Let $|\psi\rangle$ be the initial state, and let $U_t=e^{-i Ht}$ be the evolution operator, assuming a time-independent Hamiltonian. I will also assume for simplicity that we are working on a discrete basis. If you want to work with continuous variables, you can replace sums with integrals and you should mostly be fine.
Suppose we start at $t=0$, and measure the state at times $\{t_k\}_{k=1}^N$, letting it evolve freely in the intermediate times.
Measuring at $t=t_1$ gives the outcome $x$ with probability $p(x,t_1)=|\langle x|U_{t_1}|\psi\rangle|^2$, and a post-measurement state $|x\rangle$.
Write the coefficients of $|\psi\rangle$ in the basis of $|x\rangle$ as $|\psi\rangle=\sum_x c_x |x\rangle$, and define the kernel of the evolution as
$K(x,y;\delta t)\equiv\langle x|U_{\delta t}|y\rangle$.
Finally, let us define $\Delta_k\equiv t_k- t_{k-1}$.
We can then write $p(x,t_1)$ (assuming a discrete set of possible outcomes) as
$$p(x,t_1)=\left|\sum_y K(x,y;\Delta_1)c_y\right|^2.$$
Because we don't know the post-measurement state after the first measurement, we now need to switch to a density matrix formalism to take into account this classical uncertainty. We therefore write the post-measurement state as :
$$\rho_1=\sum_x p(x,t_1) \mathbb P_x,
\text{ where } \mathbb P_x\equiv |x\rangle\!\langle x|.$$
At time $t_2$, before the second measurement, the state is therefore given by
$$\tilde\rho_2=\sum_x p(x,t_1)\, U_{\Delta_2}\mathbb P_x U_{\Delta_2}^\dagger,$$
which then results in an outcome $x$ with probability
$p(x,t_2)=\sum_y |K(x,y;\Delta_2)|^2 p(y,t_1)$, and a post-measurement state
$$\rho_2=\sum_{x}p(x,t_2) \,\mathbb P_x
= \sum_{x,y} |K(x,y;\Delta_2)|^2
\Big|\sum_z K(y,z; \Delta_1)c_z\Big|^2 \, \mathbb P_x.$$
You can keep going and compute the state at each successive measurement time $t_k$.
If this reminds you of Feynman's path integral formulation, it's because it kind of is. The difference is that here you break the interference at every measurement time, and so the final state is determined by a mixture of quantum interference and classical probabilities.
Define now for ease of notation $q_k\equiv p(x,t_k)$.
What is the probability of finding at specific $x$ for the first time at the $k$-th measurement? This will equal the probability of not finding it in the previous measurements and finding it at the $k$-th, that is,
$$(1-q_1)(1-q_2)\cdots (1-q_{k-1})q_k.$$
Note that with this formalism you can also answer other questions about the probability of finding a given result once or more at specific combinations of times.
For example, the probability of measuring $x$ at least once will be given by
$$1-\prod_{k=1}^N (1-q_k).$$
I don't know if there is a nice way to write these expressions in general. Maybe, if you write probabilities back in terms of the kernels, but I haven't tried, and the post already got a bit too long.
