You are leaving a store, and there are 2 queues with one customer in each. If the waiting times are all exponential random variables, what is the probability that we will be the last to leave?
My understanding: Let $L$ be the event of being last to leave, $Q_1$ the event of waiting in queue 1, $Q_2$ the event of waiting in queue 2. Let $X$ be your waiting time, and $X_1$ and $X_2$ the waiting time for customer in queue 1 and 2, respectively.
$P(Q_1) = P(Q_2) = \frac{1}{2}$
We have:
$$P(L) = P(Q_1)P(L|Q_1) + P(Q_2)P(L|Q_2)$$
$$P(L) = \frac{1}{2}P(L|Q_1) + \frac{1}{2}P(L|Q_2)$$
$$P(L) = \frac{1}{2}P(X>X_1+X_2|X>X_1) + \frac{1}{2}P(X>X_1+X_2|X>X_2)$$
Since $X$ is exponentially distributed, by the memoryless property,
$$P(L) = \frac{1}{2}P(X>X_2) + \frac{1}{2}P(X>X_1)$$
Supposing all three waiting times are i.i.d,
$$P(L) = \frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{1}{2} = \frac{1}{2}$$
However, according to ChatGPT, the answer is $\frac{1}{3}$ since the problem is equivalent to 3 queues with one customer in each (because of the memoryless property of exponential distributions). However, I fail to understand how this is true. With 3 queues, it is possible for you to leave first, which is impossible when there are 2 queues.
ChatGPT's answer isn't correct and it's currently not advisable to trust ChatGPT in computations. Let $Y=X_2-X_1$. If you've included serving time in waiting time, $$P_1=P(L\mid Q_1)=P(X > X_2-X_1)=1-P(X\le Y)$$ and $$P_2=P(L\mid Q_2)=P(X > -(X_2-X_1))=1-P(X\le -Y)$$
$f_{-X_1}(x_1)=\lambda e^{\lambda x_1}$
$(y-x_2<0)\implies (x_2>y)$
$$\begin{align}f_Y(y)&=\int_{-\infty}^\infty f_{X_2}(x_2)f_{-X_1}(y-x_2)dx_2\\ &=\int_{0}^\infty (x_2>y)\lambda e ^ {-\lambda x_2}\lambda e ^ {-\lambda x_2} e ^ {\lambda y}dx_2\\ &=\lambda^2 e ^ {\lambda y}\int_{0}^\infty (x_2>y)e^{-2\lambda x_2}dx_2\\ &=\lambda^2 e ^ {\lambda y}\cases{\int_{0}^\infty e^{-2\lambda x_2}dx_2 & $y<0$\\ \int_{y}^\infty e^{-2\lambda x_2}dx_2 & $y\geq 0$}\\ &=-\frac{\lambda }{2}e ^ {\lambda y}\cases{-1 & $y<0$\\ -e^{-2\lambda y} & $y\geq 0$}\\ &=\frac{\lambda }{2}\cases{e^{\lambda y} & $y<0$\\e^{-\lambda y} & $y\geq 0$} \end{align}$$
$f_{-Y}(y)=f_Y(y)$
if $Z_1=(X+X_2)-X_1=X+Y$ $$\begin{align}f_{Z_1}(z_1)&=\int_{-\infty}^\infty f_X(x)f_Y(z_1-x)dx\\ &=\frac{\lambda^2 }{2}\int_{0}^\infty \left((x>z_1)e^{-\lambda x}e^{\lambda (z_1-x)}+(x\leq z_1)e^{-\lambda x}e^{\lambda (x-z_1)}\right)dx\\ &=\frac{\lambda^2 }{2}\cases{ \int_{0}^\infty e^{-2\lambda x}e^{\lambda z_1}dx & $z_1<0$\\ \int_{z_1}^\infty e^{-2\lambda x}e^{\lambda z_1}dx+\int_{0}^{z_1}e^{-\lambda z_1}dx & $z_1\geq 0$}\\ &=\cases{ \frac{\lambda}{4}e^{\lambda z_1} & $z_1<0$\\ \frac{\lambda}{4}e^{-\lambda z_1}+\frac{\lambda^2 }{2}z_1e^{-\lambda z_1} & $z_1\geq 0$} \end{align}$$
if $Z_2=(X+X_1)-X_2=X-Y$
$$\begin{align}f_{Z_1}(z_1)&=\int_{-\infty}^\infty f_X(x)f_Y(z_1-x)dx\\ &=\int_{-\infty}^\infty f_X(x)f_{-Y}(z_1-x)dx\\ &=f_{Z_2}(z_1) \end{align}$$
$f_{Z}(z)\triangleq f_{Z_1}(z)=f_{Z_2}(z)$
$$\begin{align}P(Z\leq 0)&=\int_{-\infty}^0\frac{\lambda}{4}e^{\lambda z}dz\\ &=\frac{1}{4} \end{align}$$
$$p\triangleq P_1=P_2=1-\frac{1}{4}$$
$$\begin{align}\therefore P(L)&=P_1\times P(Q_1)+P_2\times P(Q_2)\\ &=2\times \frac{p}{2}\\ &=p\\ &=\frac{3}{4} \end{align}$$