There are 16 cards (4 red, 4 blue, 4 green, 4 yellow and have numbers 1-4) and 4 player.
a) I want to calculate the probability that 2 player have 2 red cards and 2 player have 0 red cards. How can I do that?
b) I want to calculate the probability that 2 players have 1 red card each, 1 player has 2 red cards and 1 player has 0 red cards. How can I do that?
My try:
Player A: 1 red card:
$\frac{\binom{4}{1} \binom{12}{3}}{\binom{16}{4}}$
Player B: 1 red card:
$\frac{\binom{3}{1} \binom{9}{3}}{\binom{12}{4}}$
Player C: 2 red cards:
$\frac{\binom{2}{2} \binom{6}{2}}{\binom{8}{4}}$
Player D: 0 red cards:
$\frac{\binom{0}{0} \binom{4}{4}}{\binom{4}{4}}$ which is 1.
Is this correct? Then I multiply this (A * B * C *D) and then by 12 (4! / (1!*1!*2!)
First choose two specific players, I'll call them A and B, and compute the probability that A and B each get 2 red cards. The order of dealing the cards doesn't matter, so I can assume that A picked 4 random cards first. We count the number of ways A could end up with exactly 2 red cards. There are $\binom{4}{2}$ ways to pick 2 of the 4 red cards, and $\binom{12}{2}$ ways to finish the hand by picking 2 non-red cards. The total number of hands is $\binom{16}{4}$, so the probability is $$P(\text{A picks 2 red cards}) = \frac{\binom{4}{2}\binom{12}{2}}{\binom{16}{4}}$$ Then B picks his hand, but now there are only 2 red cards left, and 10 others. $$P(\text{B picks 2 red cards, given that A did}) = \frac{\binom{2}{2}\binom{10}{2}}{\binom{12}{4}}$$ Multiply those two probablities together to get $P(\text{A and B each get 2 red cards})$. Then multiply that probability by $\binom{4}{2}$, the number of distinct pairs of players. $$P(\text{2 players got 2 red cards each}) = \binom{4}{2} \cdot \frac{\binom{4}{2}\binom{12}{2}}{\binom{16}{4}} \cdot \frac{\binom{2}{2}\binom{10}{2}}{\binom{12}{4}}$$