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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!)

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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. $$P(\text{2 players got 2 red cards each}) = \binom{4}{2} \cdot \frac{\binom{4}{2}\binom{12}{2}}{\binom{16}{4}} \cdot \frac{\binom{2}{2}\binom{10}{2}}{\binom{12}{4}}$$

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  • $\begingroup$ Can you also give me a hint on b)? $\endgroup$ – 今天春天 Oct 26 '15 at 19:13
  • $\begingroup$ Follow the same general plan. First suppose that player A gets 2 red, players B and C get 1 red, and player D gets 0. Then, by letting them pick their hands in sequence, calculate the probability of this happening. Then multiply by the number of distinct ways to assign those labels A, B, C, D (distinct meaning swapping B and C doesn't matter, since they both get 1 red card anyway). $\endgroup$ – Nathaniel Mayer Oct 26 '15 at 19:16
  • $\begingroup$ You just received a lot of information that could help you with part b and the first thing you do is ask for more rather than attempt b? Make an effort. $\endgroup$ – Vincent Oct 26 '15 at 19:17
  • $\begingroup$ Thanks for these helpful tips so far. I tried to to this (edited answer above). Can this be right? $\endgroup$ – 今天春天 Oct 26 '15 at 19:26
  • $\begingroup$ Yep, looks good $\endgroup$ – Nathaniel Mayer Oct 26 '15 at 19:33
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Hint:

Start with one player $A$ having a red card allready and $3$ open spots, and $3$ other players have $4$ open spots.

Hand out a red card randomly in the sense that the open spots have equal chances of receiving that card.

Repeat this twice.

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