probability regarding three people throwing a die There are 3 players, A, B, C, taking turns to roll a die in the order ABCABC....  
What's the probability of A is the first to throw a 6, B is the second, and C is the third?
The answer said it's 216/1001, but I always got 125/1001  
The way I did it was:
Let $X_1$= number of rolls till A throw a 6
    $X_2$= number of rolls till B throw a 6
    $X_3$= number of rolls till C throw a 6
$P(X_1=i)=\frac{1}{6} (\frac{5}{6})^{i-1}$
$P(X_2=j)=\frac{1}{6} (\frac{5}{6})^{j-1}$
$P(X_3=k)=\frac{1}{6} (\frac{5}{6})^{k-1}$
therefore, $P$(A is the first to throw a 6, B is the second, and C is the third)
$=P(X_1 < X_2 < X_3)  $
= $\sum_{i<j<k}^{  } (\frac{1}{6})^3 (\frac{5}{6} )^{(i-1)+(j-1)+(k-1)}  $
$= (\frac{1}{6})^3 \sum_{i<j}^{  } (\frac{5}{6} )^{(i-1)+(j-1)} \sum_{k=j+1}^{\infty } (\frac{5}{6} )^{k-1}  $
$ = (\frac{1}{6})^3 \sum_{i<j}^{  } (\frac{5}{6} )^{(i-1)+(j-1)} \frac{(\frac{5}{6})^{j}} {1-\frac{5}{6}} $
$ =(\frac{1}{6})^2 \sum_{i<j}^{  } (\frac{5}{6} )^{(i-1)+(j-1)+j}$
$ =(\frac{1}{6})^2 \sum_{i}^{  }(\frac{5}{6} )^{(i-1)}\sum_{j=i+1}^{\infty}  (\frac{5}{6} )^{2j-1}$
$ =(\frac{1}{6})^2 \sum_{i}^{  }(\frac{5}{6} )^{(i-1)} \frac{(\frac{5}{6})^{2i+1}} {1-\frac{5}{6}} $
$ =(\frac{1}{6}) \sum_{i}^{  }(\frac{5}{6} )^{(i-1)+2i+1}$
$ =(\frac{1}{6}) \sum_{i=1}^{\infty  }(\frac{5}{6} )^{3i}$
$=\frac{125}{546}$
Could anyone help me check where went wrong? Thanks so much!

You can avoid infinite sums altogether if you use a conditioning argument. 
Let $X_1$, $X_2$, and $X_3$ be the times of the first six for each of the three players. 
We look at the two man game first. Let $q=\mathbb{P}(X_2<X_3)$. By first step analysis, we have
$q={1\over 6}+\left({5\over 6}\right)^2q$, so $q={6\over 11}$.  
For the full three man game, let $p=\mathbb{P}(X_1<X_2<X_3)$,
First step analysis gives $p={1\over 6}q+\left({5\over 6}\right)^3p$,
or ${91\over 216} p={1\over 11}$. Finally, this means $p={216\over 1001}$.  
