Probability of Rolling a 6 3 Times? You are to tosses a fair die, observing the topside of each throw, until you observe the third-six. What is the probability that you stop tossing this die on your 18th-toss? My thinking. Each roll you have a probability of 1/6 to roll a 6. So in order to reach 18 rolls you would have probability of (1/6)^18. Is this the probability of stopping by the 18th roll? Would one need to multiply by (3c1)(2c1)(1c1)to account for order?

No. You seem to have the right idea, but not quite grasped what you're counting.

In order to reach the third success on the eighteenth try, you must have _two successes_ and _fifteen failures_ in some arrangement among the first seventeen, then _one more_ success on the eighteenth try.

So count the ways to select two from seventeen, then multiply by the probability of three successes and fifteen failures.

> #$$\mathsf P(X_3=18)~=~{^{17}\mathrm C_{2}}\cdot\tfrac{5^{15}}{6^{18}}$$