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Let \(A\) be the set of five-card hands dealt from a fifty-two-card poker deck, where the denominations of the five cards are all consecutive - for example, (7 of hearts, 8 of spades, 9 of spades, 10 of hearts, jack of diamonds). Let \(B\) be the set of five-card hands where the suits of the five cards are all the same. How many outcomes are in the event \(A \cap B\) ?

Short Answer

Expert verified
There are 36 outcomes in the event \(A \cap B\).

Step by step solution

01

Identify the possible denominations

There are 13 possible denominations in a suit (from Ace to King). A consecutive sequence of five yields 9 possible sequences, as follows: A-2-3-4-5, 2-3-4-5-6, 3-4-5-6-7, 4-5-6-7-8, 5-6-7-8-9, 6-7-8-9-10, 7-8-9-10-J, 8-9-10-J-Q, and 9-10-J-Q-K.
02

Identify the possible suits

There are four possible suits: hearts, diamonds, clubs, and spades.
03

Compute for the intersection of A and B

The intersection of A and B is a hand that satisfies both conditions: i.e., the five cards are of the same suit and have consecutive denominations. Multiplying the number of possible sequences by the number of suits will give us the total number of combinations that satisfy both conditions A and B: \(9 \times 4 = 36\).

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Key Concepts

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

Combinatorics in Poker Hands
Combinatorics is at the heart of calculating probabilities in games like poker. It's a branch of mathematics that deals with counting, both as an abstract concept and in terms of calculating possibilities in various scenarios. When dealing with poker hands, we're often interested in understanding the number of ways in which certain types of hands can be formed from a standard deck of 52 cards.

In our exercise, we focused on finding how many five-card hands can be made that are both a sequence of consecutive denominations and are all from the same suit. We use combinatorics to count these possibilities. First, we identify the total number of possible consecutive sequences. There are normally 10 such sequences, but due to the Ace being able to act at the beginning of a sequence (A-2-3-4-5) but not at the end (10-J-Q-K-A), we only have 9 valid sequences.

Next, we consider the suits. Since a standard deck has four suits, and each sequence can be of any suit, the principle of multiplication kicks in. This foundational concept of combinatorics tells us to multiply the number of sequences by the number of suits to obtain the total number of distinct hands that meet the criteria. Thus, we arrive at the solution using the formula
\[9 \text{ sequences} \times 4 \text{ suits} = 36 \text{ distinct hands}.\]
Understanding combinatorics is essential for questions involving the selection or arrangement of objects (like poker cards), where order and grouping play significant roles.
Set Theory in Probability
Set theory is the foundation of probability, providing us with a language to describe collections of outcomes and events. In probability, an 'event' is a set of outcomes from a larger 'sample space,' which represents all possible outcomes.

In our poker hand problem, we define two events:
  • Event \(A\), which includes hands with consecutive denominations.
  • Event \(B\), which includes hands where all cards share the same suit.
The task is to find the probability of the intersection of these two events, denoted as \(A \cap B\). In set theory terms, this intersection represents the set of all hands that are both consecutive and of the same suit.

Understanding set theory allows us to see that the probability of this intersection can be computed by considering each set independently first and then determining how they overlap. In probability terms, this overlap represents the hands that satisfy both conditions simultaneously. Therefore, by using set theory concepts, we can work out the number of hands that are in both set \(A\) and set \(B\), which in our case is 36.
Card Sequence Probability
Card sequence probability specifically refers to the likelihood of drawing a sequence of cards in a certain order. It's a vital aspect of determining outcomes in games like poker. Unlike generic probability problems where each event is independent, in card sequence probability, each card drawn affects the probability of drawing the next card.

However, for the purpose of our exercise, we do not need to consider the order in which the cards come out, because a hand in poker is a combination of cards, not a permutation. Therefore, the probability focuses on the specific combination (unconcerned with order) of both the sequence and the suit.

In more complicated scenarios, calculating the exact probability of drawing a specific sequence would require us to take into account the changing probabilities with each card dealt. This is where understanding the difference between combinations and permutations becomes crucial, the former being relevant to our task. In essence, the probability of getting a straight flush (a sequence of five cards in the same suit) can be deduced using the combinatorial calculations discussed, reinforcing the concept that poker probabilities are multifaceted but understandable with the right mathematical tools.

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Most popular questions from this chapter

Your favorite college football team has had a good season so far but they need to win at least two of their last four games to qualify for a New Year's Day bowl bid. Oddsmakers estimate the team's probabilities of winning each of their last four games to be \(0.60,0.50,0.40\), and \(0.70\), respectively. (a) What are the chances that you will get to watch your team play on Jan. 1 ? (b) Is the probability that your team wins all four games given that they have won at least three games equal to the probability that they win the fourth game? Explain. (c) Is the probability that your team wins all four games given that they won the first three games equal to the probability that they win the fourth game?

An urn contains one white chip and a second chip that is equally likely to be white or black. A chip is drawn at random and returned to the urn. Then a second chip is drawn. What is the probability that a white appears on the second draw given that a white appeared on the first draw? (Hint: Let \(W_{i}\) be the event that a white chip is selected on the \(i\) th draw, \(i=1,2\). Then \(P\left(W_{2} \mid W_{1}\right)=\frac{P\left(W_{1} \cap W_{2}\right)}{P\left(W_{1}\right)}\). If both chips in the urn are white, \(P\left(W_{1}\right)=1\); otherwise, \(\left.P\left(W_{1}\right)=\frac{1}{2} .\right)\)

An urn contains \(w\) white chips, \(b\) black chips, and \(r\) red chips. The chips are drawn out at random, one at a time, with replacement. What is the probability that a white appears before a red?

The Alpha Beta Zeta sorority is trying to fill a pledge class of nine new members during fall rush. Among the twenty-five available candidates, fifteen have been judged marginally acceptable and ten highly desirable. How many ways can the pledge class be chosen to give a two-to-one ratio of highly desirable to marginally acceptable candidates?

Urn I contains three red chips and five white chips; urn II contains four reds and four whites; urn III contains five reds and three whites. One urn is chosen at random and one chip is drawn from that urn. Given that the chip drawn was red, what is the probability that III was the urn sampled?

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