Three. (50 points) A and B play a game. A randomly selects a positive integer pair $(k, n)(2 \leqslant k \leqslant 36,1 \leqslant n \leqslant$ $216)$, and B performs the following operation: divides $[0,36]$ into $k$ segments, $[0,36]=\bigcup_{i=1}^{k}\left[a_{i-1}, a_{i}\right]\left(a_{0}=0<a_{1}<\cdots<a_{k}\right.$ $\left.=36, a_{i} \in \mathbf{N}_{+}\right)$, for each integer $a \in\left[a_{i-1}, a_{i}\right]$, takes $f(a)=2\left(a-a_{i-1}\right)$ or $a_{i}-a$, obtaining
$$
F(k)=\sum_{i=1}^{k} \sum_{a=a_{i-1}}^{a_{i}} f(a) .
$$

If $F(k) \geqslant n$, then A wins; if $F(k)<n$, then B wins. Find the probability that A wins.
Three, first find the minimum value of $F(k)$.
For any $a \in [a_{i-1}, a_{i}](i=1,2, \cdots, k)$, take $f(a)=\min \left\{2(a-a_{i-1}), a_{i}-a\right\}$.
Let $a_{i}-a_{i-1}=b_{i}(i=1,2, \cdots, k), a-a_{i-1}=b$.
Then $f(a)=\min \left\{2 b, b_{i}-b\right\}=g(b)\left(0 \leqslant b \leqslant b_{i}\right)$.
Obviously, $g(b)=\left\{\begin{array}{ll}2 b, & 0 \leqslant b \leqslant\left[\frac{b_{i}}{3}\right] ; \\ b_{i}-b, & \left[\frac{b_{i}}{3}\right]<b \leqslant b_{i} .\end{array}\right.$
Thus, $\sum_{a=a_{i-1}}^{a_{i}} f(a) \doteq \sum_{b=0}^{b_{i}} g(b)$
$=\sum_{b=0}^{\left[\frac{b_{i}}{3}\right]} 2 b+\sum_{b=\left[\frac{b_{i}}{3}\right]+1}^{b_{i}}\left(b_{i}-b\right)=h\left(b_{i}\right)$.
It is easy to see that $h(x)=\left[\frac{x^{2}}{3}\right]\left(x \in \mathbf{N}_{+}\right)$ is convex.
Also, $F(k)=\sum_{i=1}^{k} \sum_{a=a_{i-1}}^{a_{i}} f(a)=\sum_{i=1}^{k} h\left(b_{i}\right)$,

where, $\sum_{i=1}^{k} b_{i}=\sum_{i=1}^{k}\left(a_{i}-a_{i-1}\right)=a_{k}-a_{0}=36$.
By Jensen's inequality and the adjustment method, we know that when $b_{1}, b_{2}, \cdots, b_{k}$ differ by at most 1, $F(k)$ takes its minimum value.
Let $s=\left[\frac{36}{k}\right]$. Then $36=k s+r(0 \leqslant r<k)$. Hence $F(k)_{\text {min }}=(k-r) h(s)+r h(s+1)$.
Substituting $k=2,3, \cdots, 36$ into the above formula, we get
$$
\begin{array}{l}
F(2)_{\min }=216, F(3)_{\min }=144, \\
F(4)_{\min }=108, F(5)_{\min }=85, \\
F(6)_{\min }=72, F(7)_{\min }=60, F(8)_{\min }=52, \\
F(9+i)_{\min }=45-3 i(i=0,1,2), \\
F(12+i)_{\min }=36-3 i(i=0,1, \cdots, 6), \\
F(18+i)_{\min }=18-i(i=1,2, \cdots, 18) .
\end{array}
$$

Below, we calculate the probability of player A winning.
Obviously, when player A selects the number pair $(k, n)$, and $n \leqslant F(k)_{\min }$, player A wins. Therefore, the probability of player A winning is
$$
\left(\sum_{k=2}^{36} F(k)_{\min }\right) \div(216 \times 35)=\frac{241}{1512} .
$$