Probability-throwing a die Three players $X,Y,Z$ play with die. There is order to throwing: $XYZXYZ...$.
What is the probability that the first "$6$" would be thrown by $X$, the second "$6$" by $Y$ and the third by $Z$.
I have any idea how to solve it. Do you help me? Which tools should I use?
 A: If $X$ throws the first six, then everyone throws a non-six until he throws it. So the probability is $\frac{1}{6}(1+\lambda+\lambda^2\dots)$ where $\lambda=\left(\frac{5}{6}\right)^3$. That sums to $\frac{1}{6(1-\lambda)}=\frac{36}{91}$.
Once he has thrown the six, the prob that the next six is thrown by $Y$ is the same (because $Y$ is the first to throw after $X$'s success). Similarly for $Z$. So we get $\left(\frac{36}{91}\right)^3\approx6.2\%$.
A: In the first round of $XYZ$, the respective probabilities of $X,Y,Z$ winning are:
$\dfrac16,\; \dfrac56\dfrac16, \; \dfrac56\dfrac56\dfrac16$ 
so the odds in favor are $36:30:25\;$ respectively, and will be the same in each such cycle.
Thus the respective probabilities for $X,Y,Z$ throwing the $6$ first are $\dfrac{36}{91}, \dfrac{30}{91}$ and $\dfrac{25}{91}$

Oh, you want the probability of $X$ throwing the first six, $Y$ throwing the second.....
So once $X$ has thrown the first $6$, $Y$ is now in the same position as $X$ was at start, 
so $P(Y$ throws the second $6) = \left(\dfrac{36}{91}\right)^2$
and similarly, $P(Z$ throws the third $6) = \left(\dfrac{36}{91}\right)^3$ 
