Here is an example for the p digits with L=5.
First sequence 14159 -> SUM = 20
Second sequence 26535 -> SUM = 21
Third sequence 89793 -> SUM = 36
For single digits (L=1) the sum is equal to the digits value and thus we have a probability of 1/10 for any sum of digits from 0 to 9. If w(L,S) is the probability that a chain of length L has a sum of digits equal to S, so
w(1,y)=1/10 with y=0,1,..,9
Any further distribution can be calculated recursively:
w(L,y) = w(L-1,y) + 1/10 * sum (for all i=0..y) w(L-1,y-i)
For any L the sum of digits is located between S=0 (all digits=0) and S=9*L (all digits=9). When going to longer and longer chains the min and max sums became extremely improbable because the likelihood for a single digit long run falls like 10-L.
Graph shows the distribution of the sum of digits for different lengths of chains:
Analysis started at digit: 1
Ellapsed computer time for one class: 3 min 30 sec - 4 min