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A poker hand is a set of 5 cards randomly chosen from a deck of 52 cards. Find the probability of a (a) royal flush (ten, jack, queen, king, ace in a single suit). (b) straight flush (five in a sequence in a single suit, but not a royal flush). (c) four of a kind (four cards of the same face value). (d) full house (one pair and one triple, each of the same face value). (e) flush (five cards in a single suit but not a straight or royal flush). (f) straight (five cards in a sequence, not all the same suit). (Note that in straights, an ace counts high or low.)

Short Answer

Expert verified
(a) \( 1/649,740 \), (b) \( 36/2,598,960 \), (c) \( 624/2,598,960 \), (d) \( 3,744/2,598,960 \), (e) \( 5,108/2,598,960 \), (f) \( 10,200/2,598,960 \).

Step by step solution

01

Calculate Total Possible Hands

A poker hand consists of 5 cards drawn from a standard deck of 52 cards. The total number of possible 5-card combinations is given by the combination formula \( \binom{52}{5} = 2,598,960 \). This is the sample space for all poker hands.
02

Probability of a Royal Flush

A royal flush consists of a specific sequence (ten, jack, queen, king, ace) all in the same suit. There are 4 suits, so there are 4 possible royal flush hands. The probability is therefore \( \frac{4}{2,598,960} = \frac{1}{649,740} \).
03

Probability of a Straight Flush

A straight flush does not include royal flushes but is a sequence of 5 cards in the same suit. Each suit has 9 straight flushes (Ace-2-3-4-5 through 9-10-J-Q-K, excluding the Royal Flush). Thus, there are \( 36 \) straight flushes (\( 4 \times 9 \)). The probability is \( \frac{36}{2,598,960} \).
04

Probability of Four of a Kind

Four of a kind consists of four cards of the same rank and one additional card. There are 13 ranks and \( \binom{48}{1} \) possible choices for the fifth card. Thus, there are \( 13 \times 48 = 624 \) four of a kind hands. The probability is \( \frac{624}{2,598,960} \).
05

Probability of a Full House

A full house is a hand with three cards of one rank and two of another. There are \( 13 \) ways to choose the rank for the triple (\( 13 \times \binom{4}{3} \)), and 12 remaining ranks for the pair (\( \times 12 \times \binom{4}{2} \)). Total = \( 13 \times 4 \times 12 \times 6 = 3,744 \). The probability is \( \frac{3,744}{2,598,960} \).
06

Probability of a Flush

A flush is any five card hand of the same suit excluding straight flushes. For one suit: \( \binom{13}{5} \) minus 10 straights (3 real straights and 1 royal for each suit). Total ways for a flush: \( 4 \times (\binom{13}{5} - 10) = 5,108 \). Probability is \( \frac{5,108}{2,598,960} \).
07

Probability of a Straight

A straight is five cards in sequence, not all the same suit. Total straight sequences (Ace low or high): 10. For each, can't all be one suit: different suit = \( 4^5 - 4 \). Total straights:\( 10 \times (\binom{4}{1}(5)^5 - 4) = 10,200 \). Probability is \( \frac{10,200}{2,598,960} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poker Hand Probabilities
Poker hand probabilities refer to the likelihood of being dealt a specific type of hand from a standard deck of 52 cards. Each hand consists of 5 cards. To calculate these probabilities, it's vital to understand both the number of ways to create the desired hand and the total number of possible 5-card combinations. For a standard deck, this number is calculated using the combination formula \( \binom{52}{5} = 2,598,960 \). This means there are 2,598,960 different possible hands you could draw.

Poker hands are classified into categories based on the cards' combinations—for instance, royal flush, straight flush, four of a kind, and others. Each poker hand category has a different probability based on how exclusive or common the hand type is.
  • A royal flush is the most prestigious, comprising a 10, Jack, Queen, King, and Ace all in the same suit. This leads to only 4 possible combinations in the whole deck.
  • A straight flush involves any five consecutive cards of the same suit, except for a royal flush, resulting in a slightly more common occurrence of 36 combinations.
  • Other hands, like four of a kind or full houses, arise from distinct combinations and have their own specific probabilities. The probability of each hand decreases as the hands become more specific and harder to achieve.
Combinatorics
Combinatorics is the branch of mathematics dealing with combinations, permutations, and counting techniques. These principles are essential in computing poker hand probabilities. In card games, combinatorial formulas help determine how to select and arrange cards to form specific hands.

