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I've found it doing a simulation on R, but I can't manage to find it mathematically.

StubbornAtom
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MNM
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    As an aside, I worry that you might have gone through the effort of permuting all 52 cards in your simulation, which winds up being highly inefficient. This question is no different than the question of "Two cards are drawn from a standard shuffled pack of playing cards. What is the probability that they are both the same suit?" The fact that one card is on the top and one on the bottom of the deck rather than them being the top two cards is irrelevant. – JMoravitz Mar 30 '20 at 17:12
  • I certainly agree with your comment, but then I miss the point of the assignment then. This is really strange... Thanks for your help! – MNM Mar 30 '20 at 17:15
  • The point of the assignment? To get away from thinking that some collection of cards from a randomly shuffled deck will have a different distribution than some other collection of cards from a randomly shuffled deck... For example... the probability that the second card is a queen is exactly the same as the probability that the first card is a queen... which is further exactly the same as the probability that the last card is a queen. – JMoravitz Mar 30 '20 at 17:29
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    All the cards in the middle of the deck are irrelevant and can be completely ignored as we proceed with the calculations. This is an important concept to get across... Several people get bogged down with trying to do these highly tedious calculations taking things which are irrelevant into account when there are much easier more straightforward approaches. – JMoravitz Mar 30 '20 at 17:30
  • Thank you so much for the clarification, hope you have a great day! – MNM Mar 30 '20 at 17:33

2 Answers2

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The top card will be whatever it happens to be... it doesn't matter which.

The bottom card is equally likely to be any of the remaining $51$ cards which are not the top card. $12$ of which will be the same suit as the top card.

The probability is then:

$$\dfrac{12}{51} = \dfrac{4}{17}$$

JMoravitz
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The top card is of some suit. There are 12 remaining cards of that suit, out of 51. So the probability that the bottom card (indeed any single card) is of that suit is $12/51$.

David G. Stork
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