I simplify the lottery a little bit. Let's say the players can win a fix prize $W$, no matter how many players participate in the lottery. We do not regard any fee. Each player can win this prize with a probability of $p$. If n players win, then the prize is equally distributed among the players. W.l.o.g. we calculate the expected value for 2 players. They win in total $W$ if at least one of the players win. Thus the expected value is
$E^{(2)}(W)=W\cdot \left( p^2+2\cdot p\cdot (1-p)\right)=W\cdot (1-(1-p)^2)$$$E^{(2)}(W)=W\cdot \left( p^2+2\cdot p\cdot (1-p)\right)=W\cdot (1-(1-p)^2)$$
We can generalize the result
$E^{(n)}(W)=W\cdot (1-(1-p)^n)$$$E^{(n)}(W)=W\cdot (1-(1-p)^n)$$
Now the fix prize W has to be equally distributed amoung the $n$ players. Thus the expected value for one player is $E^{}(W)=\frac{W}n\cdot (1-(1-p)^n)$. We can look at some interesting values.
$\lim\limits_{n\to \infty} \frac{W}n\cdot (1-(1-p)^n)=0$. If (theoretically) the number of players goes to infinity the expected value is $0$.
If only one player participate at the lottery then the expected value is $W\cdot (1-(1-p)^1)=p\cdot W$.
If the prize W is increasing in some way by the number of players $n$, then this can be taken into account.
