What is the probability of A winning a trivia game? We have the following problem: 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.
(a) 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.
My solution: Let A be the event that player A wins. Let n be the number of questions A answers and m the number of questions B answers. Hence we want to calculate P(A). We have a sequence of answers and questions from players A and B. In order for A to win, he or she has to get one question right with probability $p_1$ (with potentially a prior sequence of wrong questions). So the solutions should be $P(A)=(1-p_1)^{n-1}(1-p_2)^{m}p_2^0p_1^1$. 
The issue: However, my solutions does not match with the proposed solution in Blitzstein Intro to Probability, and the solution given does not look right to me. Blitzstein solution proposes the following solution (see also Slader):
Condition on the results of the first question for each player we have,
$$
\begin{align}
P(A) &=1 \cdot P(A right) +0 \cdot P(A wrong, B right)+P(A)P(A wrong,Bwrong) \\
&=p_1+(1-p_1)(1-p_2)P(A)
\end{align}
$$
Solving for $P(A)$, we get: $P(A)= \frac{p_1}{p_1+p_2-p_1p_2}$
issue (1): My solution and the book solutions is far off. The book solution seems not be solving for the case of many (n) questions. Can you see what I am missing in my understanding?
issue (2): When the book solution condition on the first question, are we not using the Law of Total probability? In that case, we should be getting the following expression (and not theirs):
$P(Aright)= \text{as, above, this is the probability of the first question being right}$
$$
\begin{align}
P(A) &= P(A|Aright, Bwrong) \cdot P(Aright, Bwrong)  \\
&+ P(A|Awrong, Bwrong) \cdot P(Awrong, Bwrong) \\
&+ P(A|Awrong, Bright) \cdot P(Awrong, Bright) \\
&+ P(A|Aright, Bright) \cdot P(Aright, Bright) \\
&=1 \cdot p_1 + 0 \cdot(1-p_1)(1-p_2) + 0 \cdot (1-p_1)p_2 + 1p_1p_2\\
&=p_1 +p_1p_2
\end{align}
$$
This is a different expression altogether. What I am missing here?
 A: 1. Your approach can give the right answer, but it will be a much more complicated solution. I'll start by trying to add detail to your solution.
First note that $m = n-1$. They take it turns and $A$ goes first, so to begin you have a pair of incorrect answers (A answers incorrectly, B answers incorrectly). This repeats until you finish the game with just $A$ answering correctly. If that's the $n^{th}$ question that $A$ was asked, then $B$ answered a total of $n-1$ questions (all incorrectly). So you have to have $m = n-1$
What you are trying to work out a formula for
$$
\mathbb{P}\bigl(\text{player A wins}\ \big|\ \text{player A answered $n$ questions}\bigr).
$$
To get
$$
\mathbb{P}\bigl(\text{player A wins}\bigr),
$$
you need to use a law of total probability in the form:
$$
\mathbb{P}\bigl(\text{player A wins}\bigr) = \sum_{n=1}^{\infty} \mathbb{P}\bigl(\text{player A wins and answers $n$ questions}\bigr) 
$$
If you worked all this out correctly you would have a big series. If you then summed that series (it will be a geometric progression), you would get the expression that your textbook has. 
2. There's sort of a rule of thumb with these things that if one way of thinking leads to a series answer that can be summed, there's probably a quick way of arriving at the answer that involves conditioning on the first question or the penultimate question or something like that. The observation your textbook uses is that if the first pair of answers are both incorrect then the game 'resets'. This means that
$$
\mathbb{P}\bigl(A\ | \text{A wrong on first question, B wrong on first question}\bigr) = \mathbb{P}\bigl( A \bigr)
$$
So
$$
\mathbb{P}\bigl(A\ | \text{A wrong, B wrong}\bigr)\mathbb{P}\bigl( \text{A wrong, B wrong}\bigr) = \mathbb{P}(A) \times (1-p_1)\times (1-p_2).
$$
Your attempt makes a few simple mistakes (such as putting 0 for this expression). Also, in the first term you should have:
$$
\mathbb{P}\bigl(A\ | \text{A right, B wrong}\bigr)\mathbb{P}\bigl( \text{A right, B wrong}\bigr) = 1 \times p_1 \times (1-p_2)
$$
(i.e. there should be a $(1-p_2)$).
A: Ok. The winning sequence good for A is the following
1) A wins at first strike: $p_1$
2) a wins at second strike (both must lose at first strike): $(1-p_1)(1-p_2)p_1$
3) A wins at third strike: $(1-p_1)^2(1-p_2)^2p_1$
Summarizing.....it's a geometric series
$$\mathbb{P}[A_{win}]=p_1\sum_{n=0}^{\infty}((1-p_1)(1-p_2))^n=$$
$=p_1 \frac{1}{(1-(1-p_1)(1-p_2))}=\frac{p_1}{1+p_1+p_2-p_1p_2}$
