The average jackpot is $\frac{M-J}{n}$ where $J$ is the amount left in the pot after the $M$ trials. This is a linear function of $J$, so given $E(J)$ the expected average jackpot is $\frac{M-E(J)}{n}$.
Let $j(x, y)$ be the expected jackpot after $x$ trials of which $y$ are wins. Since the trials are memoryless the probability that the last trial is a win is $\frac{y}{x}$, so we get a recurrence
$j(x, y) = \begin{cases} \text{undefined} & \text{if } x<\min(y, 0) \\ 0 & \text{if } x=y \\ \frac{x-y}{x} \cdot (1 + j(x-1, y)) & \text{otherwise} \end{cases}$
We can expand that as $j(x,y) = \frac{x-y}{x} \left(1+ \frac{x-1-y}{x-1}\left(1+ \frac{x-2-y}{x-2}\left(1+\ldots\left(1+\frac{1}{y+1}\right)\ldots\right)\right)\right)$
Then by induction from the inside out, $j(x, y) = \frac{x-y}{y+1}$, so $E(J) = \frac{M-n}{n+1}$ and the expected average jackpot paid out to the $n$ winners is $\frac{M-1}{n+1}$