Players A and B throw a fair die alternately, A first. A wins at once if his throw shows 6; B wins at once if his throw shows 4, 5 or 6. Given that…

Let the throws be numbered $1,2,3,\dots$ with $A$ on odd throws. $A$ wins on throw 1 with probability $\frac16$. The game reaches throw 2 with probability $\frac56$, and $B$ wins there with probability $\frac56\cdot\frac12=\frac5{12}$. The game reaches throw 3 only if both failed, with probability $\frac56\cdot\frac12=\frac5{12}$, and $A$ wins there with probability $\frac5{12}\cdot\frac16=\frac5{72}$. The game cannot end by throw 3 in any other way, so $P(\text{ends within 3})=\frac16+\frac5{12}+\frac5{72}=\frac{12+30+5}{72}=\frac{47}{72}$. $A$ wins in $\frac16+\frac5{72}=\frac{17}{72}$ of the cases. The required conditional probability is $p=\frac{17/72}{47/72}=\frac{17}{47}=0.3617\approx0.36$. Exact fraction arithmetic confirms $17/47$.