Probability of getting a number in 4 throws of a dice

Question

Lets say we are rolling a fair 6 sided dice 4 times and recording the number we get after each roll. What is the probability of getting a number $m$ with $m\in \{1,2,3,4,5,6\}$ say one or four times; and generalizing to some number of times.

I try solve this with combinations, where repetition is allowed, then with $r= $#rolls:

$\frac{(r+n-1)!}{r!(n-1)!} \Rightarrow \frac{(4+n-1)!}{4!(n-1)!} = \frac{(n+3)!}{4!(n-1)!}$

then getting the number one, four and $m$ times is just:

$\frac{(1+3)!}{4!(1-1)!}=1,\frac{(4+3)!}{4!(4-1)!}=35,\frac{(m+3)!}{4!(m-1)!},m\leq4$

And so the probabilities would be

$\frac{1}{4^6},\frac{35}{4^6},\frac{4^5(m+3)!}{3!(m-1)!},m\leq4$

Did I do it right?

Answer 1

Assuming the die is fair, the probability of getting any particular number $k$ times in $n$ rolls is binomial probability $$\binom{n}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{n-k}$$

Answer 2

First, the denominator of your fraction should be $6^4$, not $4^6$: this is the total number of possible results when throwing four ordered dice.

Second, your numerator doesn't look right, because it doesn't take into account the number of sides on a die. Also it says you are more likely to get four copies of the number you are looking for than one, which is clearly not right.

What you should have on the top is the number of ways to get $m$ copies of the number you are looking for. This would involve accounting for

• the choice of which $m$ dice have the required number, and
• what numbers appear on the other $4-m$ dice.