The Conjugacy Classes Of Involution what is the number of conjugacy classes of involutions in the two non-isomorphism double cover of Sn (2.Sn+) and (2.Sn-) 
 A: The answer is roughly $n/4$.
Presentations of the covering groups $2.S_n^+$ and $2.S_n^-$ were calculated by Schur, and can also be found in Chapter V, 25.15 of Huppert's book "Endliche Gruppen I". They are
$\langle\  x_i\, (1 \le i \le n-1),\, t \mid t^2=[t,x_i]=1,\, x_i^2=u, (x_ix_{i+1})^3=t, (x_ix_j)^2=t\, (j-i > 1)\ \rangle,$
where $u=1$ in $2.S_n^+$ and $u=t$ in $2.S_n^-$. The image of the generator $x_i$ in $S_n$ is the transposition $(i,i+1)$.
Representatives of the classes of involutions in either group $2.S_n$ consist of the the central element $t$, together with representatives of the classes of involutions in $S_n$ that lift to involutions in $2.S_n$. 
We can choose $x_1$, $x_1x_3$, $x_1x_3x_5$, etc as representatives of the $\lfloor n/2 \rfloor$ classes of involutions in $S_n$, and it is easy to calculate the orders of their inverse images in $2.S_n$ from the relations in the presentation.
Products of $4k$ and $4k+2$ disjoint transpositions lift to elements of orders $2$ and $4$ respectively in both covers $2.S_n^{\pm}$.
Products of $4k+1$ and $4k+3$ disjoint transpositions lift to elements of orders $2$ and $4$ in $2.S_n^+$ and orders $4$ and $2$ in $2.S_n^-$.
So in all cases roughly half of the classes of involutions in $S_n$ lift to involutions in $2.S_n^{\pm}$, and I will leave you to come up with an exact formula for this in terms of $n$.
(Note that for all involutions in $S_n$ that lift to involutions $g$ in $2.S_n$, it can be deduced from the relations that $g$ and $gt$ are conjugate.)
