2 Persons toss a fair cube. $X$ is the toss times when $A$ gets $1$, and $Y$ is the toss times when $B$ gets $1$. What is the probability $P(X > Y)$? Can you give me some hints and help to solve this question? I've forgotten a bit the technique.

Danny and Alex making a series of tossing a fair cube.
Let $X$ be random variable which its value is the times Danny tossed the cube until got $1$,
And Let $Y$ be random variable which its value is the times Alex tossed the cube until got $1$.
Calculate $P(X > Y)$.

Thanks in advance.
 A: We show how to solve the problem on the assumption they alternate, that A goes first, and that the game is over when a $1$ is tossed. (That was mentioned in comments, but it is not my interpretation of the wording.)
It is clear that with probability $1$ the game terminates. Let $a$ be the probability that A wins. We condition on the result of the first toss.
If the first toss is a $1$, then A has won. The conditional probability that she wins given that she did not toss a $1$ is $1-a$, for the roles of A and B are now reversed. It follows that
$$a=\frac{1}{6}+\frac{5}{6}(1-a).$$
Now we can find $a$.
Another way: The probability A wins on the first toss is $1/6$. The probability she wins on the third toss is $(5/6)^2(1/6)$, for A must toss a non-$1$, then B must toss a non-$1$, then A must toss a $1$. Calculate in the same way the probability A wins on the fifth toss, and so on, and sum the resulting series.
Remark: From the wording, it looks as if A and B keep separate counts.  So my interpretation of the problem is different from OP's interpretation. We solve the problem under this interpretation.
By symmetry, we have $\Pr(X\gt Y)=\Pr(Y\gt X)$. It follows that
$$\Pr(X\gt Y)=\frac{1-\Pr(X=Y)}{2}.$$
Now we calculate $\Pr(X=Y)$. The probability that $X=Y=1$ is $(1/6)^2$. The probability that $X=Y=2$ is $((5/6)(1/6))^2$. Continue.
