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Assume you pick cards from a deck of cards at random without replacement (i.e. don't put it back in the deck once you pick it). How many cards must be picked to guarantee you have 10 cards of the same suit?

Standard deck of 52 cards. 13 cards in each suit. The four suits are Clubs, Hearts, Diamonds, and Spades.

I believe that this is a example of the pigeon hole principle and I want some assurance that I am doing this correctly

would the answer be:

$4(10 - 1) + 1 = 37$ cards

because there are four different suits and I have to have 10 cards of the same suit?

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Yes, that's right. You can't guarantee it with fewer, since if you only take $36$ cards you could get $9$ in each suit.

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I agree with Especially Lime here, but I have a different way to think about it.

Pick a random suit in the beginning like, Diamonds. Then pick a suit different from diamonds, like Clubs. Then pick Hearts. Finally, pick Spades. Keep repeating this process until you get $10$ of some suit. This is the guaranteeing number because it takes so long to get. You will realize by doing this process, you will get $10, 9, 9, 9$ for each of the suits. So, it will be $37$ as David smith said. In general, for $n$ cards of the same suit, you will need $4n - 3$ cards to guarantee $n$ cards of the same suit.

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  • $\begingroup$ Where picking deliberately a suit's card became correct? It has to be random. Your view is confusing, might be stated it wrong. $\endgroup$ – jiten Nov 29 '20 at 0:31

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