91Ó°ÊÓ

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 $(AandB)$is

$P\left[(A∪B{)}^c\right]=1-P(A∪B)$

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

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

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

$=\frac{62}{81}$

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"] localid="1649572019674" $=\underset{i∈\mathrm{ℕ}}{∑}\left[P{\left[(A∪B{)}^c\right]}^{i}P\left(A\right)\right.$

$=\underset{i∈\mathrm{ℕ}}{∑}{\left(\frac{62}{81}\right)}^i\frac{1}{9}$

$=\frac{1}{9}\underset{i∈\mathrm{ℕ}}{∑}{\left(\frac{62}{81}\right)}^i$

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

$=\frac{9}{19}$