Game of Coin Toss - Probability

Question

Two people decided to play a coin toss game. The rules of the game are:

(i) Everyone plays 2 times in sequence; (ii) the first to obtain two head wins;

Suppose that the two individuals names are A and B.

A starts play the game and throw a coin twice times and after that B throw the coin twice and so on, the first obtain two head in sequence wins.

Knowing that is a fair coin w got the follow probabilities: $ P(K) = \frac{1}{2} $ and $ P(C) = \frac{1}{2}$.

My two questions are:

(a) Is the probability of A wins bigger than B wins? If yes, why?

(b) Is this a conditional probability?

(c) If A, who plays first, wins and B still having the change to play, A still having an advantage?

My understanding about the problem,

Let's call $ W_{a} $ the event where A wins and $ W_{a}^c $ where A losses; the same to B being $ W_{b} $ the event where B wins and $ W_{b}^c $ where B losses.

$ P(W_{a}) = 1/4 $ becuase $P(C) \times P(C) = \frac{1}{4}$

Nevertheless, the probability of B wins depends the results of A, conducting us to a conditional problem:

$ P(W_{b} \mid\ W_{a}^c) = ? $

Is my thinking right? What are the answers to the questions?

Answer 1

Your $1/4$ is the probability that $A$ wins on his first try, before $B$ even has a chance to try.

Let $p$ be the probability that $A$ wins eventually. It will be larger than $1/4$. In fact it has to be larger than $1/2$: going first is clearly an advantage. We can figure it out.

If $A$ does not win on his first try then $B$ will win the game eventually with probability $p$, since he's now the first player in a brand new game. That means $A$ wins after not winning the first time with probability $(3/4)(1-p)$. That's implicitly a conditional probability. Therefore $$ p = \frac{1}{4} + \frac{3}{4}(1-p) $$ which works out to $p=4/7$.

Answer 2

A game is a sequence of alternating turns.   Player $A$ has the very first turn of play, Player $B$ the second (if player $A$ hasn't already won), and should both players lose their first turn, the game begins as new.

Player $A$ wins the game if either: Player $A$ wins the first turn, or otherwise Player $B$ loses their first turn and Player $A$ wins the subsequent game.   Thus we have a recursion.

$${\mathsf P(W)=\mathsf P(C)^2+(1-\mathsf P(C)^2)^2~\mathsf P(W)}$$

Alternatively: Player $A$ wins the game if either Player $A$ wins their first turn, or else PLayer $B$ loses the subsequent game (where player $B$ has the first turn).

$${\mathsf P(W)=\mathsf P(C)^2+(1-\mathsf P(C)^2)~(1-\mathsf P(W))}$$