What is the optimal path between $2$ fixed points around an invisible obstructing wall? Every day you walk from point A to point B, which are $3$ miles apart. There is a $50$% chance each walk that there is an invisible wall somewhere strictly between the two points (never at A or B). The wall extends $1$ mile in each direction perpendicular to the line segment (direct path) between A and B, and its position can be at any random location between the two points. That is, it can be any distance between A and B such as $0.1$ miles away from A, $1.5$ miles away, $2.9$ miles away.... You don't know if the wall is present or where it is until you walk into it.  You must walk around the wall if present as there is no other way to circumvent it.  Assume the wall is negligible thickness (like a force field), all ground is perfectly flat, and the y coordinates at both A and B are $0$ (although I don't think the optimal path will change much if they weren't).
What strategy minimizes the average expected walk distance between A and B?  How do we know for certain this strategy yields the shortest average walking distance?
To get the $100$ bounty points, I would like a convincing argument from you as to why you feel your solution is optimal and cannot be beaten.  For those of you using computer simulation, anything reasonably close to optimal is a candidate for the bounty. 
Can anyone beat the optimal solutions submitted so far?  There are still a few days to try.  Even a slight beat (if not due to roundoff error or some other error) would help prove previous solutions are non-optimal.  You can just use the table of $31$ points (x-y coordinates) and compute that path length and then if you can find better then I will accept that answer.  I think by Sunday night I may have to award the bounty otherwise it may get auto-awarded.

At any point the ideal position would be either in the middle or at the corner, depending on if this point is the wall or not. The probability for this position $x_b$ being a wall position given that we already are at another point $x_a$ is $\frac{1}{2}\frac{1-x_a}{n}$ where $n$ is the resolution of a simulation. That is n steps in the dimension from start to goal.
So two cases for each pair of points $(x_a,x_b)$: 


*

*direction $(x_b-x_a,y-y_a)$ with probability $\frac{1}{2}\frac{1-x_a}{n}$

*direction $(x_b-x_a,0-y_a)$ with probability $1-\frac{1}{2}\frac{1-x_a}{n}$


Calculting the expected vector from each ($x_a,y_a$) as a weighted sum of the above cases would give us a set of vectors from each point which we could interpolate our way forward through.
This is probably not the best way as I probably forgot something, but at least presents one important way to think : calculate a vector field which we then use as the help to calculate a path through following the arrows.
