Randomly select two cards in sequence from a full deck of 52 cards, what i s the probability that the first one is a King given that the second one is a King. If someone can please help me with this question.
2026-04-01 14:01:46.1775052106
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Deck of Cards Stats Probability Question
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You might as well pick them in the other order and ask the chance the second is a King given that the first is a King. Given that the first is a King, you are drawing from a $51$ card deck which has ????
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What are the odds that the second card drawn is a king? We pick the second card first, and there is a $\frac{1}{13}$ chance that it is a king. Any card will do for the first card. So the total probability is $1.$
What is the probability that both cards are kings? This probability is simply $\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}.$
The conditional probability (that both cards are kings given the second is a king) then is $\frac{\frac{1}{221}}{\frac{1}{13}} = \boxed{\frac{1}{17}}.$
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By Bayes Theorem, we just need to compute the total probability the second card is a king, and then compute the fraction of that total that comes from the first card also being a king.
The total probability the second card is a king is 4/52 * 3/51 + 48/52 * 4/51. The first summand is prob(first = king) * prob(second = king | first = king). The second summand is prob(first card != king) * prob(second = king | first != king).
So the answer is (after noting we can eliminate common denominators in all these fractions) 4 * 3 / (4*3 + (48*4) = 3/51.
Not coincidentally, this is the same as the probability of the second card being king given that the first is king, due to the symmetry of the two situations. When you select two cards in order from the deck, each pair that arises has a mirror image pair. So the corresponding numbers we are concerned with are the same.
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You have one king in a known position, leaving 51 other places in the deck for the remaining 3 kings. Given that, what is the probability that one of those three kings is in a specific place?
That is: the favoured space is ways to select 2 places from 50 (and 1 from 1). The total space is ways to select 3 places from 51.
Divide and calculate.
EDIT: I misunderstood the question, corrected now.
The probability of drawing a king after you have already drawn one king is $\frac{3}{51}=\frac{1}{17}$, since there are only 3 kings and 51 cards left.