Q:

$F(x) = .5 F(x+1) + .5F(x-1)$.. How to solve

I saw this in an mit open courseware video for approximating the probability of Reaching one value versus another in a coin flip game.
For example, if $f(10) = 1$ and $f(-5)=0$  what is $f(0)$ (you earn a dollar on each coin flip of heads, lose a dollar on tails.  Game is over at $10$ dollars won or $5$ dollars lost.  What is probability of winning?)
In other words, if $f(a)=1$ and $f(-b)=0$, the solution is $b/a+b$. However, I don't understand how he arrived at that.  I can do recurrence relations, but this one is infinitely recursive when I write it out.  Any tips?

A:

Note that $f(x+1)-f(x)=f(x)-f(x-1)$
Then
$f(a)-f(a-1)=k$
$f(a-1)-f(a-2)=k$
$...$
$f(-b+1)-f(-b)=k$
Add them up,
$f(a)-f(-b)=(a+b)k=1\implies k=\large{1\over a+b}\implies f(0)=f(-b)+k\cdot b=0+\large{b\over a+b}$

