A person walks 2 meters to the left or 1 meter to the right with 50:50 chance.
Point A is 3 meters to the left and Point B is 3 meters to the Right
What is the probability of reaching poing A first?
For example, with just two consecutive moves to the left, the person can walk 4 meters reaching the 3 meter goal. Unfortunately when i try to solve this by iteration, i quickly realize there are infinite scenarios.
Imagine -1 -1 +2 -1 -1 +2 -1 -1 +2 -1 -1 +2 ... as you can see...you can keep going staying close to 0, without reaching 3 or -3 (Left or Right) This makes me believe there needs to be some infinite sum or integral to solve for this cases with smaller and smaller probability.
By using a computer program i found out that, if you raise the distance of point A and B from 3 to 10 or 100 or 1000... the probability of reaching point A first starts to go to 100%, despite the distance from the center of A and B staying equal. I imagine some log function or root function hidden in there.
As usual, the solution is to consider, for every $-3\leqslant k\leqslant3$, the probability $p_k$ to hit A first starting from $k$. Then $p_{-3}=0$, $p_3=p_4=1$ and the Markov property after one step yields $$p_2=\tfrac12(1+p_1)\quad p_1=\tfrac12(1+p_0)\quad p_0=\tfrac12(p_2+p_{-1})\quad p_{-1}=\tfrac12(p_{-2}+p_1)\quad p_{-2}=\tfrac12(0+p_0)$$ Solving this $5\times5$ affine system yields the desired probability $p_0$.
...That is, unless I am mistaken, $$p_0=\tfrac45$$