Chance on winning by throwing a head on first toss.

Problem: Players A,B and C toss a fair coin in order, the first to throw a head wins what are their respective chances of winning?

Attempt: Let X = event that A throws a head on the first toss, and Y = event that B throws a head on first toss, similarly let Z = event that C throws a head on first toss.

Then there are eight different triple for tossing a coin.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

where H is a head, and T is a tail.

Can anyone please help me? I don't know how to continue. Any feedback/help would be really appreciated. Thank you.

Consider the winners of the triples. $S = \{\underbrace{HHH}_A, \underbrace{HHT}_A, \underbrace{HTH}_A, \underbrace{THH}_B, \underbrace{HTT}_A, \underbrace{THT}_B, \underbrace{TTH}_C, \underbrace{TTT}_{\text{repeat}}\}$

Now can you tell the probabilities of each winning?

$A$ wins if the sequence of tosses is any one of the following:

$$H, TTTH, TTTTTTH, \cdots, T^{3i}H, \cdots $$ that is, Tails $3i$ times in succession, $(i = 0, 1, 2, \ldots)$ followed by a Head. and so $A$ wins with probability $$\left.\left.\frac 12 \right[1 + \left(\frac 12\right)^3 + \left(\frac 12\right)^6 + \cdots\right] = \left.\left.\frac 12 \right[1 + \frac 18 + \frac{1}{8^2} + \cdots\right] = \frac{1}{2}\times\frac{1}{1-\frac{1}{8}} = \frac 47.$$

That $B$'s win probability is exactly one-half of this is easily found by noting that the sequences of tosses that are a win for $B$ are simply those in the list above preceded by the $T$ (Tail) that is $A$'s initial loss. Similarly, $C$' win probability is exactly one-half of $B$'s win probability.