What is the probability to win? Die game You have a die. If you get one pip at any point in the game you lose. If you get two,..., six pips you start adding the number of pips to a sum. To win the sum must get greater or equal to 100. What is the probability to win the game?

The probability of winning with #2=$i$, #3=$j$, #4=$k$, #5=$l$ and #6=$m$  is given by the multinomial coefficient 
$(1/6)^{i+j+k+l+m}\ \frac{(i+j+k+l+m)!}{i!\ j!\ k!\ l!\ m!}$,  
where $i$, $j$, $k$, $l$ and $m$ are constrained by
$i\leq 50$,
$j\leq \lceil (100-2i)/3\rceil=j_{max}(i)$,
$k\leq \lceil (100-2i-3j)/4\rceil = k_{max}(i,j)$
$l\leq \lceil (100-2i-3j-4k)/5\rceil = l_{max}(i,j,k)$
$m= \lceil (100-2i-3j-4j-5l)/6\rceil = m(i,j,k,l)$.
Summing over all choices gives
$\sum_{i=0}^{50}\sum_{j=0}^{j_{max}(i)}\sum_{k=0}^{k_{max}(i,j)}\sum_{l=0}^{l_{max}(i,j,k)}(1/6)^{i+j+k+l+m(i,j,k,l)} \frac{(i+j+k+l+m(i,j,k,l))!}{i!\ j!\ k!\ l!\ m(i,j,k,l)!}$.
I cannot see a way to reduce it but it would be easy to put this into a machine to solve.
