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 :(

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.
