Let there be $n$ men and $n$ women. Consider a pairing with $k_{m,m}$ man-man pairs, $k_{w,w}$ woman-woman pairs, $k_{m,w}$ and $k_{w,m}$ opposite gender pairs. Total number of pairs $k_{m,m}+k_{w,w}+k_{m,w}+k_{w,m} = n$. Total number of men and women respectively $2 k_{m,m} + k_{m,w}+k_{w,m} = n$ and $2 k_{w,w} + k_{m,w}+k_{w,m} = n$. It follows that $k_{m,m}=k_{w,w}$. This configuration can be built by splitting $n$ men into $\binom{n}{2 k_{m,m}}$ partitions of $2 k_{m,m} + (n-2k_{m,m})$, likewise for women. $2 k_{m,m}$ men are permuted in $(2k_{m,m})!$ ways, and remaining men permuted in $(n-2k_{m,m})!$ ways. Resulting $n$ pairs can be rearranged in $\mathrm{multinom}(k_{m,m},k_{w,w}, k_{m,w}, k_{w,m})$ ways.
So we have
$ (2n)! = \sum_{k_{m,m}, k_{m,w}, k_{w,m} >=0 } \chi_{2 k_{m,m}+k_{m,w}+k_{w,m}=n} \left( \binom{n}{2 k_{m,m}} (2k_{m,m})! (n- 2k_{m,m})! \right)^2 \mathrm{multinom}(k_{m,m},k_{w,w}, k_{m,w}, k_{w,m}) $
Which is
$ 1 = \frac{1}{\binom{2n}{n}} \sum_{k_{m,m}, k_{m,w}, k_{w,m} >=0 } \chi_{2 k_{m,m}+k_{m,w}+k_{w,m}=n} \mathrm{multinom}(k_{m,m},k_{w,w}, k_{m,w}, k_{w,m}) $
Now to find the requested probability we restrict the summation to $k_{m,w}+k_{w,m} <= 30$. This gives
In[114]:= With[{n = 100}, (n!)^2/(2 n)! Sum[ Boole[2 k + m + w == n] Boole[m + w <= 30] Multinomial[k, k, m, w], {k, 0, n}, {m, 0, n}, {w, 0, n}]] Out[114]= \ 1918296922985300702931961626782543256990306077/\ 41121343590504031983862003937481222553223476334365
which approximately is $0.0000466497$.