Question

In a particular card game, each player begins with a hand of 2 cards, and then draws 6 more. Calculate the probability that the hand will contain three of a kind

(3 cards of one value, with the other cards of 5 different values).

Answer #1

total cards drawn = 8

total number of cards = 52

same kind of cards in a deck = 4

total number of different cards = 13

total number of ways 8 cards can be drawn from 52 cards = 52C8

total number of ways 3 same cards ca be drawn from any 4 = 4C3

now, the hand needs 5 more cards, completely different from each other = 4C1*4C1*4C1*4C1*4C1

there are 6 different cards in hand which can be chosen 13C6 ways

number of ways (3 cards of one value, with the other cards of 5 different values can be chosen = 13C6*4C3*4C1*4C1*4C1*4C1*4C1

P[ hand will contain 3 cards of one value, with the other cards of 5 different values ] = 13C6*4C3*4C1*4C1*4C1*4C1*4C1 / 52C8 = 4^6*1716 / 33111678600 = 2.1227 * 10^-4 = 0.00021

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