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COMBINATORICS Example 15 EX: Calculate the odds (or probabilities) of the following 5-card poker hands: a) royal flush b) four-of-a-kind c) straight-flush (excluding royal flush) d) full house e) flush (excluding straight-flush) f) straight (excluding straight-flush) g) three-of-a-kind h) two-pair i) one-pair j) high card Also, verify that the probabilities sum to unity. Assume a 52-card deck. (The number of possible 5-card hands is 52 C 5 = 2,598,960.) SOL'N: a) A royal flush is ace, king, queen, jack, and ten of the same suit. If we order the 5-card hand from highest card to lowest, the first card will be an ace. There are four possible suits for the ace. After that, the other four cards are completely determined. Thus, there are 4 possible royal flushes:
royal flushes = 4 C 1 = 4
Dividing by the number of possible hands gives the probability: € P (royal flush) =
− 6 or 1 in 649, b) A straight-flush (excluding royal flush) is all cards the same suit and showing consecutive numbers (but not the highest 5 consecutive numbers).
COMBINATORICS Example 15 (cont.) If we order the 5-card hand from highest number to lowest, the first card may be one of the following: king, queen, jack, 10, 9, 8, 7, 6, or 5. (Note: the ace may be the card above a king or below a 2, but we would have a royal flush if it were the card above the king.) There are 9 possibilities. After the first card, whose suit we may choose in 4 ways, the remaining cards are completely determined.
straight flushes = 9 4 C 1 = 9·4 = 36
Subtracting the number of royal flushes and dividing by the number of possible hands gives the probability: € P (straight - flush) =
= 1.385 ⋅ 10 −^5 or 1 in €
3
c) A four-of-a-kind is four cards showing the same number plus any other card. If we order the 5-card hand with the four-of-a-kind first, we have 13 C 1 choices for the number showing on the first four cards. Since we will have all four suits, we have only 4 C 4 = 1 way to choose the suits. The remaining card will be any of the 48 remaining cards:
4-of-a-kinds = 13 C 1 · 4 C 4 · 48 C 1 = 13·1·48 = 624
Dividing by the number of possible hands gives the probability: € P ( 4 - of - a - kind) =
= 2.401⋅ 10 −^4 or 1 in 4165 d) A full house is three cards showing the same number plus a pair. If we order the 5-card hand with the three-of-a-kind first, we have 13 C 1 choices for the number showing on the first three cards. For the choice of suits, we have three out of four possible suits or 4 C 3 = 4 possibilities. For the remaining pair, we have 12 C 1 choices for the number showing on the two cards. For the choice of suits, we have two out of four possible suits or 4 C 2 = 6 possibilities.
COMBINATORICS Example 15 (cont.) € P (straight not flush) =
= 3.925 ⋅ 10 −^3 or 1 in €
g) A three-of-a-kind is three cards showing the same number plus two cards that do not form a pair or create a four-of-a-kind. If we order the 5-card hand with the three-of-a-kind first, we have 13 C 1 choices for the number showing on the first three cards. Since we will have three out of four suits, we have 4 C 3 = 4 ways to choose the suits. The remaining two cards must show different numbers than the four-of-a- kind and each other. There are 12 C 2 choices for these numbers. The last two cards may have any of the four suits, however.
3-of-a-kind = 13 C 1 · 4 C 3 · 12 C 2 · 4 C 1 · 4 C 1 = 13·4·66·4·4 = 54,
Dividing by the number of possible hands gives the probability: € P ( 3 - of - a - kind) =
= 2.113⋅ 10 −^2 or 1 in 47.3. h) Two-pairs is two cards showing the same numbers and another two cards showing the same numbers (but not all four numbers the same) plus one extra card (not the same as any of the other numbers). If we order the 5-card hand with the two pairs first, we have 13 C 2 choices for the two numbers showing on the two pairs. Each pair will have two out of four suits. Thus, we have 4 C 2 · 4 C 2 = 6·6 = 36 ways to choose the suits. The remaining card must show a different number than the two pairs. There are 11 C 1 choices for this number. The last card may have any of four suits.
2-pair = 13 C 2 · 4 C 2 · 4 C 2 · 11 C 1 · 4 C 1 = 78·6·6·11·4 = 123,
Dividing by the number of possible hands gives the probability: € P (2 - pair) =
= 4.754 ⋅ 10 −^2 or 1 in 21.
COMBINATORICS Example 15 (cont.) i) One-pair is two cards showing the same numbers and another three cards all showing different numbers. If we order the 5-card hand with the pair first, we have 13 C 1 choices for the number showing on the pair. The pair will have two out of four suits. Thus, we have 4 C 2 = 6 ways to choose the suit. The remaining three cards must show different numbers than the pair and each other. There are 12 C 3 choices for these numbers. The last three cards may each have any of four suits.
1-pair = 13 C 1 · 4 C 2 · 12 C 3 · 4 C 1 · 4 C 1 · 4 C 1 = 13·6·220·4·4·4 = 1,098,
Dividing by the number of possible hands gives the probability: € P ( 1 - pair) =
= 4.226 ⋅ 10 −^1 or 1 in 2.37. j) High card means we must avoid higher-ranking hands. All higher-ranked hands include a pair, a straight, or a flush. We now count the number of possible high-card hands Because the numbers showing on the cards must be five different numbers, we have 13 C 5 choices for the five numbers showing on the cards. Each of the cards may have any of four suits. 13 C 5 · 4 C 1 · 4 C 1 · 4 C 1 · 4 C 1 · 4 C 1 = 1287·4^5 = 1,317, We subtract the number of straights, flushes, and royal flushes. (Note that we avoided having any pairs or more of a kind.) Dividing the difference just calculated by the number of possible hands gives the probability: € P (high - card) =
or € P (high - card) = 0.501 or 1 in 2. REF: Probability: 5-card Poker Hands , Tom Ramsey, http://www.math.hawaii.edu/~ramsey/Probability/PokerHands.html, Feb. 1, 2008.
