A token starts at the point $(0,0)$ of an $xy$-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Perform the coordinate transformation $(x, y)\rightarrow (x+y, x-y)$. Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\langle 1, -1 \rangle$, $\langle 1, 1 \rangle$, $\langle -1, -1 \rangle$, $\langle -1, 1 \rangle$ respectively. Moreover, the transformation takes the equation $|y| = |x|$ to the union of the x and y axis. Exactly half of the moves go up in the new coordinates, and half of them go down. In order to end up on the x axis, we need to go up thrice and down thrice. The number of possible sequences of up and down moves is the number of permutations of $UUUDDD$, which is just $\binom63 = 20$. The probability of any of these sequences happening is $\left(\frac12\right)^6$. Thus, the probability of ending on the x axis is $\frac{20}{2^6}$. Similarly, the probability of ending on the y axis is the same. 
However, we overcount exactly one case: ending at $(0, 0)$. Since ending on the x axis and ending on the y axis are independent events, the probability of both is simply $\left(\frac{20}{2^6}\right)^2 = \frac{25}{256}$. Using PIE, the total probability is $\frac{20}{64} + \frac{20}{64} - \frac{25}{256} = \frac{135}{256}$, giving an answer of $\boxed{391}$.
~sampai7