Question

Problem 6. (20 pts) By definition, a poker hand is a set of 5 cards from a standard French deck of 52 cards1. Solve the following two problems (a) A poker hand is said to be a four of a kind if it has four cards with the same value. For instance, four sevens, four Queens or four Aces. Compute the prob- ability of drawing a four of a kind hand. (b) A poker hand is said to be a full house if it has three cards with the one value, and two cards with a second value. For instance, three sevens and two Queens., or three Aces and two fives. Compute the probability of drawing a full house. Conclude why in the game of poker, a four of a kind beats a full house

Answer 1

Number of ways of selecting 5 cards out of 52 cards is

$\binom{52}{5}=2598960$

(a)

There are total 13 denominations and each denomination has 4 cards. So number of ways of selecting 1 denominations and then 4 cards out of 4 is

$\binom{13}{1}\binom{4}{4}=13$
And since we need 4 of same kind so remaining 1 card must come from different denomination so number of ways of selecting 1 denominations out of remaining 12 denominations and then 1 card from selected denomination is

$\binom{12}{1}\binom{4}{1}=48$
So number of ways of selecting four of a kind is :

$\binom{13}{1}\binom{4}{4}\binom{12}{1}\binom{4}{1}=13\cdot 48=624$

The required probability is

P(4 of a kind) = 624 / 2598960  = 0.0002401

Answer: 0.0002401

(b)

Number of ways of selecting 1 denominations out of 13 is C(13,1). Number of ways of selecting 3 cards out of 4 cards of selected denomination is C(4,3). And then select one denomination out of remaining 12 denominations is C(12,1) and then 2 cards from each selected denominations is C(4,2). So number of ways are there to draw a 5 card poker hand that contains 3 a kind and 2 a kind, that is full house, is

C(13,1)C(4,3)C(12,1)C(4,2) = 3744 ways

So  the probability of drawing a 5 card poker hand that contains a full house is

3744/ / 2598960 =0.0014

Answer: 0.0014