Combinatorics with playing cards

Question

I have a question that I can't seem to find the answer to:

What are the odds of pulling 3 cards from a normal deck of playing cards (52 cards, 4 suits, 13 different values) and at least 2 of the cards being of the same suit?

I've found the odds of picking 3 of a kind in a 5 card poker hand (while avoiding pairing the other 2 cards to make a full house) or 4 of a kind, but the "at least" portion of my question has been tricky for me - because it could be 2 or 3 of the same suit.

Answer 1

Hint:

$$ N(\text{2,3 same suit}) = N(\text{all combi}) - N(\text{no card from the same suit})$$

Answer 2

You want to obtain either three cards of one of the suits, or two cards of one suit and one card of another; when selecting three cards from the deck.

Answer 3

For this, you can use a binomial distribution. To do this quickly, use the $\text{binomcdf}$ function on a calculator. Depending on the calculator you use (online, TI, Casio, etc.), you may have to find the complement of the binomial distribution, or the calculator may just automatically give you the answers for at least, at most, etc.

To do it by hand, you have to evaluate:

$$\sum_{x=2}^3(^3_x)*(.25)^x*(.75)^{3-x}$$ $$\approx 0.156250$$

Note: this assumes the cards are being replaced.

Answer 4

Assuming you draw cards with replacement, there are ${52 \choose 3}$ ways of drawing 3 cards from a deck.

When at least two cards are from the same suit, we have two scenarios: 1) only 2 cards are from the same suit and the third card is from another suit, 2) all three cards are from the same suit.

The number of cards for the first scenario is: $${4 \choose 1}{13 \choose 2}{39 \choose 1}$$

The number of cards for the second scenario is: $${4 \choose 1}{13 \choose 3}$$

So the total probability is: $${{4 \choose 1}{13 \choose 2}{39 \choose 1} + {4 \choose 1}{13 \choose 3} \over {52 \choose 3}} \approx 0.602$$