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A player is randomly dealt a sequence of 13 cards from a deck of 52-cards. All sequences of 13 cards are equally likely. In an equivalent model, the cards are chosen and dealt one at a time. When choosing a card, the dealer is equally likely to pick any of the cards that remain in the deck.

  1. If you dealt 13 cards, what is the probability that the 13th card is a King?

A) 1/52

B) 1/13

C) 1/26

D) 1/12

As per me, the answer should be as follows:

If 4 Kings are in first 12 cards , then P(13th card is King) = 0

If 3 Kings are in first 12 cards , then P(13th card is King) = 1/40

If 2 Kings are in first 12 cards , then P(13th card is King) = 2/40

If 1 Kings are in first 12 cards , then P(13th card is King) = 3/40

If 0 Kings are in first 12 cards , then P(13th card is King) = 4/40

Total = 0+(1/40) + (2/40) + (3/40) + (4/40) = 1/4

However, the given answer is 1/13

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    $\begingroup$ Your solution is incorrect because you don't consider the probability that you are in each of the given cases. But it's too much work. In a well shuffled deck, each card is equally likely to be in any slot. Thus the answer is $\frac 1{13}$. $\endgroup$
    – lulu
    Commented May 29, 2022 at 11:12
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    $\begingroup$ Note: common sense should tell you that $\frac 14$ is much too high. If you were asked about the $52^{nd}$ card, your method yields an answer $>1$. $\endgroup$
    – lulu
    Commented May 29, 2022 at 11:16

1 Answer 1

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The mistake in your solution lies in the fact that when you add up the probabilities that the King is in one of the cases you mentioned, you have to multiply (before adding them up) each of these probabilities by the probability of being in that specific case.

To be clearer: Consider the last of your cases, what is the probability that $0$ kings are in the first $12$ cards? It's$$C_0=(1-\frac{4}{52})(1-\frac{4}{51})\dots(1-\frac{4}{41})$$ So, in the "Total" you evaluated, the term for this case should have been $C_0 \frac{4}{40}$. If you correct all the other terms of the sum you end up with $\frac{1}{13}$

But there's a fastest way to solve the exercise. Because the solution of the problem is equivalent to the probability that the first card is a King. What is the probability that the first card is the King? Of course it's $\frac{4}{52}=\frac{1}{13}$

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