Probability of rolling 3 sixes in 6 rolls I roll a die six times. What is the probability of rolling at least 3 sixes? Please share the formula as well, so I can figure it out myself in future.

Let $p$ = probability of getting a $6$ for each toss, then $p = \dfrac{1}{6}$, and $q = 1 - p = \dfrac{5}{6}$ is the probability of getting a number other than $6$. Let $k$ = the number of successful trials. This is a binomial experiment consisting of $6$ independent trials. So $n = 6$. Let $x$ be the number of $6$'s in $6$ tosses of the fair die, then we find $P(x \geq 3) = 1 - P(0) - P(1) - P(2)$.
$P(x = k) = \binom n k\cdot p^k\cdot q^{n-k}$. So:
$P(0) = P(x=0) = \binom 6 0\cdot \left(\dfrac{1}{6}\right)^0\cdot \left(\dfrac{5}{6}\right)^6 = 0.335$
$P(1) = P(x=1) = \binom 6 1\cdot \left(\dfrac{1}{6}\right)^1\cdot \left(\dfrac{5}{6}\right)^5 = 0.402$
$P(2) = P(x=2) = \binom 6 2\cdot \left(\dfrac{1}{6}\right)^2\cdot \left(\dfrac{5}{6}\right)^4 = 0.201$.
So: $P(x \geq 3) = 1 - 0.335 - 0.402 - 0.201 = 0.062$
