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I believe that we are determining, in a worst case scenario, how many cards must we pick to draw a two different suit. My thought process is that we must get rid of all 13 cards to deplete an entire suit before we would be able to ensure the next card drawn is of a different suit. If this is so, then would the correct answer be to draw 14 cards(13 same suit and 1 different suit) to ensure a different suit?

Followup Question for Clarification

What is the probability that we will pick two cards with two different suits? I know we are able to pick any card for first one so I believe that means we have a 52/52 possibility for the first card. For the second card would we then have a 39/51 possibility (where 39 is the remaining number of cards that are not the suit)?

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    $\begingroup$ Correct on both accounts. $\endgroup$ – Mike Earnest Aug 9 '18 at 23:25
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    $\begingroup$ You may be interested to know that your argument for guaranteeing two different suits is called the Pigeonhole Princple. $\endgroup$ – Austin Mohr Aug 10 '18 at 4:52
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Your reasoning in both parts is perfectly fine. For the first problem we need to exhaust one suit completely to be sure that we'll draw a card from a different suit. As each suit has $13$ cards we need at least $13+1=14$ draws.

For the second part you could think like this: The first card is only required to determine the suit we don't want and hence the good combination would be drawing a card from those $3$ suits in the second draw. As you've noticed that probability is $\frac{39}{51}$

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since there are 13 cards in each suite to ensure that at least one is of a different suite you must draw 14 cards.

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  • $\begingroup$ This is half of the correct answer posted almost a year ago. You'd be better off spending time on new questions that don't already have good answers. $\endgroup$ – Ethan Bolker May 2 '19 at 13:08
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We ll take the worst case that we have selected 13 cards but they are all of same suit. 14th draw will guarantee the draw of two different suits. For the second part: Two cards randomly can be chosen in C(52,2) ways. But two suits can be selected in C(4,2 ) ways. And two cards of those selected suits can be chosen in C(13,1)×C(13,1) ways. So it becomes 39/51.

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