A critical point here is to note that the total number of tickets is not set a priori. This is good, because it makes the expected utility function non-linear in $t_i$, and so permits us to proceed (half-way).

Writing $S$ for the total number of tickets and $S_{-i}$ for the total number minus the purchases of one bunny, and simplifying, the expected utility is

$$\mathbb{E}[u_i(t_i, g_i)] = \frac{t_i}{S}\cdot g_i(C,x) -pt_i \tag{1}$$

The first order condition for utility maximization of one bunny with respect to number of tickets bought is,

$$\frac {\partial \mathbb{E}[u_i(t_i, g_i)]}{\partial t_i} = \frac{S_{-i}}{S^2}g_i(C,x) - p=0$$

$$\implies t_i = \left(\frac {S_{-i} g_i(C,x)}{p}\right)^{1/2} - S_{-i} \tag{2}$$

Now the way the problem is formed, I understand that all bunnies are identical, as regards their preferences. It also appears that there is no income constraint here. So with respect to the fox's scheme, the bunnies are totally identical, and each bunny will buy the same number of tickets. So $S_{-i} = (n-1)t_i$ and $(2)$ becomes

$$t_i = \frac {n-1}{n^2}\frac {g_i(C,x)}{p} \tag{3}$$

Moreover the profits of the fox are certain: there is no probability of graver loss other than refunding one bunny. So we have with certainty,

$$\pi = (n-1)pt_i \tag{4}$$

We get

$$(3),(4) \rightarrow \pi = \left (\frac{n-1}{n}\right)^2\cdot g_i(C,x)$$

and maximum profits are where the $g$ function is maximized, which determines $x^*$. And this is where the problem lies here: ticket price remains indeterminate.

The profit function is linear in price, this is the problem. While this is standard in a perfectly competitive environment, here we have a monopoly. To fix this, one should go back to the expected utility function and change its quasi-linear form, and assume instead concave utility in $pt_i$, $v(pt_i), v'>0, v''<0$. This will maintain price as an argument of the profit function together with $x$, and maximization of profit with respect to $(p,x)$ jointly could be attempted.

A critical point here is to note that the total number of tickets is not set a priori. This is good, because it makes the expected utility function non-linear in $t_i$, and so permits us to proceed (half-way).

Writing $S$ for the total number of tickets and $S_{-i}$ for the total number minus the purchases of bunny $i$, and simplifying, the expected utility is

$$\mathbb{E}[u_i(t_i, g_i)] = \frac{t_i}{S}\cdot g_i\cdot [C-x^2+x] -pt_i \tag{1}$$

The first order condition for utility maximization of one bunny with respect to number of tickets bought is,

$$\frac {\partial \mathbb{E}[u_i(t_i, g_i)]}{\partial t_i} = \frac{S_{-i}}{S^2}g_i\cdot [C-x^2+x] - p=0$$

$$\implies t_i = \left(\frac {S_{-i} g_i\cdot [C-x^2+x]}{p}\right)^{1/2} - S_{-i} \tag{2}$$

The second-order condition is satisfied so this will be a maximum. Rearranging $(2)$ we obtain

$$S = \left(\frac {S_{-i} g_i\cdot [C-x^2+x]}{p}\right)^{1/2} \tag{3}$$

The choice of $i$ was arbitrary so we have

$$S_{-i} g_i = S_{-j} g_j,\;\;\; \forall i\neq j \implies (S-t_i)g_i = (S-t_j)g_j$$

$$\implies t_j = S - \frac {g_i}{g_j}(S-t_i), \;\;\; \forall j\neq i \tag{4}$$

Summing over $j\neq i$ we obtain

$$S-t_i =S_{-i} = (n-1)S - (S-t_i)g_i\sum_{j\neq i}g_j^{-1} $$

$$\implies (S-t_i) = \frac {n-1}{1+g_i\sum_{j\neq i}g_j^{-1}}S \tag{5}$$

Inserting $(5)$ into $(3)$ we get

$$S = \left(\frac {\frac {n-1}{1+g_i\sum_{j\neq i}g_j^{-1}}S g_i\cdot [C-x^2+x]}{p}\right)^{1/2}$$

$$\implies S = \frac {(n-1)g_i\cdot [C-x^2+x]}{\left(1+g_i\sum_{j\neq i}g_j^{-1}\right)p} \tag{6}$$

We were able to express total demand as a function of the decision variables of the fox, and the parameters/random variables of the model. Nevertheless it also hints at the problem here (multiply by $p$ to get the revenue function), but let's derive it explicitly.

For later use, from $(5)$ we also get

$$t_i = S\left(1-\frac {n-1}{1+g_i\sum_{j\neq i}g_j^{-1}}\right) \tag{7}$$

Turning to the profit function of the fox, it has certain gross revenues equal to $pS$ and then she will have to payback the amount paid by the bunny that gets to win the lottery. So with probability $t_i/S$ the fox gets $pS - pt_i$. So the expected profit function, after ticket sales have been finalized and before the lottery draw, is

$$E(\pi) = \sum_{i=1}^n \frac {t_i}{S}\left(pS - pt_i\right) = p\sum_{i=1}^n \frac {t_i(S-t_i)}{S} \tag{8}$$

Inserting $(5),(7)$ into $(8)$, we have

$$E(\pi) = p\sum_{i=1}^n \frac {S\left(1-\frac {n-1}{1+g_i\sum_{j\neq i}g_j^{-1}}\right)\left(\frac {n-1}{1+g_i\sum_{j\neq i}g_j^{-1}}S\right)}{S} $$

$$= pS\sum_{i=1}^n \left(1-\frac {n-1}{1+g_i\sum_{j\neq i}g_j^{-1}}\right)\left(\frac {n-1}{1+g_i\sum_{j\neq i}g_j^{-1}}\right)$$

Using also $(6)$ we get, after simplification

$$E(\pi) = \frac {(n-1)^2g_i\cdot [C-x^2+x]}{1+g_i\sum_{j\neq i}g_j^{-1}}\cdot \sum_{i=1}^n \left[\frac {\left(g_i\sum_{j\neq i}g_j^{-1}-n+2\right)}{\left(1+g_i\sum_{j\neq i}g_j^{-1}\right)^2}\right] \tag{9}$$

Equation $(9)$ reveals the issues here: While from $(3)$ the fox can infer all actually realized $g_i$'s, ex ante, the expected profit function looks like a very complicated function of $n$ Uniform $(0,1)$ random variables.

But most importantly, profit does not depend on price (since to begin with Total Revenue does not depend on price). While this is standard in a perfectly competitive environment (where market equilibrium determines price), here we have a monopoly. To fix this, one should go back to the expected utility function and change its quasi-linear form, and assume instead concave utility in $pt_i$, $v(pt_i), v'>0, v''<0$. This will maintain price as an argument of the profit function together with $x$, and maximization of profit with respect to $(p,x)$ jointly could be attempted.