Matching red to blue dots I have two red points, $r_1$ and $r_2$, and two blue points, $b_1$ and $b_2$. They are all placed randomly and uniformly in $[0,1]^2$. 
Each dot points to the closest dot from another colour; closest is defined wrt the Euclidean distance. We use $x \to y$ to indicate dot $x$ points to dot $y$.
If $r_1 \to b_1$, what is the probability that $r_2 \to b_1$ too?
NOTE it must be larger than 1/2 because $r_1 \to b_1$ tells us in a way that $b_1$ is likely to have a centric location, and thus is likely than it is closer to $r_2$ too than $b_2$.
 A: From what you have written, J assume that all random dots in the question are distributed independently. 
Now, suppose $b_1 = (x_1, y_1)$ and $b_2 = (x_2, y_2)$ are fixed. Thus for a random dot $r_1$, uniformly distributed on $[0; 1]^2$, the probability, that it lies closer to $b_1$, than to $b_2$ is $\mu(\{(x, y) \in [0;1]^2| (x_1 + x_2 - 2x)(x_2 -x_1) + (y_1 + y_2 - 2y)(y_2 -y_1)>0\})$, where $\mu$ stands for Lebesgue measure. Now, as $r_1$ and $r_2$ are independent, then the probability, that both $r_1$ and $r_2$ lie closer to $b_1$, than to $b_2$ is $(\mu(\{(x, y) \in [0;1]^2| (x_1 + x_2 - 2x)(x_2 -x_1) + (y_1 + y_2 - 2y)(y_2 -y_1)>0\}))^2$.
Now as $b_1$ and $b_2$ are also independent and uniformly distributed, we can conclude, that in our initial problem  $$P(r_2 \to b_1|r_1 \to b_1) = \frac{\int_0^1 \int_0^1 \int_0^1 \int_0^1 (\mu(\{(x, y) \in [0;1]^2| (x_1 + x_2 - 2x)(x_2 -x_1) + (y_1 + y_2 - 2y)(y_2 -y_1)>0\}))^2dx_1dy_1dx_2dy_2}{\int_0^1 \int_0^1 \int_0^1 \int_0^1 \mu(\{(x, y) \in [0;1]^2| (x_1 + x_2 - 2x)(x_2 -x_1) + (y_1 + y_2 - 2y)(y_2 -y_1)>0\}))dx_1dy_1dx_2dy_2}$$
