If we play a game where I start with 2 dollars and you start with one dollar, and I have a probability of 1/3 of winning a dollar from you and you have a 1/3 probability of winning a dollar from me, what is the probability that I win all of your money?
I drew a Markov chain with 3 dollars and 0 dollars as the absorbing states. My initial starting state is $2.
Let S2 be the starting state from 2 dollars and S1 be the intermediate state at 1 dollar.
The equations I formed are:
P(S0) = 0, P(S3) = 1
P(S2) = $\frac 13$ *P(S3) + $\frac 13$ * P(S1)
P(S2) = $\frac 13$ *1 + $\frac 13$ *P(S1)
Now P(S1) = $\frac 13$*P(S0) + $\frac 13$*P(S2)
P(S1) = $\frac 13$*0 + $\frac 13$*P(S2)
P(S1) = $\frac 13$*P(S2)
Now plugging this value back in the previous equation: P(S2) = $\frac 13$ + $\frac 13$*$\frac 13$*P(S2)
(8/9)*P(S2) = $\frac 13$
P(S2) = $\frac 38$
Thus getting $\frac 38$ as my answer. However, the answer given by my friend was $\frac 23$.
Is my answer or approach right?
The missing detail in equations like $$P(S_2) = \frac13 P(S_3) + \frac13 P(S_1)$$ is also the missing detail in the description "I have a probability of $\frac13$ of winning a dollar from you and you have a $\frac13$ probability of winning a dollar from me". Namely: what happens in the other $\frac13$ of cases?
The implication, I think, is that if neither player wins a dollar from the other in a given round, then there's a tie and the amount of money doesn't change. In other words, there is a $\frac13$ probability of going from state $S_2$ back to state $S_2$. The equation for $P(S_2)$ should be $$P(S_2) = \frac13 P(S_3) + \frac13 P(S_2) + \frac13 P(S_1)$$ and if you make the same change to $P(S_1)$ then you will get the correct answer of $\frac23$.