Early 5's

A player has a fair 6-sided die with faces numbered 1 to 6. The player rolls the die repeatedly, recording the outcomes in the order they occur. The player continues rolling until either two 5s appear (not necessarily consecutive) or both a 4 and a 6 have been rolled. Determine the probability that the player stops rolling due to rolling two 5s.

Answer

Define the following probabilities:

\begin{align*} p_{00} &= \text{Probability of eventually getting two 5s first,}\\ &\quad \text{given we have not yet rolled a 4, a 6, or any 5},\\[6pt] p_{10} &= \text{Probability of eventually getting two 5s first,}\\ &\quad \text{given we have exactly one 5, but no 4 or 6},\\[6pt] p_{01} &= \text{Probability of eventually getting two 5s first,}\\ &\quad \text{given we have a 4 or a 6, but no 5},\\[6pt] p_{11} &= \text{Probability of eventually getting two 5s first,}\\ &\quad \text{given we have a 4 or a 6, and exactly one 5}. \end{align*}

Using the law of total probability for each state:

p_{00} \;=\; \frac{3}{6}\,p_{00} \;+\; \frac{2}{6}\,p_{01} \;+\; \frac{1}{6}\,p_{10},

p_{10} \;=\; \frac{3}{6}\,p_{00} \;+\; \frac{2}{6}\,p_{10} \;+\; \frac{1}{6}\,p_{11},

p_{01} \;=\; \frac{4}{6}\,p_{01} \;+\; \frac{1}{6}\,p_{10} \;+\; \frac{1}{6}\,p_{11},

p_{11} \;=\; \frac{1}{2}.

Solving this system gives:

p_{01} \;=\; \frac{1}{4}, \quad p_{10} \;=\; \frac{3}{8}, \quad p_{00} \;=\; \frac{7}{18}.

Notes and comments

Comment 1: Note that when we are in state $p_{01}$ or $p_{11}$ , we can lose in only one way (by getting the other number), not two. Keep that in mind when setting up the transitions.

Comment 1: Also note that this question even though it is a bit similar in text with the above one, it is a widely different question conceptually.