Problem source

An examination consists of several papers, which are marked independently.
The mark given for each paper can be an integer from $0$ to $m$ inclusive, and the total mark for the examination is the sum of the marks on the individual papers. In order to make the examination completely fair, the examiners decide to allocate the mark for each paper at random, so that the probability that any given candidate will be allocated $k$ marks $(0\leqslant k\leqslant m)$ for a given paper is $(m+1)^{-1}$.
If there are just two papers, show that the probability that a given candidate will receive a total of $n$ marks is 
\[
\frac{2m-n+1}{\left(m+1\right)^{2}}
\]
for $m< n\leqslant2m$, and find the corresponding result for $0\leqslant n\leqslant m$. 
If the examination consists of three papers, show that the probability that a given candidate will receive a total of $n$ marks is 
\[
\frac{6mn-4m^{2}-2n^{2}+3m+2}{2\left(m+1\right)^{2}}
\]
in the case $m< n\leqslant2m$. Find the corresponding result for $0\leqslant n\leqslant m$, and deduce the result for $2m< n\leqslant3m$.

Solution source

In order to receive $n$ marks over the two papers, where $m < n \leq 2m$ the student must receive $k$ and $n-k$ marks in each paper. Since $n > m$, $n-k$ is a valid mark when $n-k \leq m$ ie when $n-m\leq k$, therefore the probability is:

\begin{align*}
\sum_{k = n-m}^m \mathbb{P}(\text{scores }k\text{ and }n-k) &= \sum_{k=n-m}^m \frac{1}{(m+1)^2} \\
&= \frac{m-(n-m-1)}{(m+1)^2} \\
&= \frac{2m-n+1}{(m+1)^2}
\end{align*}

If $0 \leq n \leq m$ then we need $n-k$ marks in the second paper to be positive, ie $n-k \geq 0 \Rightarrow n \geq k$, so

\begin{align*}
\sum_{k = 0}^n \mathbb{P}(\text{scores }k\text{ and }n-k) &= \sum_{k = 0}^n \frac{1}{(m+1)^2} \\
&= \frac{n+1}{(m+1)^2} 
\end{align*}

On the first paper, they can score any number of marks, since $n > m$, so we must have:

\begin{align*}
\sum_{k=0}^m \mathbb{P}(\text{scores }k\text{ and }n-k) &= \frac{1}{m+1} \sum_{k=0}^m \mathbb{P}(\text{scores }n-k\text{ on second papers}) \\
&= \frac{1}{m+1}\l \sum_{k=0}^{n-m} \frac{2m-(n-k)+1}{(m+1)^2} +\sum_{k=n-m+1}^m \frac{n-k+1}{(m+1)^2}\r
\end{align*}