The primary formula used in calculating poker hands is the combination formula, denoted as \( \binom{n}{r} \), which represents the number of ways to choose \( r \) cards from \( n \) cards without regard to the order of selection. For the poker problem, the formula \( \binom{52}{5} \) helps determine the total number of poker hands possible.

Combinatorial strategies are also used to find specific hand types. For instance, determining combinations for a full house requires selecting ranks and combinations of suits through several iterative steps. It involves choosing ranks for triples and pairs separately, which makes use of the multiplication principle in combinatorics. Understanding these fundamental approaches allows players to assess probabilities more effectively during the game.
Card Games Probability
Card games probability is the study of predicting the likelihood of events within card games, informed by a player's hand, the deck, and any communal cards. In poker, evaluating probabilities can significantly affect decision-making and strategy.

Card game probabilities hinge on calculating how a particular hand type can occur given the constraints of the deck and the rules of the game. This complex process uses the principles of combinatorics to calculate the total possible hands and the count of a specific hand type.
  • For example, a flush probability involves excluding straight flush combinations and observing only suits.
  • Similarly, four of a kind requires knowing the count of four matching cards, then choosing an additional distinct card.
Effective probability assessment in poker empowers a player to better anticipate potential outcomes and adjust their strategy accordingly. By recognizing which hands are more likely, a player can make more informed decisions about betting and folding.
Mathematical Problem Solving
Mathematical problem solving in the context of poker hands involves applying systematic approaches to calculate the likelihood of various hands. Solving these problems requires several key steps.

First, outline the problem by identifying what probability to find. Next, use combinatorial principles to determine the number of ways specific hands can be drawn. This often involves breaking down the hand into parts, such as choosing ranks or suits, and ensuring consideration of all possible combinations.

Lastly, to finalize the probability, divide the number of successful outcomes by the total number of possible outcomes, represented by the total combinations of poker hands. This step-by-step method is not only crucial in solving poker probability problems but also applies to a wide range of mathematical challenges.

Throughout the process, careful planning and execution of mathematical principles lead to accurate solutions, demonstrating the impact of solving techniques on understanding complex probability scenarios. These strategies provide a solid foundation for developing strategic thought in poker and other areas where probabilistic evaluations are critical.

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Most popular questions from this chapter

Given any ordering \(\sigma\) of \(\\{1,2, \ldots, n\\},\) we can define \(\sigma^{-1},\) the inverse ordering of \(\sigma,\) to be the ordering in which the \(i\) th element is the position occupied by \(i\) in \(\sigma\). For example, if \(\sigma=(1,3,5,2,4,7,6)\), then \(\sigma^{-1}=(1,4,2,5,3,7,6)\). (If one thinks of these orderings as permutations, then \(\sigma^{-1}\) is the inverse of \(\left.\sigma .\right)\) A fall occurs between two positions in an ordering if the left position is occupied by a larger number than the right position. It will be convenient to say that every ordering has a fall after the last position. In the above example, \(\sigma^{-1}\) has four falls. They occur after the second, fourth, sixth, and seventh positions. Prove that the number of rising sequences in an ordering \(\sigma\) equals the number of falls in \(\sigma^{-1}\).

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Baumgartner, Prosser, and Crowell are grading a calculus exam. There is a true-false question with ten parts. Baumgartner notices that one student has only two out of the ten correct and remarks, "The student was not even bright enough to have flipped a coin to determine his answers." "Not so clear," says Prosser. "With 340 students I bet that if they all flipped coins to determine their answers there would be at least one exam with two or fewer answers correct." Crowell says, "I'm with Prosser. In fact, I bet that we should expect at least one exam in which no answer is correct if everyone is just guessing." Who is right in all of this?

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