A player rolls four 20-sided dice, takes the lowest value, ignores the rest. What is the probability of this value at least 7?

Question

I'm designing a tabletop game, and I need to figure out how to calculate a few probabilities:

1. Roll 3 20-sided dice, take the highest value. What is the probability of it being 7 or higher? 15 or higher?
2. 1. Roll 4 20-sided dice, take the highest value. What is the probability of it being 7 or higher? 15 or higher?
3. 1. Roll 3 20-sided dice, take the lowest value. What is the probability of it being 7 or higher? 15 or higher?
4. 1. Roll 4 20-sided dice, take the lowest value. What is the probability of it being 7 or higher? 15 or higher?

How can I do this? Could you explain to me how this works, or even better - give me a simple formula?

Answer 1

If you roll one $20$-sided die, then the probability that it is $7$ or higher is $\frac{14}{20}$. In general, to get a $k$ or higher, the probability is $\frac{21 - k}{20}$.

If you have $n$ $20$-sided dice and you take the highest value, then the probability that this max is $7$ or higher is the probability that at least one die is also $7$ or more. The complement is that all of the dice are $6$ or less, so we have: $$ 1 - \left( \frac{6}{20} \right)^n $$

Likewise, if you have $n$ $20$-sided dice and you take the lowest value, then the probability that this min is $7$ or higher is the probability that all of the dice are $7$ or higher, so we have: $$ \left( \frac{14}{20} \right)^n $$

---

So to generalize...

Roll $n$ $20$-sided dice, take the highest value. What is the probability of it being $k$ or higher?

$$ 1 - \left( \frac{k - 1}{20} \right)^n $$

Roll $n$ $20$-sided dice, take the lowest value. What is the probability of it being $k$ or higher?

$$ \left( \frac{21 - k}{20} \right)^n $$