How to find the probability that 5 appears twice before both 4 and 6 appear when a fair six-sided die is rolled

Question

This is from QuantGuide(Doubly 5 II):
Jenny has a fair 6−sided die with numbers 1−6 on the sides. Jenny continually rolls the die and keeps track of the outcomes in the order they appear. Jenny rolls until she sees two 5s (not necessarily consecutive) OR both 4 and 6. Find the probability Jenny stops rolling due to seeing two 5s.
My Approach: The following cases to be considered are:
When Jenny stops after seeing 2 5s before 4 and 6. Its probability will be: \begin{equation} P = \Sigma_{n \ge 2}\frac{1}{6^2}\binom{n-1}{1}\binom{2}{1}(\frac{2}{3})^{n-2}=0.5 \end{equation} When Jenny stops after seeing 4 and 6 and no 5 being rolled. \begin{equation} P = \Sigma_{n \ge 2}\frac{1}{6^2}\binom{n-1}{1}\binom{2}{1}(\frac{2}{3})^{n-2}=0.5 \end{equation} When Jenny stops after seeing 4 and 6 and one 5 is rolled. \begin{equation} P = \Sigma_{n \ge 3}\frac{1}{6^3}\binom{n-1}{1}\binom{n-2}{1}\binom{2}{1}(\frac{2}{3})^{n-3}=0.25 \end{equation} Hence the conditional probability will be $P=\frac{0.5}{0.5+0.5+0.25}=0.4$. But this answer doesn't work.
Please suggest where I'm going wrong.

Answer 1

Your calculation is a bit strange - it's not clear why you think your series represent values you describe, and how you get probability greater than $1$ in denominator.

Define success as "having two fives before having both $4$ and $6$", and denote

• $p_{00}$ - probability of success before we have any 4,5 or 6
• $p_{10}$ - probability of success after having one 5, but no 4 or 6
• $p_{01}$ - probability of success after having one from 4 or 6, but no 5s (it doesn't matter if we have 4 or 6)
• $p_{11}$ - probability of success after having one from 4 or 6, and one 5

From symmetry, $p_{11} = \frac{1}{2}$.

The rest, checking what we can roll $$p_{00} = \frac{3}{6} \cdot p_{00} + \frac{2}{6} \cdot p_{01} + \frac{1}{6} \cdot p_{10}$$ $$p_{10} = \frac{3}{6} \cdot p_{10} + \frac{2}{6} \cdot p_{11} + \frac{1}{6} \cdot 1$$ $$p_{01} = \frac{4}{6} \cdot p_{01} + \frac{1}{6} \cdot p_{11} + \frac{1}{6} \cdot 0$$

From this, we can find $p_{01} = \frac{1}{4}$, $p_{10} = \frac{2}{3}$ and $p_{00} = \frac{7}{18}$.