Edit 3. Fixed question to be more clear and include current solution
Problem
Two players player 1 and player 2 plays a game of fair coin flipping. Player 1 starts with $A$ coins and Player 2 with $B$ coins. Let $P_n$ defines the probability of player 1 winning all coins in a game when player 1 has $n$ coins. Find the recurrence relation of $P_n$ and then find its closed formula.
This formula can then be used to know the probability of player 1 winning if he has, i.e. twice the coins player 2 has.
Solution?
I know that $P_0 = 0$, because if she has $0$ coins, then she can't win, what about $P_{A+B}?$
$$P_n = 0.5P_{n+1} + 0.5P_{n-1}$$ Which gives the characteristic equation: $$x^2-2x+1 = (x-1)^2=0 \implies x=1\ (double\ root)$$ This can then be put into the general solution for linear homogenous recurrence relations: $P_{hn} = C_1+nC_2$ and here I'm stuck once again. To find these two values I'd need two boundary points
Say you have k coins. Then after the first flip, you are equally likely to have k-1 or k+1.
So the probability of winning all from a starting point of $k$ is $P_k = 0.5 P_{k+1} + 0.5 P_{k-1}$
It's fairly easy from here to verify that $P_k = k/(n_1+n_2)$.