What is the probability of rolling at least one one AND no sixes with $n$ six-sided dice?

Question

I gather that the probability of rolling at least one 1 when $n$ six-sided dice are rolled is equivalent to $1-(5/6)^n$ and conversely $(5/6)^n$ for rolling no sixes...

But how do you calculate the odds for both occurring at the same time?

Answer 1

Problems like this can be handled combinatorially: Out of the $6^n$ possible rolls, there are $5^n$ rolls for which no sixes occur, but that includes $4^n$ rolls for which no ones occur either, so the answer is

$${5^n-4^n\over6^n}$$

Note, this can be rewritten in various ways, including $\left(5\over6\right)^n\left(1-\left(4\over5\right)^n\right)$.

Answer 2

If $A$ is the event of rolling $1$ at least once and $B$ is the event of rolling no sixes then $$P(A\cap B)=P(B)P(A\mid B)=P(B)[1-P(A^{\complement}\mid B)]$$

You allready know $P(B)=\left(\frac56\right)^n$.

Can you find $P(A^{\complement}\mid B)$ yourself?

(in words: the probability of rolling no $1$ under the condition that all $n$ rolls give an element of $\{1,2,3,4,5\}$)

Answer 3

If $A$ is the event that you roll no $6$'s, and $B$ is the event that you roll at least one $1$, then $$P(A\cap B) = P(A) - P(A\cap B^c).$$ We know $P(A)$, so only $P(A\cap B^c)$ is left (where $B^c$ is the complement of $B$).

But $A\cap B^c$ is the event of rolling no $6$'s and rolling no $1$'s, which has the probability $(\frac{4}{6})^n$.