Two players throw a die until the sequence $1,2,3$ appears, and the winner is the one who roll $3$. What is the probability the second player wins? 
Alameda and Belisario alternate turns throwing a fair die. Alameda plays first and they continue throwing, one at a time, until the sequence $1$-$2$-$3$ appears. Whoever throws the $3$ is the winner. What is the probability that Belisario wins?

Hmmm - probabilities is not my strong domain!
First let's see what the chances are to get a $1$-$2$-$3$ regardless who gets it.
Is it $1$ in $6\cdot 6\cdot 6$?
Then if this probability is $p$, my understanding is that the probability for Belisario to win is smaller, but I can't compute it :(
 A: Let's use states.
We'll label a state according to how much of the $1,2,3$ chain has been completed and according to who's turn it is.  Thus you start from $(A,\emptyset)$, and the other states are $(B,\emptyset),(X,1),(X,1,2)$ Win and Loss (Where $X\in \{A,B\}$.  In a given state $S$ we let $p_S$ denote the probability that $B$ will win.  Thus the answer you want is $P_{A,\emptyset}$.  In this way we have $6$ variables (since the probability from the Win, Loss are clear).  Of course these variables are connected by simple linear equations.
For instance $$P_{A,\emptyset}=1-P_{B,\emptyset}$$ and, more generally, $$P_{A,s}=1-P_{B,s}$$ where $s$ is any part of the sequence.  Thus we are down to $3$ variables.
(Why?  Well, In the state $(A,\emptyset)$, A is in the exact same position that $B$ is in in the state $(B,\emptyset)$.  Thus $A$ has the same probability of winning from $(A,\emptyset)$ as $B$ has of winning from $(B,\emptyset)$.  Same with any $s$)
Considering the first toss we see that $$P_{A,\emptyset}=\frac 16\times P_{B,1}+\frac 56\times P_{B,\emptyset}$$
(Why?  Well, $A$ either throws a $1$ or something else.  The probability of throwing a $1$ is $\frac 16$ and if that happens we move to $(B,1)$.  If $A$ throws something else, probability $\frac 56$, then we move to $(B,\emptyset)$)
Similarly:  $$P_{B,1}=\frac 16\times P_{A,1,2}+\frac 16\times P_{A,1}+\frac 46\times P_{A,\emptyset}$$  and $$P_{B,1,2}=\frac 16\times 1+\frac 16\times P_{A,1}+\frac 46\times P_{A,\emptyset}$$
(Why?  Similar reasoning.  Consider the possible throws $B$ might make and what states they each lead to).
Solving this system we get the answer $$\boxed {P_{A,\emptyset}= \frac {215}{431}\approx .49884}$$
Note:  I used Wolfram alpha to solve this system but it's messy enough so that there could certainly have been a careless error.  I'd check the calculation carefully.
Sanity check:  Or at least "intuition check".  Given that this game is likely to go back and forth for quite a while before a winner is found, I'd have thought it was likely that the answer would be very close to $\frac 12$.  Of course, $A$ has a small advantage from starting first (it's possible that the first three tosses are $1,2,3$ after all), so I'd have expected an answer slightly less than $\frac 12$.
Worth remarking:  sometimes intuition of that form can be a trap. After all, the temptation is to stop checking as soon as you get an answer that satisfies your intuition.  In fact, the first time I ran this, I got an answer of $.51$ which seemed wrong.  Worse, that solution showed that $P_{A,1,2}$ was about $.58$ which seemed absurd (how could $B$ have a strong advantage when $A$ is one toss away from winning?).  So, I searched for and found the careless error.  Second trial gave all plausible results so I checked casually and stopped. But you should do the computation again to be sure.
A: We have three probabilities to consider, all from the point of view of the player who is about to roll.
$p_0$ is the probability of winning if no part of the winning sequence has been rolled.  (This is Alameda's situation at the beginning of the game.)  Call this situation state $0$.
$p_1$ is the probability of winning if the opponent has just rolled at $1$.  Call this situation state $1$.
$p_2$ is the probability of winning if the opponent has just rolled $2$ and the roll immediately before that was $1,$ so that rolling a $3$ will win.  Call this situation state $2$.
Suppose no part of the sequence has been rolled.  Then if you roll anything but a $1,$ your opponent will be in state $0$ and you will win if he loses; that is, you win with probability $1-p_0.$  If you roll a $1,$ your opponent will be in state $1,$ and again, you win if he loses.  That is  $$p_0=\frac56(1-p_0)+\frac16(1-p_1)$$ 
Similar considerations give $$p_1=\frac46(1-p_0)+\frac16(1-p_1)+\frac16(1-p_2)$$ and $$p_2=\frac46(1-p_0)+\frac16(1-p_1)+\frac16$$  where the last $\frac16$ is the case where the roller wins by rolling $3$.  We can write these equations more neatly as $$\begin{align}
11p_0+p_1+0p_2&=6\\
4p_0+7p_1+p_2&=6\\
4p_0+p_1+6p_2&=6
\end{align}$$
(Sorry about the $0p_2$ in the first equation.  I couldn't figure out how to format things.)
Anyway, solve the system will for $p_0,p_1,p_2.$  Belasario's probability of winning is $1-p_0,$ which turns out to be $${215\over431}$$
EDIT
I started typing this before lulu's answer was posted, but I'm such a slow typist that his answer had been up for a while before I finished.  I'll leave it for a bit before deleting it, just so you can check if we have the same equations.
