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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?

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This is a Binomial distribution: en.wikipedia.org/wiki/Binomial_distribution –  1015 Dec 27 '12 at 19:19

1 Answer 1

up vote 5 down vote accepted

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 $(\lfloor n/6 \rfloor, \lfloor n/6 \rfloor, \lfloor n/6 \rfloor, \lfloor n/6 \rfloor, \lfloor n/6 \rfloor, n-5 \lfloor n/6 \rfloor)$

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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
1  
@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

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