times given that 1,2 have occurred n_1,n_2 times}}\\ \times \underbrace{\dbinom{n-n_1-n_2-n_3}{n_4} \left(\dfrac16 \right)^{n_4}}_{\text{Probability 4 occurs n_4 times given that 1,2,3 have occurred n_1,n_2,n_3 times}}\\ \times \underbrace{\dbinom{n-n_1-n_2-n_3-n_4}{n_5} \left(\dfrac16 \right)^{n_5}}_{\text{Probability 5 occurs n_5 times given that 1,2,3,4 have occurred n_1,n_2,n_3,n_4 times}}\\ \times \underbrace{\dbinom{n-n_1-n_2-n_3-n_4-n_5}{n_6} \left(\dfrac16 \right)^{n_6}}_{\text{Probability 6 occurs n_6 times given that 1,2,3,4,5 have occurred n_1,n_2,n_3,n_4,n_5 times}}\\ = \dfrac{n!}{n_1!n_2!n_3!n_4!n_5!n_6!} \dfrac1{6^n}$$

The mistake when you write the probability as $\dfrac1{\dbinom{6+n-1}6}$ is that you are assuming that all the possible combinations of $(n_1,n_2,n_3,n_4,n_5,n_6)$ are equally likely to occur, which is not the case. For instance, $(n_1,n_2,n_3,n_4,n_5,n_6)=(0,0,0,0,0,n)$ is a highly unlikely event compared to $(\lfloor n/6 \rfloor, \lfloor n/6 \rfloor, \lfloor n/6 \rfloor, \lfloor n/6 \rfloor, \lfloor n/6 \rfloor, n-5 \lfloor n/6 \rfloor)$

-
No, this is incorrect. I am looking for the probability each $j$ appears exactly $n_j$ times. Your solution represents for one $j$. –  idealistikz Dec 23 '12 at 9:02
Thank you. What is the first solution incorrectly assuming? –  idealistikz Dec 23 '12 at 9:15
@idealistikz How did you obtain the first solution? It is not clear to me what argument gives you that answer. –  user17762 Dec 23 '12 at 9:18
The denominator is the total combinations of creating $n$ subsets with replacement. –  idealistikz Dec 23 '12 at 9:22
@idealistikz In your first case, you are assuming that all possible combinations of $(n_1,n_2,n_3,n_4,n_5,n_6)$ are equally likely to occur, which is not the case. For instance, $(n_1,n_2,n_3,n_4,n_5,n_6) = (0,0,0,0,0,n)$ is a highly unlikely event compared to $(n/6,n/6,n/6,n/6,n/6,n/6)$. –  user17762 Dec 23 '12 at 9:47