A fair die is thrown again and again. Find the probability that the first six comes on the fourth…

Step 1. First six on the fourth throw. Count strings of four throws: the first three have $5$ choices each and the last is a six. So $\dfrac{5^3 \times 1}{6^4} = \dfrac{125}{1296} \approx 0.0965$.

Step 2. No six in six throws. $\left(\dfrac{5}{6}\right)^6 = \dfrac{15\,625}{46\,656} \approx 0.3349$.

Step 3. The smallest number of throws. The chance of at least one six in $n$ throws is $1 - \left(\dfrac{5}{6}\right)^n$. For $n = 12$ it is about $0.8878$, which is below $0.9$. For $n = 13$ it is about $0.9066$, which is above.