[Math] Probability of drawing a king of hearts from 8 cards

probability

We have a well-shuffled deck of 52 cards and the top eight cards are turned over.

a. What is the probability that one of the eight cards is a king of hearts?

b. A second deck is shuffled and its top eight cards are turned over. What is the probability that a visible card from the first deck matches a visible card from the second deck?

I'd like to say that I have no background in probability at all so I'm pretty lost on this. But I have some thoughts. For (a), the probability that the first card is a king of hearts is 1/52. I am guessing that we don't just multiply 1/52, 1/51, 1/50, etc. but I don't know the procedure. Do we assume that the first card being the King of hearts is independent from the second card being a king of hearts?

Best Answer

a) Any one of the fifty-two cards in the shuffled deck is equally likely to be the king of hearts, and there are eight cards showing.   So the probability that the $\rm K\heartsuit$ is one of the cards showing is:

$$\dfrac 8{52}$$

b) Remember that $\binom nk$ counts the ways to select $k$ items from a set of $n$. $$\dbinom nk~=~\dfrac{n!}{k!(n-k)!}~=~{}^n\mathrm C_k$$

We want the probability for not selecting all 8 cards from the 44 cards in the second deck that are not those showing in the first deck, when selecting 8 from 52 cards.   Using the rule of complementary probability, the probability for at least one match among the cards showing in both decks is:

$$1-\dfrac{\binom{44}{8}}{\binom{52}{8}}$$

Or we can take the long route. The probability that the first card showing in the first deck is not showing in the second deck is $\tfrac{44}{52}$. The conditional probability that the second card showing in the first deck is not showing in the second deck given that the first card of the first deck is not either is $\tfrac {43}{51}$.   And so on; the probability that a particular card is not showing in the second deck given that $k$ other cards are also not showing is $\tfrac{44-k}{52-k}$.   We put these iterated conditional probabilities together via multiplication, giving:

$$1-\dfrac{44\cdot 43\cdot 42\cdot 41\cdot 40\cdot 39\cdot 38\cdot 37}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47\cdot 46\cdot 45}$$

Which is the same thing.

Rule of Complementary Probability: The probability of an event equals one minus the probability of its complement.

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