Probability of rolling above a certain number across multiple 20 sided dice. [closed]

What is the probability of rolling a $15$ or higher on any of a set of $20$-sided dice.

For example. If I have $3$ $20$-sided dice, what are the odds that if I roll all three of them the result on at least one of them will be a $15$ or higher.

• Your thoughts? What did you try?
– 5xum
Oct 20, 2017 at 8:13
• I have been all over the board with this. But it's been so many years since I looked into probability I'm not sure where to start. Based on a few different places I mainly thought of 1-(14/20)^n But this was based on the formula for rolling a 6 on n dice. "P(a 6 in two rolls) = 1-(5/6)^2 similarly for six rolls it will be 1-(5/6)^6" Oct 20, 2017 at 8:21
• @MiuKujo What you wrote in your comment (efforts and eventually despair) should be part of your question. Oct 20, 2017 at 8:31
• Odds and probability do not mean the same thing. Oct 20, 2017 at 9:17

First, consider for $1$ dice. Let $X$ be the random variable that refers to the value of the dice. \begin{align} P(X\geq 15) &= P(X = 15 \lor X = 16 \lor X = 17 \lor X = 18 \lor X = 19 \lor X = 20) \\&= P(X=15) + P(X=16) + P(X=17) + P(X=18) + P(X=19) + P(X=20) \\&= 6\times\frac{1}{20} \end{align}

Now, consider for $n$ dices. The probability of the all values be $\geq 15$ is \begin{align} P(X_{1}\geq 15 \land X_{2}\geq 15 ... \land X_{n}\geq 15) &= \Pi_{i=1}^{n}P(X_{i}\geq 15) \\&= \Pi_{i=1}^{n}\left(6\times\frac{1}{20}\right) \\&= \left(6\times\frac{1}{20}\right)^{n} \end{align}

The probability of there exists at least on dice with value $\geq 15$ is \begin{align} P(\exists i:X_{i}\geq 15) &= 1- P(\forall i:X_{i}<15) \\& = 1- \Pi_{i=1}^{n}P(X_{i}< 15) \\& = 1- \Pi_{i=1}^{n}\left(1-P(X_{i}\geq 15)\right) \\& = 1- \Pi_{i=1}^{n}\left(1-\frac{6}{20}\right) \\& = 1- \left(\frac{14}{20}\right)^{n} \end{align}

• This is the probability that all rolls are above $15$. OP is asking for the probability of at least one roll.
– 5xum
Oct 20, 2017 at 8:26

Let $A_i$ denote the event "the $i$-th dice rolls $15$ or higher".

Then, what you want $P(A_1\lor A_2\lor A_3)$ but this is annoying to calculate.

It's easier to calculate $P(\neg(A_1\lor A_2\lor A_3))$ and then use the fact that $P(\neg X)=1-P(X)$.

We first write $P(\neg(A_1\lor A_2\lor A_3)) = P(\neg A_1 \land \neg A_2\land \neg A_3)$ and then use the fact that the rolls of the dice are independent, so $$P(\neg(A_1\lor A_2\lor A_3)) =P(\neg A_1) \cdot P(\neg A_2) \cdot P(\neg A_3) = \left(\frac{14}{20}\right)^3$$

So the number you are looking for, the probability that at least one die rolls $15$ or higher, is $$1-\left(\frac{14}{20}\right)^3=1-0.7^3=0.657=65.7\%$$