# Combinatorics with playing cards

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.

• Just divide it into cases. Do three of the same suit, then two of one suit and on of another. – Ross Millikan Jul 31 at 2:40
• As Karn states below, it might be easier to calculate the probability of getting three different suits. The probability that the second card does not match the suit of the first card is $3\cdot 13/51$. Now you just need to calculate the probability that the third card does not match the suit of the first two cards. – irchans Jul 31 at 3:30
• Am I following your hint correctly that this would get me the answer: $1 - ( (52/52) ⋅ (3⋅13/51) ⋅ (2 ⋅ 13/50) )$ – Stephen Dean Jul 31 at 6:59

Hint:

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

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.

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.

• IMHO, your formula is correct if you replace each card after it is drawn, but I believe the original poster wanted to get the answer assuming all three cards are drawn without replacement. – irchans Jul 31 at 3:26
• Since the cards are being selected without replacement, this is a hypergeometric distribution problem. – N. F. Taussig Jul 31 at 3:27
• Ah, I'll add that to the answer. – N. Bar Jul 31 at 3:55

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: $${{4 \choose 1}{13 \choose 2}{39 \choose 1} + {4 \choose 1}{13 \choose 3} \over {52 \choose 3}} \approx 0.602$$

• Thanks, this is very helpful. How would it change if there is no replacement? – Stephen Dean Jul 31 at 19:53