Problem
There are $d$ different cities. Last year Alice and Bob did NOT live in the same city. The probability that Alice moved to a new city since last year is $m$. This probability is the same for Bob. Each city has the same probability of being chosen. Let's assume that the movement of Bob and Alice (as well as their choice of destination) are independent.
What is the probability that Alice and Bob live in the same city this year?
My Thoughts
Possibility 1
One moved but not the other. The one who moved went to the city of the onw who did not move. This happens with probability $2m(1-m)\frac{1}{d-1}$
Possibility 2
Both moved and end up in the same city. For this to be true, the one who moved must go in any city other than the one where the other was at first place and the second one has to go to the same city. I think this should occur with probability $m^2\frac{d-2}{d-1}\frac{1}{d-1}$.
My Answer
The probability that Alice and Bob still live in the same city is
$$2m(1-m)\frac{1}{d-1} + m^2\frac{d-2}{d-1}\frac{1}{d-1} = \frac{m(2d-md-2)}{(d-1)^2} $$
Is it correct?
Just for fun, here's another solution. Instead of saying that each person moved with probability $m$, we could say that with probability $m\cdot\frac d{d-1}$ they chose one of the $d$ cities to live in and with probability $1-m\cdot\frac d{d-1}$ they stayed put. The probability that they now live in the same city is $\frac1d$ times the probability that at least one of them made a choice, i.e.
$$ \frac1d\left(1-\left(1-m\cdot\frac d{d-1}\right)^2\right)=\frac{m(2d-md-2)}{(d-1)^2}\;, $$
which I believe coincides with your result if you correct the mistake in the last step.
Of course this isn't quite rigorous for the case $m\cdot\frac d{d-1}\gt1$, but since it's really just a rearrangement of the calculation, the formula must be the same in that case.