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: Understanding the Concept
Call a roll of 3 or 5 an A-roll (probability \(\frac13\)) and any other roll a B-roll (probability \(\frac23\)). Player A wins on two A-rolls in a row. Player B wins on two B-rolls in a row.

Step 2: States
Let \(a\) be the probability that A eventually wins when the last roll was an A-roll, and \(b\) when the last roll was a B-roll.
\[ a=\frac13\cdot1+\frac23\,b,\qquad b=\frac23\cdot0+\frac13\,a \]

Step 3: Solve
\[ b=\frac a3\Rightarrow a=\frac13+\frac{2a}{9}\Rightarrow\frac{7a}{9}=\frac13\Rightarrow a=\frac37,\ b=\frac17 \]

Step 4: First roll
\[ P(\text{A wins})=\frac13a+\frac23b=\frac17+\frac2{21}=\frac{5}{21} \]
This is option (B).

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