Question

A fair die is rolled indefinitely. Player A wins if two consecutive rolls show 3 or 5, and player B wins if two consecutive rolls show 1 or 2 or 4 or 6. The probability that player A wins in the long run is

• A. \(\frac{2}{3}\)
• B. \(\frac{5}{21}\)
• C. \(\frac{1}{7}\)
• D. \(\frac{2}{21}\)

Hint:

Set up two equations for A winning given the type of the last roll.

Answer 1

The Correct Option is B

Step 1: Approach
Think of the game as a Markov chain on the type of the last roll and compute the winning chance by first-step analysis in a different order.

Step 2: First two rolls
Possible pairs: AA with probability $\dfrac19$ (A wins), BB with probability $\dfrac49$ (B wins), AB with $\dfrac29$, BA with $\dfrac29$. After an AB the process restarts from a B-roll, and after BA from an A-roll.

Step 3: Equation
Let $W$ be the winning probability from the very start. Using the states above $W=\dfrac19+\dfrac29b+\dfrac29a$ with $a=\dfrac37$, $b=\dfrac17$ obtained from the standard recursion $a=\dfrac13+\dfrac23b$, $b=\dfrac13a$:
\[ W=\frac19+\frac{2}{63}+\frac{6}{63}=\frac{7}{63}+\frac{8}{63}=\frac{15}{63}=\frac5{21} \]
Option (B).

Final Answer:
Solving the two recursion equations gives 5/21 for the chance that A wins, option (B). \[ \boxed{\frac{5}{21}} \]