We have two players, A and B, who play a game in several rounds. The game stops when one of them wins two more rounds than the other. What is the probability that B wins if in each round the probability of B winning is $1-p$?
Here's my solution which gives me an incorrect answer. But I don't see what is wrong.
Let $A_i$ denote the event that player A won round $i$ and similarly for $B_i$. Then the game will end in the following scenarios:
$A_1, A_2$ (here A wins)
$A_1, B_2, A_3$ (here A wins)
$A_1, B_2, B_3$ (here B wins)
$B_1, A_2, A_3$ (here A wins)
$B_1, A_2, B_3$ (here B wins)
$B_1, B_2$ (here B wins)
Now let $\beta$ denote the event that $B$ wins the game. Then $$P(\beta)=p P(\beta|B_1) + (1-p) P(\beta | A_1)$$
Let's calculate $P(\beta | A_1)$ and $P(\beta | B_1)$ separately.
First, let's calculate $P(\beta | A_1)$. We have
$$P(\beta | A_1)= p P(\beta | A_1A_2) + (1-p)P(\beta | A_1B_2) = p\cdot 0 + (1-p)[P(\beta | A_1B_2A_1)p + P(\beta | A_1B_2B_3)(1-p)] = 0 + (1-p)[0\cdot p + 1\cdot (1-p)] = (1-p)^2$$
Next, let's calculate $P(\beta | B_1)$. We have
$$P(\beta | B_1) = pP(\beta |B_1A_2)+(1-p)P(\beta | B_1B_2) = p[pP(\beta | B_1A_2A_3) + (1-p)P(\beta| B_1A_2B_3)] + (1-p)\cdot 1 = p[p\cdot 0 + (1-p) \cdot 1] + (1-p) = p(1-p) + (1-p) = (1-p)(p+1)$$
Lastly,
$$P(\beta)= p\cdot (1-p)^2 + (1-p)\cdot (1-p)(p+1)$$
This is different that the presumably correct answer, which is $(1-p)^2/(1-2p(1-p))$. What did I do wrong?
Probably the most natural way to think about this is as a biased random walk $S_n$ with length $n$ starting at $0$ and ending at $\pm 2.$ In this case, player $A$ wins if we end at $2$. Otherwise, player $B$ wins. You move to the right with probability $p$ and to the left with probability $q$ where $p + q = 1.$ Applying the well known result that $Z_n = (\frac{q}{p})^{S_n}$ is a martingale and Optional Stopping Theorem, we have $$\mathbb{P}(\text{Player A wins}) = \frac{1 - (\frac{q}{p})^{-2}}{(\frac{q}{p})^2 - (\frac{q}{p})^{-2} }.$$ So the probability player $B$ wins is $$\boxed{1 - \frac{1 - (\frac{q}{p})^{-2}}{(\frac{q}{p})^2 - (\frac{q}{p})^{-2} }}.$$