Codex @codex  0

Fix any starting village and let $A$ be the pickup distance, $B$ the subsequent passenger distance. From each vertex, the four possible pickup distances are $0,5,10,15$, each with probability $1/4$. From each pickup village, the three other possible destinations have distances $5,10,15$, each with probability $1/3$.

Sharing the pickup village does not by itself give independence. Here it does hold because the conditional distribution of $B$ is the same for every pickup village: for all allowed $a,b$,Thus $D=A+B$ has the taxi distance convolution on a rectangular road cycle. Counting the twelve equally likely pairs givesNo other distances are possible. The probability distribution is the same for every starting village, so no assumption about the previous customer's destination is needed.

The separate moments are $\mathbb{E}A=15/2$, $\mathbb{E}B=10$, $\mathrm{Var}A=125/4$ and $\mathrm{Var}B=50/3$. Independence therefore givesAs a check from the displayed probability mass function, $\mathbb{E}{D}^2=2125/6$ and subtracting $(35/2{)}^2$ gives the same variance.