Question goes like this:
Two players, A and B, alternatively toss a fair coin (A goes first, then B). The game ends when there is a head followed by a tail (HT), and the person who tosses the tail wins. What is the probability that A wins?
Suggested solution:
Let $P(A)$ be the probability of A winning and $P(B) = 1 - P(A)$, be the probability for B winning.
Using conditional probability, $$P(A) = P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} H)\cdot P(A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} H) + P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} T)\cdot P(A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} T)=\frac12\cdot P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} H) + \frac12\cdot P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} T) $$
If A's first throw is T, B effectively goes first and thus, $P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} T) = P(B) = 1 - P(A)$
If A's first throw is H, conditioning on B's first throw we get, $$P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} H) = P(A| B\hspace{1mm} throws\hspace{1mm} T)\cdot P(B\hspace{1mm} throws\hspace{1mm} T) + P(A| B\hspace{1mm} throws\hspace{1mm} H)\cdot P(B\hspace{1mm} throws\hspace{1mm} H) = 0\cdot \frac12 + (1 - P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} H))\cdot \frac12$$ $$P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} H) = \frac13$$
Combining all this information we get, $$P(A) = \frac12\cdot \frac13 + \frac12\cdot (1 - P(A))$$ $$P(A) = \frac49$$
I'm unsure how the solution makes the claims: $$P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} T) = P(B) = 1 - P(A)$$ $$ P(A| B\hspace{1mm} throws\hspace{1mm} H) = (1 - P(A| A's\hspace{1mm}1st\hspace{1mm} throw\hspace{1mm} is\hspace{1mm} H))$$
For the two final claims, the probability that $A$ wins given $A$'s first throw is a tails is the same probability as $B$. We can imagine that before anyone flips, our first flip is tails. Then, after another tails flip by $A$, we have the same game, but now $B$ goes first. Hence, our players have swapped and the probability that $A$ wins given the first flip is tails is the probability that $B$ wins, which is $1 - \mathbb{P}(A)$ since they are exclusive events.
Same concept for the second one, we have just switched positions.
The general idea here is that you should only have memory of the last flip, and if the current flip is the same as the last, we have essentially flipped the order of the players.