Short answer
I think the formula for the expected successes is this:
\begin{align}
E &= n \cdot \frac{3d - t - 2e + 1}{e-1}, &\text{where } & 1 ≤ t ≤ e ≤ d
\end{align}
While the variance could be this (not tested):
\begin{align}
V = n \cdot \left(\frac{d-t+1}{d-1} - \frac{(e-t)^2-(d-e+1)^2}{(d-1)^2}\right)
\end{align}
Here is what all the variables mean:
\$d\$ ... number of sides a single die has (in Shadowrun \$d = 6\$ - we roll plain old six-sided dice)
\$n\$ ... number of such dice in the pool (usually \$n = Attribute + Skill\$ in Shadowrun)
\$e\$ ... minimum roll for a die to explode (\$e = 6\$ in Shadowrun - only the 6 explodes)
\$t\$ ... minimum roll for a success (\$t = 5\$ in Shadowrun - 5 and 6 are successes)
\$h\$ ... number of hits, i.e. dice with a result \$≥t\$ in the roll (not needed here)
Knowing the average spread (from the variance) is nice too, because you'll also want to know if it is still a frequent occurrence to get, I don't know, 12 successes on a roll of just 16 dice, or if 8 hits is already very unlikely. I.e. with a lower explosion threshold, higher hit counts become more likely. However, the expectation value might be very similar to that of a lower hit-threshold \$t\$ at higher explosion-threshold \$e\$.
The Math behind
Exploding on 6 only:
If you want formulae, I thought I might give a brief summary of my question about exploding die pools and its answers. You can show the formulae below to be true for probabilities of exactly \$h\$ hits, the expectation values of hits \$E\$ and their variances \$V\$:
\begin{align}
p^\text{non-exp}_{d,n,t,h} &= \binom{n}{h}\left(\frac{d-t+1}{d}\right)^h\left(1-\frac{d-t+1}{d}\right)^{n-h}\\
E^\text{non-exp}_{d,n,t} &= n\ \frac{d-t+1}{d}\\
V^\text{non-exp}_{d,n,t} &= n\ \frac{(d-1)(d-t+1)}{d^2}\\
%
p^\text{exp}_{d,n,t,h} &= \frac{(t-1)^n}{d^{n+h}} \sum_{k=0}^{\max(h,n)} \binom{n}{k}\binom{n+h-k-1}{h-k}\left[\frac{d(d-t)}{t-1}\right]^k\\
E^\text{exp}_{d,n,t} &= n\ \frac{d-t+1}{d-1}\\
V^\text{exp}_{d,n,t} &= n\ \frac{t\,(d-t+1)}{(d-1)^2}\\
\end{align}
The ideas for proofs can be found on math stackexchange. Now this assumes, that dice only explode at the maximum roll of 6 in your case. So it can't tell you anything about rolls where dice explode e.g. on 5 and 6. Except, that it stands to reason that a a roll of a six sided where 1 and 2 are no successes, 3 and 4 are successes without re-rolls and 5 and 6 are successes with explosion is equal to a roll of three-sided dice where 1 is not a success, 2 is a success without re-roll and 6 is an exploding success.
I've put together a small web-page (useful for Shadowrun or the oWoD) for this and tested it with a simulation:
Arbitary explosion thresholds:
The formulae should be fairly easy to modify for arbitrary explosion thresholds with the same reasoning used in my link. Let's call the explosion threshold \$e\$. So if the roll explodes on 5 and 6, then \$e = 5\$ in this case (for Shadowrun we'd have \$e = d = 6\$). The expectation value \$E_1\$ of a single die has to fulfill this equation:
$$ E_1 = 0 \cdot \frac{t-1}{d} + 1 \cdot \frac{2d-t-e}{d} + (E_1+1) \cdot \frac{d-e-1}{d}$$
Zero successes with a probability \$\frac{t-1}{d}\$, on success and no explosions with a probability of \$\frac{2d-t-e}{d}\$ and in case of exploding dice we have a probability of \$\frac{d-e-1}{d}\$ to get \$E_1\$ more successes.
This can be solves for \$E_1\$. Now the expectation value for \$n\$ dice is just \$n\$ times that for one dice (\$E = n E_1\$):
\begin{align}
E &= n \cdot \frac{3d - t - 2e + 1}{e-1}, &\text{where }& 1 ≤ t ≤ e ≤ d
\end{align}
Note, that while the formulae for exploding on the highest value are thoroughly tested, I did not test the above formula.
