Question

A gin hand consists of 10 cards from a deck of 52 cards. Find the probability that a gin hand has

(a) all 10 cards of the same suit.
(b) exactly 4 cards in one suit and 3 in two other suits.
(c) a 4, 3, 2, 1, distribution of suits.

Answer 1

The probability of a gin hand having all 10 cards of the same suit is 4 * (13C10 / 52C10). The probability of a gin hand having exactly 4 cards in one suit and 3 cards in two other suits is (4 * (13C4) * (39C3)) / 52C10. The probability of a gin hand having a 4, 3, 2, 1, distribution of suits is (4 * (13C4) * (13C3) * (13C2) * (13C1)) / 52C10.

Step-by-step explanation:

(a) Probability of all 10 cards of the same suit:

There are 4 suits in a deck of cards. Since we want all 10 cards to be of the same suit, we can choose any one of the 4 suits. The number of ways to choose 10 cards of the same suit is 13C10 (combination).

Therefore, the probability is:

P(all 10 cards of the same suit) = 4 * (13C10 / 52C10)

(b) Probability of exactly 4 cards in one suit and 3 cards in two other suits:

There are 4 suits in a deck of cards. We can choose any one of the 4 suits for the 4 cards and any two of the remaining 3 suits for the 3 cards each. The number of ways to choose these cards is (13C4) * (39C3).

Therefore, the probability is:

P(exactly 4 cards in one suit and 3 cards in two other suits) = (4 * (13C4) * (39C3)) / 52C10

(c) Probability of a 4, 3, 2, 1, distribution of suits:

There are 4 suits in a deck of cards. We can choose any one of the 4 suits for the 4 cards, any one of the remaining 3 suits for the 3 cards, any one of the remaining 2 suits for the 2 cards, and any one of the remaining 1 suit for the remaining 1 card. The number of ways to choose these cards is (13C4) * (13C3) * (13C2) * (13C1).

Therefore, the probability is:

P(a 4, 3, 2, 1, distribution of suits) = (4 * (13C4) * (13C3) * (13C2) * (13C1)) / 52C10