There's a gambler with initial money $k=100\$$. He plays till bankruptcy or till having $N=500\$$. In every game he wins $100\$$ with probability $p=\frac{1}{2}$, loses $100\$$ with probability $q=\frac{1}{4}$ or the game ends in tie with probability $r=\frac{1}{4}$.
(a) Calculate the probability of racking up 500\$,
(b) Calculate the mean time of the game.
I've seen exercises with gamblers ruin, with $p$ and $q=1-p$, but in my case there's also $r$. Does it change much, or do I just "forget" about the r and calculate the ruin of the gambler? I'm obviously referring to "Unfair coin flipping" -> https://en.wikipedia.org/wiki/Gambler%27s_ruin. There is a formula, and then I just have to solve a differential equation.
According to (b) I have no idea how should I calculate the above.
Please help with both sub-points. Any help will be much appreciated.
Best Answer
Lulu has given you a method for the probabilities of reaching $500$ before reaching $0$: dividing the values by $100$ to give you labels, you want to find the solution to $$\psi_i=\tfrac 12 \psi_{i+1} +\tfrac 14 \psi_i +\tfrac 14 \psi_{i-1} \\ \text{ i.e. } 2\psi_{i+1} -3 \psi_i + \psi_{i-1}=0$$ with the boundary conditions $\psi_0=0$ and $\psi_5=1$. In particular you are trying to find $\psi_{1}$. As Rushabh Mehta says, you would get the same answer ignoring ties by starting with $\psi_i=\frac 23 \psi_{i+1} +\frac 13 \psi_{i-1}$
For the expected time, you do something similar but this times have $$\tau_i=1+\frac 12 \tau_{i+1} +\frac 14 \tau_i +\tfrac 14 \tau_{i-1} \\ \text{ i.e. } 4+2\tau_{i+1} -3 \tau_i + \tau_{i-1}=0$$ with the boundary conditions $\tau_0=0$ and $\tau_5=0$, and in particular you are trying to find $\tau_{1}$. My comment was that if you ignore ties by changing this to $t_i=1+\frac 23 t_{i+1} +\frac 13 t_{i-1}$ then you would get $\tau_i = \frac43 t_i$