Mario Party 3 Mini-game Probability Question

Question

I have a question about a mini-game in Mario Party 3. I have extracted the mathematical information from the game below.

Setup: Four players $A,B,C$, and $D$ line up in some order. There are $12$ cards which are face-down on a table in front of the first person in line. The cards are as follows: $6$ blue cards, $4$ green cards, and $2$ gray cards.

Gameplay: The players take turns, based on their order in line, flipping over a card. If the card is blue, the player is safe and goes to the back of the line and the card is discarded. If the card is green, the player is out, removed, and the green card is discarded. If the card is gray, the player remains in the game and the order of the remaining players (including the player that drew the gray card) in the line is randomly shuffled; the gray card is discarded. The game ends when there is only one person left.

My Question: Let's say the initial order of the line-up is $A, B, C, D$. I was wondering if there was a way to determine which player has the best, second best, third best, and worst chance of winning the game. I'm not sure if this is possible since the two gray cards shuffle the line-up when they are flipped over. I would also be interested in solutions of the simplified version of the game with $10$ cards ($6$ blue and $4$ green) in which the gray cards have been removed.

Thanks!

Answer 1

As mentioned in the comments, once a gray card is drawn, all remaining players have the same winning probability. That makes it straightforward to determine the winning probabilities by enumerating all permutations of the $12$ cards. Since the $12$-th card will never be drawn, a slight optimisation is to consider only the permutations of the different multisets of $11$ cards. Here's code that performs the computation. The winning probabilities for the alphabetical original lineup with $A$ drawing first are

\begin{align} p_A&=\frac{1831}{9240}\approx19.8\%\;,\\ p_B&=\frac{2131}{9240}\approx23.1\%\;,\\ p_C&=\frac{2759}{10395}\approx26.5\%\;,\\ p_D&=\frac{2543}{8316}\approx30.6\%\;. \end{align}

Answer 2

This is not a complete answer and just add a supplementary mathematical formulation based on Markov chain / recurrence relation .

Let $p(m, n, b, g, r)$ be the probability of winning for the player lined up at the $m$th position, with a total of $n \geq m$ players remaining, and $b$ blue cards, $g \geq n$ green cards and $r$ gray cards are remained on the table.

Then we have the recurrence relationship $$ p(m, n, b, g, r) = \frac {b} {b + g + r} p(m-1, n,b-1,g,r) + \frac {g} {b + g +r} p(m-1,n-1,b,g-1,r) + \frac {r} {b + g + r} \frac {1} {n}, m \in \{2, \ldots, n\}$$ $$ p(1, n, b, g, r) = \frac {b} {b + g + r} p(n, n,b-1,g,r) + \frac {r} {b + g + r} \frac {1} {n} $$

and by solving this we should be able to obtain the answer. (not verified yet)

joriki has already provided a computational solution, and probably the computational complexity of solving this is similar to his algorithm.