Two Car Meeting At a Point Probability Question There is a car at point A. It can go either right or down. There is a another car at point B. This car can go either up or left. Obviously with this restrictions cars cannot go in reverse directions. If car at the point A goes to point B, and car at point B goes to point A with the same speed, what is the probability that these two cars meet at point Z.



(As an example one of the path that car at point A will follow is AKZMB or, one of the path that car at point B will follow is BMDNA.)

I have couple of answers but I can't be sure about which one is true. I need an explanation for right answer. Thank you for your time.
i think by counting, we have 6 ways A to B, 6 ways B to A
so Sample space =6*6=36 
after 4 steps then A to B, and B to A. so two cars meet at point Z when they goes 2 steps.
A - Z - B have 4 way 
B - Z - A have 4 way
so we have 16 cases  A goes to point B, and car at point B goes to point A, meet at point Z
 P = 16 /36 =4/9
I hope this solution is correct

edit

"assuming that at each intersection, the driver chooses the way to turn by flipping a fair coin." is not correct. I can Choose another distribution but the way A to B is unchanged, but with  your assuming , answer change.
Cool question, I like this problem!

You could consider the arrangements of points and lines as nodes and edges, and then consider the hole arrangement as a graph (as in graph theory).
Doing this, allows you to then construct a tree, featuring all possible ways for both cars to get from A to B. 
When I did this, I found that there are 12 possible ways for each car to get to the other side (A to B, B to A), 10 from those ways cross the node Z, at some point during there way to the other side. Considering this, and the fact that you need both cars to take a route with Z being an equal distance fron there starting point, you will find that you have a 1/2 chance of both cars Meeting each other at Z. 
This means a 50% change!

I hope this helped...
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