Let's first try to solve an easier version of the same problem, i.e for some $$$N$$$ and $$$X$$$, find the probability that the required sum is equal to $$$X$$$. Let's just focus on the $$$N^{th}$$$ roll for now (since each roll is independent). What are the possibilities? We see that if get $$$k$$$ ($$$1 \le k \le 6$$$) on the $$$N^{th}$$$ roll, then to get a sum $$$X$$$ we have to find the probability that sum was $$$X-k$$$ in the previous roll (i.e the $$$N-1 ^{th}$$$ roll). So if we have stored the previous answer (in this case answer to state $$$N-1,X-k$$$) then we can use it to find the current answer. Thus we can use this observation to get the dp recurrence as $$$ dp[i][j] = \frac{\sum_{k=1}^{6} dp[i-1][j-k]}{6} $$$ (base cases when $$$N=1$$$). However, we are not done yet. Since we need the answer for required sum to be atleast $$$X$$$, thus the final answer is $$$= \sum_{i=x}^{6N} dp[N][i]$$$ (the total sum can go till maximum $$$6N$$$). Total Time Complexity = $$$O(6NX)$$$ which is enough to pass the given constraints.