probability of rolling at least N of C when rolling X dice I have been looking this up but am struggling to find a straightforward answer.
I am still confused about the process.
So basically what is the probability of rolling at least three of a particular number say X when rolling five dice.
I believe the way to go is work out the probability of rolling 5 of the same number then add it too the probability of 4 of the same number and so on.
so 1/6^5 + 1/6^4 + 1/6^3 + 1/6^2.
Is this correct or am I leaving something out?
I originally phrased my question wrong asking the probability of getting x of the same number.
A good example of what I mean is 
The probability of rolling at least three 2's when rolling five dice
 A: We assign a probability distribution to what we want to count. First define the random variable $X_1$ to be the number of $1$'s we throw, $X_2$ to be the number of $2$'s we throw, etc. Then $X_i$ is binomially distributed $B(\frac16,5)$. To compute the probability of throwing exactly $n$ of a particular number, say $N$, we have to find $$P(X_N=n)=P(X_1=n)$$ since $X_1$ has the same distribution as all the other $X_i$'s. Then $$P(X_1=n)={5\choose n}\frac1{6^n}\frac{5^{5-n}}{6^{5-n}}$$as described by the other answer. Then to find the probability of rolling at least $3$ of this number, you work out $$P(X_1=5)+P(X_1=4)+P(X_1=3)$$
A: First choose one to be shown $3$ times. There are $6\choose{1}$ ways to choose this die. Then use the standard binomial to get the probability of that selected die showing up exactly $3$ times.
$${6\choose{1}}\cdot{5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2 \approx .193$$
And for the same number being shown exactly $4$ times:
$${6\choose{1}}\cdot{5\choose{4}}\cdot\frac{1}{6}^4\cdot\frac{5}{6}^1 \approx .0193$$
And for the same number being shown exactly $5$ times:
$${6\choose{1}}\cdot{5\choose{5}}\cdot\frac{1}{6}^5\cdot\frac{5}{6}^0 \approx 7.713\cdot10^{-4}$$
Then summing these $3$ results we get $p \approx 0.213$
For a fixed value:
We take away the $6\choose{1}$'s since we are no longer choosing $1$ of the $6$ die to be shown at least $3$ times. We have
$${5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2 + {5\choose{4}}\cdot\frac{1}{6}^4\cdot\frac{5}{6}^1 + {5\choose{5}}\cdot\frac{1}{6}^5\cdot\frac{5}{6}^0 \approx 0.03549$$
