Calculate the maximum probability of a result of rolling n dice of varying number of faces Disclaimer: I'm a computer programmer more than a mathematician, so reading text like that of the answer to this question is a little (read: a lot) over my head.
I've written an algorithm (brute-force, O(nk) running time, where n is the number of dice, and k is the number of sides) that calculates the probability of rolling each possible value of a specified number of dice, each of may have a different number of faces.  For example, rolling d8 + 2d20 (1 8-sided die and 2 20-sided dice) has a minimum roll of 3 and a maximum roll of 48, each of which have a probability of 0.0003125.  The values 25 and 26 are most likely to show up, each with a probability of 0.045.
I'm wondering if there's a way, without doing the full calculation, to determine--or make a conservative estimate of--this maximum value of 0.045 in polynomial time.  I know that this is also the probability of rolling the middle value, or ceiling((max - min) / 2) + min.
I've written a probability distribution graph using jQuery Flot.  Since this is such a long running calculation for large numbers of dice with large numbers of sides (e.g. 4d100), I calculate the distribution in chunks, and update the graph periodically.  I'd like to have a maximum probability calculated ahead of time so the axes don't change as I make updates to the data.
Calculating the total number of permutations is obviously easy--multiple each die's number of sides together--so calculating the probability of any one permutation is 1 / totalPermutations.  That's about as far as I got.  Help?
 A: The probability generating function for the sum of $n$ dice of $d_1, \ldots, d_n$ sides respectively is 
$$ G(x) = \prod_{j=1}^n \sum_{i=1}^{d_j} \dfrac{x^i}{d_j}$$
The probability of a sum of $s$ is the coefficient of $x^s$ in this polynomial.
You can also write $$\sum_{i=1}^d x^i = x \dfrac{x^{d} - 1}{x - 1}$$
For example, for four dice, each with $100$ sides,
$$ G(x) = 100^{-4} x^4 \left(\dfrac{x^{100} - 1}{x - 1}\right)^{4}$$
Consider the numerator and denominator 
$$\eqalign{N(x) &= 100^{-4} x^4 (x^{100} - 1)^4 = 100^{-4} (x^{404} - 4 x^{304} + 6 x^{204} - 4 x^{104} + x^4)\cr
D(x) &= (x - 1)^4 = x^4 - 4 x^3 + 6 x^2 - 4 x + 1\cr}$$
Now $G(x) = \sum_{s=4}^{400} p_s x^s$ where $p_s$ is the probability of getting a sum of $s$.  Since $G(x) D(x) = N(x)$, the coefficient of $x^s$ in $N(x)$ is 
$$ r_s = p_{s} - 4 p_{s-1} + 6 p_{s-2} - 4 p_{s-3} + p_{s-4}$$
which gives you a recurrence relation you can use to calculate all the $p_s$, starting with $p_0 = \ldots = p_3 = 0$, where $r_s = 0$ except for $$r_4 = r_{404} = 10^{-8}, r_{104} = r_{304} = -4 \times 10^{-8}, r_{204} = 6 \times 10^{-8}$$ 
