# Pigeonhole Principle Problem with a standard deck of cards

If I have a standard deck of cards, how man cards must I draw to ensure that I get three cards of the same kind. How many cards must I draw ensure that I get 5 cards of the same suit.

I am new to this topic but I think that in the first case the pigeonhole would be the the kind of cards (example: aces) and in the second case the kind of cards would be the suit of course.

Can someone please offer guidance as to how to go about solving this problem.

For each question, aim to "fill up" all the pigeon holes. Do this by counting all distinct possibilities. Your answer will then be one more than this count.

In a 52 card deck, we have 13 possible kinds $\times$ 4 possible suits.

• To ensure two of a suit, draw 5. (We're filling 4 holes. At the least, we have one of each suit. Drawing one more ensures a repetition)

• To ensure three of a suit, draw 9. (We have eight holes filled. At the least, there are two of each suit. The ninth ensures that at least one suit will have three cards.)

• If you need two of a kind (any kind, not a specific card like a Jack), then draw 14 cards. The first 13 ensure that you have everything from Ace to King. The next draw gives two of a kind. For three of a kind, you'll need $2 \times 13 +1$. – zahbaz Oct 30 '14 at 17:59
You have $\color{green}4$ slots, and you want $\color{red}3$ (or $\color{blue}5$) cards from the same slot. The pigeonhole principle says that if you draw at least $\color{red}2\times \color{green}4 + 1$ (or $\color{blue}4\times \color{green}4 + 1$) then you are sure to success.