What is the probability that a person is the second person to roll a 6 for the first time? Suppose $A$, $B$, and $C$ roll a die in turns. What is the probability that $A$ is the second person to roll a $6$ for the first time?
 A: Avoiding explicit infinite sums:
Let's try a simpler question: two people X and Y are playing, with X rolling first.  What is the probability $p_x$ that X is the first to get a $6$?
Considering what might happen on the first roll (either X rolls a $6$ and is first to do so or X does not roll a $6$ with Y becoming the first to roll next), you have $p_x = \frac16 + \frac56 p_y$ and $p_y = \frac56 p_x$.  You can solve these simultaneous equations to get  $p_x=\frac6{11}$ and $p_y= \frac5{11}$
Now let's try your question with $p_a, p_b, p_c$ being the respective probabilities of being second to roll a $6$ in the three-player game.  You want to find $p_a$ 
Again considering what might happen on the first roll,  you  have $p_a = \frac56 p_c$ and $p_b=\frac16 p_x + \frac56 p_a$ and $p_c=\frac16 p_y + \frac56 p_b$.  You can solve these, using the earlier results, to get $p_a=\frac{300}{1001}$, $p_b=\frac{341}{1001}$, $p_c=\frac{360}{1001}$
A: So consider A about to succeed at the event.  For some positive integer $k$, everybody has rolled $k$ times.  A has gotten no sixes (which can happen in $5^k$ ways), one of the other players has already rolled one or more sixes (the player can be chosen in 2 ways and their rolls in $6^k-5^k$ ways), and the third player has not yet rolled a 6 (which can happen in $5^k$ ways).  With that set up, A then needs to roll a 6.  The probability of all of this is $$\frac{1}{6}\sum_{k=1}^\infty\frac{5^{2k}\cdot2(6^k-5^k)}{6^{3k}}=\frac{300}{1001}\approx0.2997$$
A: Since you only consider the cases where players roll a $6$ or not, you may write each turn as a tripple $(Y,Y,N)$, meaning A rolled a 6, B rolled a 6 but C did not. Your game then becomes a sequence of tripples of the previous form, i.e.,
$$(Y,Y,N), (Y,N,Y),...,(Y,N,N),...$$ 
Now the probability that A rolls the second 6 for the first time (call this event E) translates to any finite sequence of the form
$$(N,N,N), (N,N,N),...,(N,Y,N),...,(Y,N,N)$$
or
$$(N,N,N), (N,N,Y),...,(N,N,N),...,(Y,N,N)$$ 
where all but one tripple have the form $(N,N,N)$. You can write this now via a geometric random variable $X_A\sim G(1/6)$, namely
$$P[E] = \sum_{n=2}^{\infty} P[E|Number\;of\;rounds = n]P[Number\;of\;rounds = n]$$
$$\sum_{n=2}^{\infty} P[E|X_A = n]P[X_A = n]$$ 
Now, assuming independence of the players and the rolls, we can describe the number of $6$ which player B, and analogously C, rolls in $n$ rounds by a binomial $B_n\sim Binom(n,1/6)$, or $C_n\sim Binom(n,1/6)$. The term $P[E|X_A = n]$ has then the form
$$P[E|X_A = n] = P[B_n = 1]P[C_n = 0]+P[B_n = 0]P[C_n = 1] = 2P[B_n = 1]P[C_n = 0].$$
From here, you put everything together in $\sum_{n=2}^{\infty} P[E|X_A = n]P[X_A = n]$ and should arrive at 
$$P[E] = \dfrac{2(1/6)^2}{(1-1/6)}\sum_{n=1}^{\infty}n(1-1/6)^{(3n-1)}=\dfrac{2(1/6)^2}{(1-1/6)}\dfrac{(1/6)^2}{(1-(1/6)^3)^2}.$$
