The problem setup is as follows: Suppose we have a game of 12 cards with 4 different symbols, $a,b,c$ and $x$. If we enumerate the cards, we have the set $\{(a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,x_1,x_2,x_3)\}$ Each card has a weight $w^\alpha_i$ for $\alpha\in\{a,b,c,,x\}$ and $i\in\{1,2,3\}$. Suppose the cards are placed face down, and we flip over a single card in succession determined from the given weights. The game concludes once $3$ matching cards are flipped with the exception of card $x$ (i.e., if $3$ $x$'s are flipped over first, the game continues). What is the probability that the winning match is $a,b,$ or $c$, respectively? What is the probability that three $x$'s are matched first and the winning match is $a, b,$ or $c$, respectively?
I stumbled across this problem in regards to a game, where probabilities were listed presumably from simulating the game (the weights were explicitly given). I thought I could derive an explicit solution myself, but my discrete probability theory has proven too rusty. Any help on the problem would be greatly appreciated.
EDIT: As per the suggestion in the comments, let me expand on the problem as I have interpreted it (which may or may not be valid).
Let $X_j$ denote the random variable which denotes the $j$th flip. Then the probability of the first flip being card $\omega\in\{a,b,c,x\}$ is given by $$P(X_1=\omega)=\frac{\sum_{i=1}^3w_i^\omega}{\sum_{i,\alpha}w_i^\alpha},\qquad\omega\in\{a,b,c,x\}.$$ After the first flip, if the chosen card corresponded to weight $w_{i_1}^{\alpha_1}$, then we would have the probability of the following flip given by $$P(X_2=\omega)=\frac{1}{\sum_{(i,\alpha)\neq(i_1,\alpha_1)}w_i^\alpha} \begin{cases} \sum_{i\neq i_1}w_i^\alpha&\text{ if }\alpha=\alpha_1\\ \sum_iw_i^\alpha&\text{ if }\alpha\neq\alpha_1 \end{cases}.$$
Continuing in this fashion does not seem to be the correct method for obtaining the desired result, and has brought me here for further guidance. Any reading or topics related to this or a similar problem would be beneficial as well.
The process is similar to that of the multivariate Wallenius' noncentral hypergeometric distribution where each card has multiplicity 1. Unfortunately, the top search result (PDF) for this distribution suggests that the calculation for even the simpler univariate-with-multiplicity case faces a dilemma between the inefficiency of the brute-force recursive / Markov chain method and numerical issues with more clever methods.
From a pragmatic perspective, with only 12 unique cards, the number of possible game states (i.e. whether each of the 12 cards has been drawn) is $2^{12} = 4096$, and the game is guaranteed to conclude within 10 steps. So the brute-force method of simply applying the Markov chain transition 10 times is well within the reach of any modern PC, as well as being flexible and numerically stable.