Suppose that the vertices of a regular graph are coloured red and blue with an equal number of each (we call this a balanced colouring). Since the graph is undirected, the number of edges from a red vertex to a blue vertex is clearly the same as the number of edges from a blue vertex to a red vertex. However, if instead of edges we count walks of length 2 which do not stay within their starting colour class, then this symmetry disappears. Our aim in this paper is to investigate how extreme this asymmetry can be. Our main question is: Given a $d$-regular graph, for which pairs $(x,y)\in[0,1]^2$ is there a balanced colouring for which the probability that a random walk starting from a red vertex stays within the red class for at least $2$ steps is $x$, and the corresponding probability for blue is $y$? Our most general result is that for any $d$-regular graph, these pairs lie within the convex hull of the $2d$ points $\left\{\left(\frac{l}{d},\frac{l^2}{d^2}\right),\left(\frac{l^2}{d^2},\frac{l}{d}\right) :0\leq l\leq d\right\}$. Our main focus is the torus for which we prove both sharper bounds and existence results via constructions. In particular, for the $2$-dimensional torus, we show that asymptotically, the region in which these pairs of probabilities can lie is exactly the convex hull of: \[ \left\{\left(0,0\right),\left(\frac{1}{2},\frac{1}{4}\right),\left(\frac{3}{4},\frac{9}{16}\right),\left(\frac{1}{4},\frac{1}{2}\right),\left(\frac{9}{16},\frac{3}{4}\right),\left(1,1\right)\right\} \]