Q. 3.74 - A First Course in Probability

Question

$A$ and $B$ alternate rolling a pair of dice, stopping either when $A$ rolls the sum $9$ or when $B$ rolls the sum $6$. Assuming that $A$ rolls first, find the probability that the final roll is made by $A$. 

Step-by-Step Solution

Answer

Probability that the final roll made by $A$ is $\frac{9}{19}$ where assuming that $A$ rolls first

Step1: Find the probability of A rolls then B rolls

A- $A$ rolls $9$ times in a certain order.

B- $B$ rolls $6$ dice in a certain order.

Probabilities:

$P\left(A\right)=\frac{4}{36}=\frac{1}{9}, A=\left\{\right(3,6),(4,5),(5,4),(6,3\left)\right\}$

$P\left(B\right)=\frac{5}{36}, B=\left\{\right(1,5),(2,4),(3,3),(4,2),(5,1\left)\right\}$

$A$ and $B$ can only appear in separate rolls and are thus distinct.

The likelihood that neither $A$ nor $B$ will roll their winning number in one turn $(A and B)$ is

$P\left[(A\cup B{)}^c\right]=1-P(A\cup B)$

$=1-\left[P\right(A)-P(B)-P(AB\left)\right]$

$=1-\left[P\right(A)-P(B)-P(A\left)P\right(B\left)\right]$

$=1-\left[\frac{4}{36}+\frac{5}{36}-\frac{4}{36}\times \frac{5}{36}\right]$

$=\frac{62}{81}$

Step2: Find the Probability that the final roll made by A

The likelihood of $A$ finishing the game is the sum of the probabilities of mutually exclusive situations in which $A$ rolls $9$ after $i=0,1,2,3,...$ turns in which neither player wins. Each turn's outcome is distinct.

$P$["$A$ ends the game"] $=\sum _{i\in \mathrm{\mathbb{N}}}\left[P{\left[(A\cup B{)}^c\right]}^{i}P\left(A\right)\right.$

$=\sum _{i\in \mathrm{\mathbb{N}}}{\left(\frac{62}{81}\right)}^i\frac{1}{9}$

$=\frac{1}{9}\sum _{i\in \mathrm{\mathbb{N}}}{\left(\frac{62}{81}\right)}^i$

$=\frac{1}{9}\frac{1}{1-\frac{62}{81}}$

$=\frac{9}{19}$