Probability of getting a particular value or higher on at least two of the dice when rolling three six-sided dice?

Question

I've calculated the probability for rolling a particular value or higher on both dice when rolling 2d6, as follows.

1 or higher: 100% 2 or higher: 69.44% 3 or higher: 44.44% 4 or higher: 25% 5 or higher: 11.11% 6 or higher: 2.778%

If I add a third dice to this, how does it affect the probability if I still only need the particular value or higher on two or more of the dice?

Answer 1

This can be treated as a binomial: each die either rolls the value $v$ or higher, or it does not; the dice roll independently; there is a fixed number $n$ of trials (dice); and the probability is the same for all dice.

The general binomial formula is $$P(X=k) = {n \choose k}\cdot p^k \cdot (1-p)^{n-k}$$ where $p$ is the probability of a single die rolling at or above value $v$, which is $\frac{7-v}{6}$, $n=3$, and $k = 2$.

Answer 2

If you require $2$ or more success from $3$ dice, then visualize the 'failures'.

For throwing a $2$ or higher at least twice, we need to avoid combinations such as $116$, $151$ or $411$, and obviously $111$. There are $3$ 'legs' of dimensions $1\times1\times5$, and a cube dimension $1$.

We can repeat this pattern for the other cases, and this leads to a general formula:

$$3k^2(6-k)+k^3$$ $$=2k^2(9-k)$$

There are $216$ possible throws in total, so:

$$\text{P}(\text{two dice greater than k})=1-\frac{k^2(9-k)}{108}$$.

Hand-written cases:

1.1.5.3+1=16
2.2.4.3+8=56
3.3.3.3+27=108
4.4.2.3+64=160
5.5.1.3+125=200