Kait rolls a fair 6-sided die until she rolls a 6. If she rolls a 6 on the \(N\)th roll, she then rolls the die \(N\) more times. What is the probability that she rolls a 6 during these next \(N\) times?
Solution
Answer: \(\frac6{11}\)
Consider the complementary event. If the first \(6\) occurs on turn \(i\), the previous \(i-1\) rolls and the next \(i\) rolls must all be different from \(6\). The probability of this is \(\left(\frac56\right)^{2i-1}\cdot\frac16\).
Sum over all \(i\geq1\). This is a geometric series with initial term \(\frac5{36}\) and common ratio \(\frac{25}{36}\):
\[\Pr(\text{no 6 in the next }N\text{ rolls})=\frac{5/36}{1-25/36}=\frac5{11}.\]
Hence the requested probability is \(1-\frac5{11}=\frac6{11}\).