The problem is saying there is this person who throws dice until he gets a 6. What's the probability that he throws at least four dice. I can not simplify this problem without it having a huge number of additions.
$(\frac{5}{6})^3 (\frac{1}{6})^1$ + $(\frac{5}{6})^4 (\frac{1}{6})^1$ + $(\frac{5}{6})^5 (\frac{1}{6})^1$ ...
Is there any other way to solve this problem? Thank you for your time!
Hint: Try computing the probability that he throws less than four dice instead.