189. Two players take turns tossing a coin. The one who gets heads first wins. What is the probability that the game will never end? What is the probability that the starting player will win? What is the probability that the second player will win?
189. Let's denote:

the probability of event $A$: "the starting player wins" - by $x$, the probability of event $B$: "the second player wins" - by $y$, the probability of event $C$: "the game never ends" - by $z$. Events $A, B, C$ are pairwise mutually exclusive and form a complete system, therefore

$$
x+y+z=\mathbf{P}(A)+\mathbf{P}(B)+\mathbf{P}(C)=1
$$

The probability of the second player winning is equal to the probability that tails appears on the first flip (this probability is $1 / 2$), multiplied by the probability of the second player winning given this condition (property 6). But then the second player becomes the starting player, so this conditional probability is equal to $x$. Therefore,

$$
y=\frac{1}{2} x
$$

The probability of the first player winning is found using the law of total probability. Consider the events:

$D$: "heads appears on the first flip",

$F$: "tails appears on the first flip".

Events $D$ and $F$ are mutually exclusive and form a complete system; by property 7,

$$
x=\mathbf{P}(A)=\mathbf{P}(D) \mathbf{P}(A / D)+\mathbf{P}(F) \mathbf{P}(A / F)
$$

Obviously, $\mathbf{P}(D)=\mathbf{P}(F)=1 / 2, \mathbf{P}(A / D)=1$. To calculate $\mathbf{P}(A / F)$, note that if tails appears on the first flip and the second player becomes the starting player, then the first player is in the position of the second and the probability of his winning is $y$. Therefore,

$$
x=\frac{1}{2}+\frac{1}{2} y
$$

Solving equations (2) and (3) simultaneously, we find $x=\frac{2}{3}, y=\frac{1}{3}$. From relation (1) we get $z=0$.

The fact that the probability of the game continuing indefinitely without a result is zero can also be obtained by direct calculation (as was done for the probability of not rolling a six with a die).