# 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!

• You have two mistakes which combine to give you half the right answer. First, in your last step, any suit will work if you have $4$ different suits in the first $4$ draws -- so $12/48$ makes your answer $1/4$ of what it "should be". Second, the method of choosing one card of each suit and then an arbitrary fifth card over-counts by a factor of two, because any hand, say 3S 3H 5D 7C 9C, gets counted twice, once with 7C in the first group of $4$ and once with 9C in the first group. $2*(1/4) = 1/2$
– Ned
May 12 '20 at 14:36

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.

• I see now, thank you! May 12 '20 at 10:23

You could have picked either of the cards of the same suit first, so you’re counting each admissible hand exactly twice.

• So, because theoretically I could have picked the card that has the same suit first, and not last I need to multiply everything by 2? May 12 '20 at 9:47
• @OUR: No, the opposite: Because of that, you need to divide by $2$, since you double-counted. May 12 '20 at 10:04
• But then I get an answer that is smaller by a factor of 4 :( May 12 '20 at 10:07
• @OUR You should still multiply the result by the number of suits (see also my answer).
– user
May 12 '20 at 10:09
• @OUR I added the explanation where your way fails.
– user
May 12 '20 at 10:20

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%