Probability of picking 4 different valued and suited cards from pack I want to know the probability of picking 4 different valued and suited cards from a standard pack of $52$ cards. I reasoned that first you pick a card, and from the remaining $51$ cards you can pick $52-(13+3)$ cards and from the $50$ remaining cards you can pick $52-2(13+3)$ cards and from the remaining $49$ cards you can pick $52-3(13+3)$ cards so the probability is
$$\frac{36\cdot20\cdot 4}{51\cdot50\cdot49}=\frac{96}{4165}$$
Is my reasoning correct? Thanks for any comments.
 A: I would rather try to count the number of hands of 4 cards that fullfill your conditions:


*

*you have to pick one spade card: $13$ choices

*you have to pick one heart card: $12$ choices (it should be different from the spade value)

*for the diamond you have $11$ possible choices

*for the club, there remain $10$ choices


Finally there are $13 \cdot 12 \cdot 11 \cdot 10 = 17160$ possible hands.
The total number of hands is ${52 \choose 4} = 270725$, and finally your probability is
$$p={17160 \over 270725} = {264 \over 4165}$$
A: There are certainly $36$ cards that are a different suit and value from the first one. After two are chosen, there are $22$ good options for the third card, because $11$ of $13$ cards in each of the two remaining suits dodge the already-selected values. For the last card, it has to be in the one remaining suit, and there are three values it cannot be, so there are $10$ good options for it. Thus:
$$\frac{36}{51}\cdot\frac{22}{50}\cdot\frac{10}{49}=\frac{12\cdot 11\cdot 2}{17\cdot 5\cdot 49}=\frac{264}{4165}$$
When you subtracted $2(13+3)$, that was a bit too much, because part of each $3$ is part of the other $13$, so you counted $2$ cards twice.
