Draw 4 cards where: 3 cards same suit and remaining card of different suit

Question

Four cards are drawn from a standard 52-card deck without replacement. Find the probability that exactly 3 cards are of the same suit and the remaining card is of a different suit.

What I did: (4C1)(13C3)(3C1)(13C1) divided by (52C4)

But the final right answer is 0.375. Can you please tell me how to solve it and why mine is wrong? Thank you.

Answer 1

There are 4 ways to choose the suit we want three of. There are then 3 ways to choose the remaining suit (we have one of). Then there are ${13 \choose 3}$ ways to choose the three card from the first suit, and $13$ ways to choose the card from the final one.

We divide this by ${52 \choose 4}$ ways to pick four cards, so we have as our probability:

$$\frac{4 \times 3 \times {13 \choose 3} \times 13}{{52 \choose 4}} \approx 0.1648$$

Answer 2

Consider that there are 13 cards per suit. Then, to fulfill the stated conditions, we need

Then $P$($3$ cards same suit and $1$ card other suit| $4$ suits) = $4$ * $\dfrac{\binom{13}{3}\binom{39}{1}}{\binom{52}{4}}$ = $4$ * $\dfrac{286*39}{270,725}$ = $0.1648$

@Henno, @David, and @calculus: You are correct, there are four ways (suits) for this occur.

Answer 3

Another approach is to draw the cards one by one.

Draw 3 cards of the same suit and 1 card of another suit:

$yyyx$

In total there are 4 ways.

And the same kind of drawings can be made with the other 3 suits. Thus we have in total $4\cdot (1+3)=16$ ways.

The probability of drawing cards one way is $\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}\cdot \frac{39}{49}$

All together the probabilitiy is $16\cdot \frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}\cdot \frac{39}{49}\approx 0.1648$