Taking the initial algorithm Errata gave.
Quote:
For 20d6 drop 8:
Odds of the lowest roll being 6 is (1/6)^20.
Odds of the lowest roll being 5 is (2/6)^20 - (1/6)^20.
Odds of the lowest roll being 4 is (3/6)^20 - (2/6)^20 etc.
So take each case and stipulate temporarily that that is the lowest roll. Then multiply whatever result you get with that assumption by the odds of that being true.
If you assume the lowest roll is 4, you know you can't roll lower than 4, so they're not really d6 anymore. You rolled at least one 4. So in this case, it's the odds of the lowest being 4 odds of the result being the same as 19d3 drop 7 + 12*3
The degenerate case of the lowest being 6 is just 12*6.
Using N continuous (0-1) dice temporarily, so there is a M'th least die'. We can skip straight to that.
The odds of the m die being less than an X is x^m*(1-x)^(n-m)*(n!)/(m!)/(n-m!)
[note that n&m are constant for any one 'game']
If the die are all greater than X, then we can consider the situation where they can roll between X&1 which is easy to calculate the average = (x+1)/2
The average overall is thus just the integral of
x^m*(1-x)^(n-m)*(n!)/(m!)/(n-m!) * (x+1)/2
Which has to be pretty near doable, though you may need to break it into sections and approximate each.
For discrete die you'd be able to sum exactly, but then you'd need to deal with die on the 'border'
I'll have to put some more thought in it