Problem:
Alice and Bob are playing a game together. They play a series of rounds until one of them wins two more rounds than the other. Alice wins a round with probability $p$. What is the probability that Bob wins the overall series?
I attempted this problem and got it wrong, but I would like a second opinion in interpreting the solution so that I can learn from this experience. As the title implies, the probability of Bob winning the entire series can be found using the Law of Total Probability. I need some verification that I'm interpreting the constituent probabilities correctly.
If we let $C$ be the probability of winning the series overall, and each $B_i$ be the probability of Bob winning each round for matches $i = 0, 1, 2$, then we have the following:
$P(C) = P(C|B_2)P(B_2) + P(C|B_1)P(B_1) + P(C|B_0)P(B_0)$
In principle, this makes sense, however the next step establishes the following:
$P(B_2) = q^2$
$P(B_1) = 2pq$
$P(B_0) = p^2$
$P(C|B_2) = 1$
$P(C|B_1) = P(C)$
$P(C|B_0) = 0$
The first probability, $P(B_2)$ seems pretty straightforward, it's the probability of two consecutive wins, where $q = 1 - p$.
I'm uncertain about the second, $P(B_1) = 2pq$. I think this represents the situations where Bob wins-Alice wins or Alice Wins-Bob Wins, which would be $pq + qp = 2pq$. Does that seem correct?
I'm also shaky on the third one, $P(B_0) = p^2$. I think this represents the situation in which Bob wins zero games, meaning Alice has one two (consecutive?) games to put her two over Bob for the win, hence the $p^2$.
For $P(C|B_2) = 1$, this must be the probability that Bob wins the overall series given he's already won two matches to put him over Alice, so this can only be 1 because of the "pre-condition".
But $P(C|B_1) = P(C)$ is a bit confusing for me. I think this represents the situation in which a couple games have been played, but the players have ended up tied, so it "resets", and its existence is meant to represent the recursive nature of the game, which keeps going until one of the two players wins two more over the other. Does that sound correct?
And the last one must be the probability that Bob wins the overall series given he won zero games, which of course makes that 0.
Put this all together and solve for $P(C)$ gives us $\frac{q^2}{1-2qp}$
It all checks out, except for how you describe the events.
I interpret $B_i$ as the event that Bob wins $i$ among the first $2$ consecutive rounds, for $i\in\{0,1,2\}$.
$B_0$ is the event that Bob wins none of the first two rounds, therefore certainly loses the game (Alice won). Thusly, $\mathsf P(B_0)= p^2$ , and $\mathsf P(C\mid B_0)=0$ .
Likewise, $B_2$ is the event that Bob wins two of the first two rounds, therefore certainly wins the game. Thusly, $\mathsf P(B_2)= (1-p)^2$ , and $\mathsf P(C\mid B_2)=1$ .
Finally, $B_1$ is the event that Bob wins exactly one among the first rounds, so $\mathsf P(B_1)= 2p(1-p)$ , because to do so, he must win either the first or second and Alice the other. For a reality check, recall the Binomial Expansion Theorem: $$\mathsf P(B_0\cup B_1\cup B_2) {= p^2+2p(1-p)+(1-p)^2 \\=(p+(1-p))^2\\=1}$$
Now, should $B_1$ occur, both Bob and Alice shall have won the same amount of rounds and then begin the third round in the same state as they began the game. So $\mathsf P(C\mid B_1)=\mathsf P(C)$.
The rest is just LoTP and algebra. $$\begin{align}\mathsf P(C)&=\mathsf P(B_0)\mathsf P(C\mid B_0)+\mathsf P(B_1)\mathsf P(C\mid B_1)+\mathsf P(B_2)\mathsf P(C\mid B_2)\\ &=p^2\cdotp 0+2p(1-p)\cdot \mathsf P(C)+(1-p)^2\cdotp1\\(1-2p(1-p))\mathsf P(C)&=(1-p)^2\\\therefore\qquad\mathsf P(C)&=\dfrac{(1-p)^2}{1-2p(1-p)}\\&=\dfrac{(1-p)^2}{p^2+(1-p)^2}\end{align}$$
Alternatively, consider this: The game shall be won the first ocassion where an even number of rounds are played and the same player won the last two of these. Then the probability that Bob will be that player is:$$\dfrac{\mathsf P(B_2)}{\mathsf P(B_0)+\mathsf P(B_2)}=\dfrac{(1-p)^2}{p^2+(1-p)^2}$$