Binomial or something else?

Question

Players A and B take turns in answering trivia questions, starting with player A answering the first question. Each time A answers a question, she has probability $p_1$ of getting it right. Each time B plays, he has probability $p_2$ of getting it right.

Suppose that the first player to answer correctly wins the game (with no predetermined maximum number of questions that can be asked). Find the probability that A wins the game.

Can someone post step-by-step solution? This is a question from Joe Blitzstein's Stat 110 homework question

Answer 1

Solution #$1$:

Let $p$ be the probability that $A$ wins the game.

Then $$p = p_1 + (1-p_1)(1-p_2)p$$ Solve for $p$.

Explanation:

Either $A$ wins on the first question (probability $p_1$), or if not (probability $(1-p_1)$), then if $B$ doesn't win on his first question (probability $(1-p_2))$, player $A$ is effectively at the start of a brand new game, hence the probability of winning from that point is exactly $p$.

Solution #$2$:

$A$ wins on the first question with probability $p_1$.

$A$ wins on $A's$ second question with probability $(1-p_1)(1-p_2)p_1$.

$A$ wins on $A's$ third question with probability $(1-p_1)^2(1-p_2)^2p_1$.

. . .

Thus, the probability that $A$ wins is $$\sum_{n=0}^\infty(r^n)p_1$$ where $r=(1-p_1)(1-p_2)$.

It's the sum of a geometric series. Can you finish it?

Answer 2

It can be solved considering just the first two rounds.

Odds in favor of $A = p_1 : (1-p_1)p_2$

Convert odds to probability, $Pr = \dfrac{p_1}{p_1+(1-p_1)p_2}$

$\underline{Added\; explanation}$

Probabilities for every subsequent two rounds will just get multiplied by some factor,
which means that the odds in favor (which is a ratio) remains unchanged