Gambler's ruin alternative solution

Question

Consider the gambler's ruin problem. Suppose there are two gamblers $A$ and $B$ and each time, they bet $\$ 1$. If $A$ starts with $\$ i$, calculate the probability that $A$ wins the game (by getting all $N$ dollars). Assume that the probability of winning (resp. losing) a single bet (in any round) is $p$ (resp. $1-p$).

The usual solution defines $P_i = P(A \text{ wins } | A \text{ starts at } \$ \ i)$ and solves for $P_i = p P_{i+1} + q P_{i-1}$ with initial conditions $P_0 = 0, P_N = 1$.

I am curious why we assume $P_i = P(A \text{ wins } | A \text{ starts at } \$ \ i)$ and not $P_i = P(A \text{ wins}, A \text{ starts at } \$ \ i)$. To do this, let's assume $Q_i = P(A \text{ wins}, A \text{ starts at } \$ \ i)$ and everything else remains as above.

Then \begin{align*} Q_i &= P(A \text{ wins}, A \text{ starts at } i) \\ &= P(A \text{ wins}, A \text{ starts at } i, A \text{ wins current round})+ P(A \text{ wins}, A \text{ starts at } i, A \text{ loses current round}) \\ &= Q_{i+1} + Q_{i-1} \end{align*}

The initial conditions are $Q_0 = 0, Q_N = 1$. But this certainly does not give a similar solution. Where did I go wrong in constructing the recurrence relation?

Answer 1

The notation P(A wins|A starts at \$i), abbreviating as $P(A|i)$, means $$ P(A|i)={P(A,i)\over P(i)} $$ It is the probability of A winning the game given the probability of A having i dollars. $P(A,i)$ is the joint probability of player A having $i$ dollars and winning the game. Since we are generally not interested in how many dollars the players have, but whether they will win the game given they start at a particular point, $P(A|i)$ is the correct notation here.

You have not supplied any details of your calculation, but if you state that you are using $Q(A,i)$ then you must also supply the probability of the player having that amount of money at the start of the game.

The initial conditions are Q0=0,QN=1.

No, the initial conditions in your system must contain whether the person has the money or not. You have to supply the probability of them having $N$ dollars as well as the probability of them winning if they have $N$ dollars. That's why your system is not appropriate.