# How many cards must be drawn unseen from a set of 52 playing cards to guarantee that at least 2 of them are the same suit?

I am having trouble starting this. I know a can use a $nCr$ method but I don't know how to apply it here.

• If you want to guarantee that two cards are from the same suit, you need only draw five cards (since there is no way you can draw five cards without at least two of them being the same suit). Dec 5 '14 at 0:36
• sigh that's what I thought and then second-guessed my self because of the simplicity. Thanks. Is there a mathematical way of showing this? Dec 5 '14 at 0:38
• Yes. Here Edwin Dec 5 '14 at 0:39
• You mean, what's the most cards you can draw, and have them all in different suits? My guess is $4$: one spade, one heart, one diamond, one club. So the way you phrased the question the answer is $5$. Am I right?
– bof
Dec 5 '14 at 0:42
• Since it is clear, that $4$ cards are not sufficient and $5$ cards are, I wonder why a "mathematical method" should be searched. That is like shooting rockets on ants. Dec 5 '14 at 0:42

## 1 Answer

You draw one... Then you draw another one.. In the worst of the cases it's not the same suit as the first one. You draw another one..., in the worst of the cases, it's not the same suit as the other two. You draw another one..., in the worst of the cases, it's not the same suit as the other three. You draw another one... It must be of some of the four suits you already have in the other 4 cards.

This comes from the Pigeonhole principle.