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Problem: Out of a pack of 52 playing cards, 3 cards are drawn one after the other without replacement. If a fourth card is drawn, what is the probability that it is a king?

One of my friends has given this as the solution.

`The probability that the 4th card is a king when the cards are drawn one by one without replacement is same as the probability that the fourth card obtained is a king when the 4 cards are drawn simultaneously.

This is the same as the probability that the first card drawn is a king when the four cards are drawn simultaneously, because when all the cards are drawn simultaneously, order doesn't matter. This probability is given by 4 C 1/ 52 C 1, hence the required probability is 4/52 = 1/13.`

But I don't think it is correct. Since chances of getting King in 4th trial will depend on the number of kings drawn in the first 3 draws? Am I correct?

So, solution would be

(4/52)(48/51)(47/50)(3/49) + (4/52)(3/51)(48/50)(2/49) + (4/52)(3/51)(2/50)*(1/49)

Is my solution correct? If not why not? If yes, then is there a faster way to get to the answer?

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    $\begingroup$ The probability is indeed $4/52$. The probability there is a king in fourth place, or $47$-th place, is the same as the probability there is a king in first place. $\endgroup$
    – André Nicolas
    Commented Sep 25, 2015 at 18:15
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    $\begingroup$ One can also do it the long way. There is a King in fourth place if we draw non-Kings $3$ times and then a King. Or if we draw one King and two non-Kings in the first three draws, and then a King. And so on. Calculate, add up, simplify. You will get (if there is no error) $1/13$. Good exercise, great incentive to solve the simple way. $\endgroup$
    – André Nicolas
    Commented Sep 25, 2015 at 18:19

5 Answers 5

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A priori the $k^{\rm th}$ card of the deck is a king with probability ${1\over13}$, whatever $k\in[52]$. Since we are not told the values of the first three cards drawn this probability is still the same, namely ${1\over13}$, at the moment we are asked the question about the fourth card.

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For the first 3 cards, there are 4 cases possible:

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CASE 1: If the first 3 cards doesn't contain any king, then $P_1 = \frac{48}{52}.\frac{47}{51}.\frac{46}{50}.\frac{4}{49}$

CASE 2: If the first 3 cards contain exactly 1 king, then $P_2 = 3.\frac{4}{52}.\frac{48}{51}\frac{47}{50}.\frac{3}{49}$

CASE 3: If the first 3 cards contain exactly 2 kings, then $P_3 = 3.\frac{4}{52}.\frac{3}{51}\frac{48}{50}.\frac{2}{49}$

CASE 4: If the all the first 3 cards are kings, then $P_4 = \frac{4}{52}.\frac{3}{51}.\frac{2}{50}.\frac{1}{49}$

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where $P_i$ is the probability of $4^{th}$ card being a King in $i^{th}$ case.

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$$P(Fourth\ card\ is\ King) = P_1 + P_2 + P_3 + P_4 = \frac{499800}{6497400} = \frac{4}{52} = 0.07692$$

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No number answer:

shuffle the cards, splay them out: Have your friend pull any card and re-insert it 4th from the top.

Now before you deal out the cards, ask yourself, "what is the probability it will be one of the four kings out of a deck of fifty-two" ?

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As kings are nor specially favoured, the probability that one comes fourth (or indeed anywhere in sequence) is the same as that for any other denomination. 13 denominations, total probability 1 (some card must appear), required probability 1/13.

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As Christian has answered, one way to think of this experiment would be to have you and your friend together pick all $4$ cards instantaneously. The fourth card is equally likely to be one of four kings from $52$ cards. Hence, the unconditional probability is $4/52$.

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