Probability - Calculating the probability of choosing 5 cards with 4 different suits fail I am trying to calculate the probability of taking from a regular deck of cards (52) 5 cards with 4 different suits, my go:  
first we think about picking 'diamond' (just to make it more simple) it is: $\frac{13}{52}$
second we think about picking 'heart': $\frac{13}{51}$
same thing for the two other suits: $\frac{13}{50}$ and $\frac{13}{49}$ respectively.
The fifth card needs to be the same one of the suits, so: $\frac{12}{48}$ because we took each card from each suit, so we have only $12$ remaining in that same suit.
But we have to consider the order that we draw the cards, so we multiply by $5!$ 
And we have my final answer of: $5! \cdot \frac{13}{52} \cdot \frac{13}{51} \cdot \frac{13}{50} \cdot \frac{13}{49} \cdot \frac{12}{48}$ 
Sadly, this answer is only half the real answer: $\approx 0.26375$  mine gives:  $\approx 0.13187$ 
Where does the $ \cdot 2$ need to come from? I don't see another orientation other than the suit we pick the second card to be of, so we don't need to multiply by $2$ but by $4C2 = 6$ and that gives me an answer that times $3$ larger than the real one, so in that case, where does the $\frac{1}{3}$ need to come from?  
Thank you!
 A: You could have picked either of the cards of the same suit first, so you’re counting each admissible hand exactly twice.
A: Obviously, the "correct" hand should contain two cards of a suit and one card of each of the other 3 suits. There are $\binom41$ ways to choose the suit with the double, and 
$\binom{13}2$ ways to choose the double out of the suit. Therefore the overall number of "correct" hands is:
$$
\binom41\binom{13}2\binom{13}1^3.
$$
Dividing this by the overall number of hands: $\binom{52}5$ you obtain the correct probability.
You way fails because of two reasons: first you double-count the hands and second your miss the factor 4 because the "last" card may be of any of 4 suits.
A: I was thinking that the problem could be solved by using the hypergeometric multivariate distribution 
P(x,y,z,t) = [ 13Cx 13Cy 13Cz 13Ct ] / 52C5
where x,y,z,t are the random variables associated to the four disjoint sets of the suites
as the only 4 possible configurations of the random variables compatible with the requested probability are the following
x  y  z  t 
2  1  1  1
1  2  1  1
1  1  2  1
1  1  1  2
we find:
P = P(2,1,1,1)+P(1,2,1,1)+P(1,1,2,1)+P(1,1,1,2) = [ 13C2 13C1 13C1 13C1 ] x 4 / 52C5 = 26%
