Mario Party 3 Mini-game Probability Question

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}