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Five cards are drawn one by one from a standard deck of $52$ cards. What is the probability of drawing a king immediately after an ace?

The number of ways for taking $5$ cards, one by one, from a deck of $52$ is $52.51.50.49.48$

The number of ways for taking the Ace of spades and, immediately after, the king of spades, is $50.49.48.4$?? If this is correct, how to generalize for any ace and king??

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    $\begingroup$ This question is asking about sequences of draws, so your sample space is permutations of five cards rather than combinations of five cards. $\endgroup$ – N. F. Taussig May 4 '18 at 22:49
  • $\begingroup$ Thank you. Now I am trying something like $\frac{50.49.48.4.16}{52.51.50.49.48}$ Because there are 16 ways of arranging A and K as its said. But I am not sure $\endgroup$ – GEW6 May 5 '18 at 1:46
  • $\begingroup$ In your attempts, you are counting sequences with two ace-king pairs twice. $\endgroup$ – N. F. Taussig May 5 '18 at 13:41
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The number of sequences of five cards is $$P(52, 5) = 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48$$ If an ace immediately precedes a king, there are four positions in the ace-king subsequence can begin, four possible suits for the ace, four possible suits for the king, and $50 \cdot 49 \cdot 48$ ways of selecting the remaining cards in the hand, which gives $$4 \cdot 4 \cdot 4 \cdot 50 \cdot 49 \cdot 48$$ However, we have counted hands with two places in which an ace immediately precedes a king twice, once for each way of designating each such subsequence as the one in which an ace immediately precedes a king. We only want to count such hands once, so we must subtract them from the total.

If there are two subsequences in which a king appears immediately after an ace, there are three objects to arrange, the two ace-king subsequences and the other card. There are $\binom{3}{2}$ ways to choose the positions of the ace-king subsequences, four ways to choose the suit of the ace and four ways to choose the suit of the king in the first such subsequence, three ways to choose one of the remaining aces and three ways to choose one of the remaining kings for the second such subsequence, and $48$ ways to choose the other card. Hence, there are $$\binom{3}{2} \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 48$$ such hands.

Therefore, the probability of drawing a king immediately after an ace among five cards drawn from a standard deck is $$\frac{4 \cdot 4 \cdot 4 \cdot 50 \cdot 49 \cdot 48 - \binom{3}{2} \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 48}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}$$

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