Rolling a 5 sided uneven die 20 times, what is the probability of getting a sum above 20? 
A 5-sided die has the sides of value [0;1;2;3;4] each with the probability of [50%;25%;8.33%;8.33%;8.33%] respectively.

If I roll 20 of those dice, what is the probability that the sum will be 21 or greater?
The way I've thought about this, is listing out a large sample space of permutations between the possible results that corresponds with a value, but it feels tedious, and I would like to know if there is a way of calculating this instead of having to write it all down by hand.
 A: For integers $n,s$ with $n\ge 0$, let $p(n,s)$ be the probability of obtaining a total score of at least $s$ using $n$ dice.

Then $p$ satisfies the recursion
$$
p(n,s)=
\begin{cases}
\text{if}\;\,s\le 0\;\text{then}\\[4pt]
\qquad\;1\\[4pt]
\text{else if}\;n=0\;\text{then}\\[4pt]
\qquad\;0\\[4pt]
\text{else}\\[4pt]
\qquad
\frac{1}{2}p(n-1,s)
\\[4pt]
\qquad
+\frac{1}{4}p(n-1,s-1)
\\[4pt]
\qquad
+\bigl(\frac{1}{12}\bigr)
\Bigl(
p(n-1,s-2)+p(n-1,s-3)+p(n-1,s-4)
\Bigr)
\\
\end{cases}
$$
Applying the recursion in Maple yields
$$
p(20,21)
=
\frac{29130345217554983}{64925062108545024}
\approx
.4486764321
$$
A: An approximation using the Normal distribution:
The mean score of one roll is $1$ and the variance is $1.666$, so the mean score of $20$ rolls is $20$ and the variance is $20 \times 1.666 = 22.32$. By the Central Limit Theorem, the score of $20$ rolls is approximately Normal with $\mu = 20$ and $\sigma = \sqrt{22.32} = 5.773$. If $X$ is the total score, then $Z = (X-\mu)/\sigma$ is approximately Normal(0,1), so (applying a correction for continuity)
$$P(X \ge 20.5) \approx P \left( Z \ge \frac {20.5-\mu}{\sigma} \right) = P(Z \ge 0.0866) = 0.465$$
