Counting Il 5 Five-Card Draw is a card game in which each player is initially dealt a hand, a subset of 5 cards ( Then the game gets complicated, but let's not worry about that. The number of different hands in Five-Card Draw is the number of 5-element subsets of a 52-element set, which is 52 choose 5: total#of hands=/52\ 2.598.960 Let's get some counting practice by working out the number of hands with various special properties 3.1 Hands with a Four-of-a-Kind A Four-of-a-Kind is a set of four cards with the same value. How many different hands contain a Four-of-a-Kind? Here a couple examples: 8·,8◇,Q,8 As usual, the first step is to map this question to a sequence-counting problem. A hand with a Four-of-a-Kind is completely described by a sequence specifying 1. The value of the four cards 2. The value of the extra card 3. The suit of the extra card Thus, there is a bijection between hands with a Four-of-a-Kind and sequences consist- ing of two distinct values followed by a suit. For example, the three hands above are associated with the following sequences (8,Q,9)→{8h,8◇,89 (2,A,4)…{2,2,2◇,24,A·} Now we need only count the sequences. There are 13 ways to choose the first value, 12 ways to choose the second value, and 4 ways to choose the suit. Thus, by the generalized Product Rule, there are 13. 12.4= 624 hands with a Four-of-a-Kind. This means that only 1 hand in about 4165 has a Four-of-a-Kind; not surprisingly, this is considered a very� � Counting III 5 Five­Card Draw is a card game in which each player is initially dealt a hand, a subset of 5 cards. (Then the game gets complicated, but let’s not worry about that.) The number of different hands in Five­Card Draw is the number of 5­element subsets of a 52­element set, which is 52 choose 5: 52 total # of hands = = 2, 598, 960 5 Let’s get some counting practice by working out the number of hands with various special properties. 3.1 Hands with a Four­of­a­Kind A Four­of­a­Kind is a set of four cards with the same value. How many different hands contain a Four­of­a­Kind? Here a couple examples: { 8♠, 8♦, Q♥, 8♥, 8♣ } { A♣, 2♣, 2♥, 2♦, 2♠ } As usual, the first step is to map this question to a sequence­counting problem. A hand with a Four­of­a­Kind is completely described by a sequence specifying: 1. The value of the four cards. 2. The value of the extra card. 3. The suit of the extra card. Thus, there is a bijection between hands with a Four­of­a­Kind and sequences consist￾ing of two distinct values followed by a suit. For example, the three hands above are associated with the following sequences: (8, Q, ♥) ↔ 8♠, 8♦, 8♥, 8♣, Q♥ (2, A, ♣) ↔ { 2♣, 2♥, 2♦, 2♠, A♣ } { } Now we need only count the sequences. There are 13 ways to choose the first value, 12 ways to choose the second value, and 4 ways to choose the suit. Thus, by the Generalized Product Rule, there are 13 · 12 · 4 = 624 hands with a Four­of­a­Kind. This means that only 1 hand in about 4165 has a Four­of­a­Kind; not surprisingly, this is considered a very good poker hand!