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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.

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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.)

Extrapolating this pattern should get your answers.

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  • $\begingroup$ I understand the second part which ask about the same suit. However the first part asks about the same kind like 3 jacks or three tens $\endgroup$ – dreamin Oct 30 '14 at 17:43
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    $\begingroup$ 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 $2 \times 13 +1$. $\endgroup$ – zahbaz Oct 30 '14 at 17:59
  • $\begingroup$ I am unsure whether or not it would be 42 . For example if we are trying to get 3 Jacks. Worst case scenario we draw all the other non Jack cards, 39, then we would need to draw 3 more cards. $\endgroup$ – dreamin Oct 30 '14 at 18:17
  • $\begingroup$ Just to clear things up, are you trying to get 3 of ANY kind or 3 of a SPECIFIC kind? If you just want to make sure that you have 3 of ANY kind, you need 27 cards. The first thirteen cards ensure that, at a minimum, you have one of each. The next thirteen ensures that, at a minimum, you have two of each. The final and 27th card then ensures you have three of some kind. You may have MORE than 3 of a kind, but that's irrelevant. The goal is to determine the minimum necessary to ensure 3 of a kind. Now if you want a SPECIFIC 3 of a kind, that's a different story. $\endgroup$ – zahbaz Oct 30 '14 at 21:49
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    $\begingroup$ Thank you. That is what I was looking for. $\endgroup$ – dreamin Oct 30 '14 at 22:04
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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.

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  • $\begingroup$ to get three of kind (example: 3 kings) wouldn't one have to draw 42 cards? $\endgroup$ – dreamin Oct 30 '14 at 17:50
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    $\begingroup$ @jdreamin in this case you have 13 slots. $\endgroup$ – mookid Oct 30 '14 at 17:53

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