Question

$A$ and $B$ alternately throw a pair of dice. $A$ wins if he throws a sum of 5 before $B$ throws a sum of 8, and $B$ wins if he throws a sum of 8 before $A$ throws a sum of 5. The probability, that $A$ wins if $A$ makes the first throw, is

• A. $\frac{8}{19}$
• B. $\frac{9}{19}$
• C. $\frac{8}{17}$
• D. $\frac{9}{17}$

Answer 1

Answer: (B)

$\frac{9}{19}$

Answer 2

Let $p$ be the probability that $A$ wins starting with his turn.

• $A$'s turn: $A$ wins immediately with a sum of 5, which occurs with probability $\frac{4}{36} = \frac{1}{9}$. With probability $\frac{8}{9}$, he gets neither a 5 nor stops the game.
• $B$'s turn: $B$ wins if he rolls an 8, with probability $\frac{5}{36}$. If $B$ does not roll an 8 (with probability $\frac{31}{36}$), the game resets to $A$'s turn.

Thus, we write:

$$p = \frac{1}{9} + \frac{8}{9} \times \frac{31}{36} \times p$$

Solve for $p$:

$$p = \frac{1}{9} + \frac{248}{324} \, p \quad \Rightarrow \quad p\left(1 - \frac{248}{324}\right)= \frac{1}{9}$$ $$1 - \frac{248}{324} = \frac{324-248}{324} = \frac{76}{324} = \frac{19}{81}$$

Thus,

$$p \times \frac{19}{81} = \frac{1}{9} \quad \Rightarrow \quad p = \frac{1}{9} \times \frac{81}{19} = \frac{9}{19}$$