Ryfujk

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 $6*6*6$?

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 $Xin {A,B}$. 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 $$P_{A,emptyset}=frac 16times P_{B,1}+frac 56times P_{B,emptyset}$$

(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: $$P_{B,1}=frac 16times P_{A,1,2}+frac 16times P_{A,1}+frac 46times P_{A,emptyset}$$ and $$P_{B,1,2}=frac 16times 1+frac 16times P_{A,1}+frac 46times P_{A,emptyset}$$

(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.

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 $${215over431}$$

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.