Suggestions for dealing with the "timed" balls-into-bins model

Question

Definitions: Let $T$ (for "time") be a random variable $T \sim \text{Exp}(\lambda)$ and $\Delta t$ is a realization (or called an observed value) of $T$. Let $D$ (for "delay") be a random variable $D \sim \text{Exp}(\mu)$. All the random variables (including those involved below) are mutually independent.

Timed Balls-into-Bins Model: There are $n$ bins. Consider two robots $R_1$ and $R_2$ which can produce multiple balls instantaneously.
At time 0, robot $R_1$ (1) produces $n$ balls instantaneously; (2) Immediately, these $n$ balls are independently sent to the $n$ bins, one ball per bin; (3) The delays for the balls going from the robot to its destination bin are IID with the same distribution as $D$ defined above.
At time $\Delta t$ (defined above), robot $R_2$ independently does exactly the same thing as robot $R_1$ does (i.e., (1), (2), and (3) for robot $R_1$ above).

Consider the time point $t$ when half (more precisely, $\lfloor n/2 \rfloor + 1$) of the $n$ bins have received the balls from $R_2$ (no constraints on the balls from $R_1$), and denote the set of these $\lfloor n/2 \rfloor + 1$ bins by $B$.

Question: What is the probability $\mathbb{P}$ of the event $E$ that there exists a bin $b_{\Box} \in B$ which receives a ball from $R_1$ before receiving a ball from $R_2$?

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Have similar balls-into-bins models been studied in the literature?
Any suggestions (not a complete solution) towards a closed form/approximation/numerical results (by using mathematical systems)/(even) simulations of the probability $\mathbb{P}$?
It is OK to simplify this model for possible tractability, for example, by using $D \sim U(a,b)$.

Answer 1

(I use the notation from my comments). By symmetry, $$\mathbb{P}(E^c)={n \choose m}\,\mathbb{P}(E^c, B=\{1,\ldots,m\})\;\;.$$ If $D_1=M_m\;\;$ we have to compute $$I_1:=\mathbb{P}(D_1^\prime>t+M_m,D_2^\prime>t+D_2,\ldots,D_m^\prime>t+D_m, D_{m+1}>M_m,\ldots,D_n>M_m)$$ Conditioning on $M_m=D_1,D_2,\ldots,D_m$ and using the independence assumptions gives $$I_1=\int e^{-\mu_1(t+s)}\left(\prod_{i=2}^m e^{-\mu_1(t+x_i)}\right) e^{-\mu_2(n-m)s} f(s,x_2,\ldots,x_m)\,dx_2\ldots dx_m\,ds$$ where $f(s,x_2,\ldots,x_m)=1_{[0,\infty[}(s) \mu_2e^{-\mu_2s} \prod_{i=2}^m \mu_2e^{-\mu_2x_i}1_{[0,s]}(x_i)$. After carrying out the $x_i$ integrals $$I_1=e^{-m\mu_1t}\mu_2^m\int_0^\infty e^{-(\mu_1+\mu_2)s}\left({1-e^{-(\mu_1+\mu_2)s}\over \mu_1 +\mu_2}\right)^{m-1} e^{-\mu_2(n-m)s}\,ds$$ and the substitution $y=1-e^{-(\mu_1+\mu_2)s}$ gives $$I_1=e^{-m \mu_1 t}\alpha^m\,B(m,\alpha (n-m)+1)$$ Finally, by symmetry the cases $D_2=M_m$,...,$D_m=M_m$ give the same result, so that$$\mathbb{P}(E^c)=m{n\choose m} e^{-m\mu_1t}\alpha^m\,B(m,\alpha(n-m)+1)=e^{-m\mu_1t}{\alpha^m\,B(m,\alpha(n-m)+1)\over B(m,n-m+1)}$$

Answer 2

Note: I am writing my own attempt as a possible approach. Please correct me if it does not make sense.

First, for simplicity, I fix $\Delta t = \frac{1}{\lambda}$, the expected value of the random variable $T$.

For any bin $b_\square \in B$, let $D_{R_1}$ denote the time it takes for the ball sent from $R_1$ to arrive at the bin $b_{\square}$, $D_{R_2}$ denote the time it takes for the ball sent from $R_2$ to arrive at the bin $b_{\square}$, and $D_{\lfloor n/2 \rfloor + 1}$ denote the $(\lfloor n/2 \rfloor + 1)$-order statistic of $n$ random variables $D_i$ ($i = 1, 2, \ldots, n)$, all of which are identically and independently distributed with the same distribution as $D$.

In terms of the notations above, for any bin $b_\square \in B$, it receives a ball from $R_1$ before receiving a ball from $R_2$ if and only if $D_{R_1} - \Delta t \le D_{R_2} \le D_{\lfloor n/2 \rfloor + 1}$. Therefore, we have

\begin{align} \mathbb{P}(E) = 1 - \left (1 - \mathbb{P}(D_{R_1} - \Delta t \le D_{R_2} \le D_{\lfloor n/2 \rfloor + 1})\right)^{\lfloor n/2 \rfloor + 1} \end{align}