# combinatorics & probability problem

There are numbered cards 1 to 13 each of colour red, green, yellow and white. And four players have been distributed 4 each of these cards randomly. What is the probability that each player gets at least one card from each colour? My answer is this: $$\frac{13^412^411^410^4}{{52\choose 16}4!}$$ Am I right?

There are $\dfrac{52!}{(36!)(24)^4}$ ways to give out the cards, since the order in which they are given to one particular person does not matter.
How many ways are there in which each person gets cards of different color? there are $13^4$ ways to select the first players cards, $12^4$ the second player $11^4$ the third and $10^4$. So the final answer is
$$\frac{13^4\cdot12^4\cdot11^4\cdot10^4}{\frac{52!}{(36!)(24)^4}}=\frac{(13^4\cdot12^4\cdot11^4\cdot10^4)(36!)(24)^4}{52!}\approx .0011\%$$
• Most of the expression is right, just a problem with the $4!$. – André Nicolas Jul 3 '14 at 4:36
• My answer and the one by Bananarama are the same, though derived omewhat differently. Maybe you can see another way by imagining the ards are dealt frt to A, then B, then C, then D. The probability A gets $4$ different colours is $39/51)(26/50)(13/49)$. I do not see a reason for a denominator of $\binom{52}{16}$. Indeed we are choosing $16$ cards, but order matters. – André Nicolas Jul 3 '14 at 11:51