gambler's ruin problem - probability of ruin

Question

I am analyzing the problem:

If we play a game where I start with \$2 and you start with \$1, and I have a probability of 1/3 of winning \$1 from you and you have a 1/3 probability of winning \$1 from me, what is the probability that I win all of your money?

The proposed solution is built on the two equations below

1. Probability of opponent winning from $2 =
          P(opponent wins | opponent wins dollar) + 
          P(opponent wins | opponent ties for dollar) + 
          P(opponent wins | opponent loses dollar) 

2. Probability of opponent winning from $1 =
        P(opponent wins | opponent wins dollar) + 
        P(opponent wins | opponent ties for dollar) + 
        P(opponent wins | opponent loses dollar) 

$$1. \quad P_2 = (1/3) + (1/3)P_2 + (1/3)P_1$$ $$2. \quad P_1 = (1/3)P_2 + (1/3)P_1 + (1/3) (0)$$

so that $P_1=1/3, \ \ P_2 = 2/3$

My understanding is that infinitely many scenarios might occur (is that correct?).
However I don't understand the construction of the equations (1. and 2.).
Could anybody explain in simple terms the logic used? Do the equations reflect the infinite scenarios?

Answer 1

In short, yes, the equations reflect the infinite scenarios because they use recursion: $P_2$ and $P_1$ are in both sides of the equation, which represents the repetitive nature of these infinite scenarios.

I am pretty sure your original equations are wrong. It should be like this:

1. Probability of opponent winning from $2 =
        P(opponent wins dollar)*P(opponent wins | opponent wins dollar) + 
        P(opponent ties for dollar)*P(opponent wins | opponent ties for dollar) + 
        P(opponent loses dollar)*P(opponent wins | opponent loses dollar) 

2. Probability of opponent winning from $1 =
        P(opponent wins dollar)*P(opponent wins | opponent wins dollar) + 
        P(opponent ties for dollar)*P(opponent wins | opponent ties for dollar) + 
        P(opponent loses dollar)*P(opponent wins | opponent loses dollar) 

I am assuming you see how I did this, although if you don't, you can ask me in the comments. This is important because the probabilities I just added are all $1/3$, which is where all of the $1/3$s in the second set of equations come from.

Now, let's look at equation 1.

• P(opponent wins | opponent wins dollar) is $1$ because if the opponent wins another dollar, then they have all three dollars, so they win the game.
• P(opponent wins | opponent ties for dollar) is the scenario where the opponent is in the same place they just were because nothing changed. Thus, the probability they win is the same, which is $P_2$.
• P(opponent wins | opponent loses dollar) is the scenario where the opponent has only one dollar left, so their probability of winning is $P_1$.

Thus, from our original equation and the above reasoning, we get: $$P_2=1/3\cdot 1+1/3\cdot P_2+1/3\cdot P_1$$

Equation 2 is similar, but somewhat different:

• P(opponent wins | opponent wins dollar) is the scenario where the opponent now has two dollars, so the probability of winning is $P_1$.
• P(opponent wins | opponent ties for dollar) puts the opponent in the same place they just were because nothing changed. Thus, the probability they win is the same, which is $P_1$.
• P(opponent wins | opponent loses dollar) is the scenario where the opponent loses because they have zero dollars left. Thus, the probability of them winning is $0$.

From our original equation and the above reasoning, we get:

$$P_1=1/3\cdot P_1+1/3\cdot P_2+1/3\cdot 0$$

Hopefully, this answer helps you understand how infinite scenarios can be describes using linear equations like this because the infinite scenarios are just leaping between the opponent having one dollar and the opponent having two dollars, which is just $P_1$ and $P_2$.