What is the formula that the sum of three independent uniform discrete random variables on the interval $[0, N]$ is exactly $N$? Let $X_1, X_2, X_3$ be independent random variables taking values in the integers $k$ such that $0 \leq k \leq N$, with uniform distribution. What is the probability that $X_1 + X_2 + X_3 = N$? 
I'm looking for an asymptotic formula, but an exact one would be great too. 
My guess is that it should be asymptotically $\frac{1}{2N}$: there's about a $\frac{1}{2}$ chance that $X_1 + X_2 \leq N$, and then for each $X_1, X_2$, there's a $\frac{1}{N}$ chance that $X_3 = N - X_1 - X_2$. This feels a bit heuristic, and I'm not quite sure how to make it rigorous. 
 A: You did most of the work already!
Notice that $X_1+X_2 \in \{0,\dots,2n\}$ and $P(X_1+X_2=n)=P(X_2=n-X_1) \overset*=\frac{1}{n+1}$.
By symmetry $X \mapsto n-X$, you also have $P(X_1+X_2<n)=P(X_1+X_n>n)$, and since these three events partition all possibilities, the two last probabilities equal $\frac{n}{2n+2}$.
Therefore, $P(X_1+X_2 \leq n)=\frac{n+2}{2n+2}$.
Finally, $P(X_1+X_2+X_3=n)=P(X_1+X_2 \leq n \text{ and } X_3=n-X_1-X_2) \overset*= \frac{n+2}{2n+2} \cdot \frac{1}{n+1} = \frac{n+2}{2(n+1)^2}$
The equalities with star involve conditional probabilities given $X_1$ or given $X_1,X_2$ followed by an expectation, which I'm omitting for brevity.
A: Yes, you can compute the exact answer that way, just taking care to compute everything exactly.
We can write by law of total probability and independence $$ P(X_1+X_2+X_3=N) = \sum_{n=0}^NP(X_1+X_2 = n, X_3 = N-n )=  \sum_{n=0}^NP(X_1+X_2 = n)P(X_3 = N-n )$$ and then since $P(X_3 = N-n) = \frac{1}{N+1},$ we have $$ P(X_1+X_2+X_3=N)  = \frac{1}{N+1}\sum_{n=0}^NP(X_1+X_2 = n) = \frac{1}{N+1}P(X_1+X_2\le N).$$
Then we just need to compute $P(X_1+X_2 \le N)$ which you are right should intuitively be about $1/2.$ Actually we can do exactly the same thing as before and write $$ P(X_1+X_2 = m) = \sum_{n=0}^mP(X_1 = n)P(X_2 = m-n) = \frac{m+1}{(N+1)^2} $$ so that $$ P(X_1 + X_2 \le N) = \sum_{m=0}^NP(X_1+X_2 = m) = \frac{1}{(N+1)^2}\frac{1}{2}(N+1)(N+2) = \frac{1}{2}\frac{N+2}{N+1}.$$
So the exact answer is $$ \frac{1}{2}\frac{N+2}{(N+1)^2}$$
