## Use a Tree Diagram

Because I am bad at remembering formulas and bad at probabilities in general I like to use a tree diagram. The other answers provide good straight forward solutions you can apply but I hope to show why those methods work.

If we roll a d20 4 times, the tree will have four levels with two branches at each level: You either roll a 7 or higher, or you don’t. At each branch we assign a probability for that individual outcome: in this case, a 6/20 for rolling a 1-6, and a 14/20 for rolling a 7-20. In this diagram I haven’t filled out the branches where we roll a 1-6 because they’re not relevant for this question.

To calculate the final probability, we need each roll to result in a 7-20. For this, we take the probability of each branch and multiply them together, so we get

$$frac{14}{20} cdot frac{14}{20} cdot frac{14}{20} cdot frac{14}{20} = left(frac{14}{20}right)^4 approx 24% $$

For 15 or higher in this case, we can substitute 14/20 for 6/20 and hence get

$$frac{6}{20} cdot frac{6}{20} cdot frac{6}{20} cdot frac{6}{20} = left(frac{6}{20}right)^4 approx 0.8% $$

Similarly if we were to look at 3 dice rolls instead, the tree would only have 3 levels, and hence we would only end up multiplying the probability 3 times, so we would get

$$frac{14}{20} cdot frac{14}{20} cdot frac{14}{20} = left(frac{14}{20}right)^3 approx 34% $$

Now if we were interested in finding the probability of getting 7 if we took the highest roll each time we would have to consider the likelihood of getting a 7-20 once anywhere in the tree, not just the branch where we get it 4 times in a row, which would be more complex. Luckily there’s a trick for these kinds of situations: Instead of considering the probability of getting at least 1 7-20 in 4 rolls, we can consider the probability of *never* getting a 7-20 in 4 rolls and our result will be 1 minus that probability.

So to do that we must first find the likelihood of getting a 1-6 4 times in a row, and we would see that it is as above

$$ frac{6}{20} cdot frac{6}{20} cdot frac{6}{20} cdot frac{6}{20} = left(frac{6}{20}right)^4 approx 0.8% $$

Therefore we must take 1 (or 100%) and subtract the probability we found which gives us a 99.2% chance of rolling at least a 7 in 4 rolls. Likewise for 3 rolls we get 97.3%