The Corporation must pick exactly $5$ of our numbers. Which $5$? These can be chosen in $\binom{6}{5}$ ways.
For each of these ways, there are $\binom{53}{1}$ ways for the Corporation to choose a sixth number that doesn't match ours.
It follows that there are $\binom{6}{5}\binom{53}{1}$ different Corporation choices that give us a second prize.
For the probability, divide by $\binom{59}{6}$.
Here's a quick derivation which gives the same solution as that provided by gar, although a slightly different recursion.
A winning $k$ ball combination consists of integer sequences $0<x_1<\cdots<x_k$ and $0<y_1<\cdots<y_k$ so that $|x_i-y_i|\le1$.
Let $A_k(n)$ be the number of ways to pick winning $k$ ball combinations so that $x_k,y_k\le n$. This is the same as $T(n+1,k)$ defined by gar.
Similarly, let $B_k(n)$ be the number of ways to pick $k$ ball combinations so that $x_k\le n$ and $y_k\le n+1$. We'd get the same number of instead we require $x_k\le n+1$ and $y_k\le n$ due to symmetry.
Obviously, $A_k(n)=B_k(n)=0$ for $n<k$. Otherwise, $A_0(n)=B_0(n)=1$.
Let's first consider $A_k(n)$. If $(x_k,y_k)=(n,n)$, what remains is picking the remaining $k-1$ balls with $x_{k-1},y_{k-1}\le n-1$, which can be done in $A_{k-1}(n-1)$ ways. If $(x_k,y_k)=(n,n-1)$, the remaining $k-1$ balls should have $x_{k-1}\le n-1$ and $y_{k-1}\le n-2$, which can be done in $B_{k-1}(n-2)$ ways; the same applies to $(x_k,y_k)=(n-1,n)$. Otherwise, $x_k,y_k\le n-1$, which leaves $A_k(n-1)$ alternatives. Thus,
$$
A_k(n)=A_{k-1}(n-1)+2B_{k-1}(n-2)+A_k(n-1).
$$
We derive a recursion for $B_k(n)$ in a similar way. If $(x_k,y_k)=(n,n+1)$, the remaining $k-1$ balls require $x_{k-1}\le n-1$, $y_{k-1}\le n$, leaving $B_{k-1}(n-1)$ alternatives. Otherwise, $x_k,y_k\le n$, leaving $A_k(n)$ alternatives. Thus,
$$
B_k(n)=B_{k-1}(n-1)+A_k(n).
$$
Plugging in $k=6$, $n=50$ gives $A_6(50)=8000567708$. Since each of the two integer sequences are drawn randomly with probability $1/\binom{n}{k}$, this gives the probability of winning
$$
\frac{A_6(50)}{\binom{50}{6}^2}
=\frac{8000567708}{15890700^2}\approx 0.00003168361647.
$$
Best Answer
To win you must select exactly 4 of the 5 'matching' numbers, and 1 of the 35 'nonmatching' numbers. Count the ways to do this and divide by the count of ways to select any 5 of the 40 numbers.
$$\frac{{5\choose 4}{35\choose 1}}{40\choose 5}$$