You have 2 people starting on opposite sides of an octahedron, each move to a random vertex each turn. you win by moving onto the vertex of the other person, what's the probability the guy who goes first wins.
How I currently attacked it
A=Probability the first person wins
P(A)=P(A|Distance of one) (as person A will always be a distance of one from B after this move)
P(A|Distance of one)= 1/4*0+1/4*(1-P(A))+1/2(1-P(A|Distance of one))
I'm not too sure about how to go about solving it from here.
You definitely have the right idea. I will let $M_k$ be the event the the person to move wins, given a distance of $k$. Then the probability that the first person wins is $$ P(M_2) = 1-P(M_1) $$ and $$ P(M_1) = \frac14 \cdot 1+ \frac12 \cdot(1-P(M_1))+\frac 14\cdot(1-P(M_2)) $$ Using the first equation gives: $$ 1-P(M_2) = \frac14 \cdot 1+ \frac12 \cdot P(M_2)+\frac 14\cdot(1-P(M_2)) $$ which we can solve to $P(M_2)=\frac25$.