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Chapter 2: Problem 43

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

Short Answer

Expert verified
Straight with 10 high: 0.0394%; Any straight: 0.392%; Straight flush: 0.0015%.

Step by step solution

01

Understand Poker Straight

A straight in poker consists of 5 consecutive cards of any suit. High card 10 indicates the straight includes cards like 6-7-8-9-10, with 10 as the highest value card.
02

Count the Possible Straights with High Card 10

For a straight with high card 10, possible sequences are: 6-7-8-9-10, each offering 4^5 (since there are 4 suits for each card) combinations. Therefore, there are \[4 \times 4 \times 4 \times 4 \times 4 = 1024\] possible hands for a straight with a high card 10.
03

Calculate Total Possible Poker Hands

The total number of 5-card combinations from a deck of 52 cards is \[ \binom{52}{5} = 2,598,960 \] hands.
04

Calculate Probability of a Straight with High Card 10

The probability of a straight having 10 as the high card is calculated by dividing the number of favorable outcomes by the total possible hands: \[ \frac{1024}{2,598,960}\approx 0.000394 \] or about 0.0394%.
05

Calculate Total Possible Straights

A straight can start with an ace, 2, 3, 4, 5 up to a 10 high (10 being the highest start). For each, there are 4^5 possible combinations. Therefore: \[ 10 \times 1024 = 10,240 \] total possible straights.
06

Calculate Probability of Any Straight

The probability of being dealt any straight is: \[ \frac{10,240}{2,598,960} \approx 0.00392 \] or 0.392%.
07

Calculate Probability of a Straight Flush

A straight flush is both a straight and a flush. For each starting card, there is only 1 way to have all cards in the same suit, hence 10 total possible straight flushes. \[ \frac{40}{2,598,960} \approx 0.000015 \] or 0.0015%.

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Key Concepts

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

Poker Probability
Poker is a popular card game that not only requires skill, but also a good grasp of probability. When playing five-card poker, understanding the probability of being dealt specific hands can give you an advantage. Probability, in this context, measures the likelihood of receiving a certain combination of cards from a standard 52-card deck.
A standard deck includes cards from four suits: clubs, diamonds, hearts, and spades, and each suit contains cards from Ace to King (13 in total). The concept of probability is used to calculate the chances of being dealt a certain hand, such as a "straight," which is a sequence of consecutive card ranks, like 5-6-7-8-9, regardless of the suit.
To determine the probability of obtaining a certain poker hand, you divide the number of possible favorable hands by the total number of possible five-card poker hands you can form from the deck. This total number is calculated using combinations, symbolized as \( \binom{52}{5} \), which represents how to choose 5 cards out of 52. This results in 2,598,960 different possible hands.
  • Learning and applying these calculations helps in making strategic decisions during gameplay.
  • Probability offers a mathematical approach to estimating which hands are likely more or less advantageous.
Card Combinations
Card combinations are at the heart of poker strategy. Each poker hand is a combination of 5 cards, and each combination has a certain rank, with "straight" being one of them.
To calculate possible combinations, we use the concept of mathematical combinations. For example, when considering a straight in poker, specifically a straight with high card 10 (like 6-7-8-9-10), each card can be of any suit. Because there are four suits, there are \(4^5 = 1024\) ways to arrange a specific straight.
It's crucial to understand that while 10 different starting points exist for a straight (starting from Ace to 10 high straight), each straight sequence has multiple combinations due to the suit variability.
  • The total possible straight combinations are \(10 \times 1024 = 10,240\). This includes all potential straights beginning with any rank from Ace to 10.
  • Combinatorial principles help in counting the number of ways a specific hand can be achieved.
Understanding card combinations allows better estimation of certain hands, aiding both poker playing and probability calculations.
Straight Flush Probability
A straight flush is a particularly rare and valuable hand in poker, which combines the elements of a straight and a flush. A straight flush means five consecutive cards of the same suit.
Unlike calculating a regular straight's probability, where each card can be any of four suits, a straight flush restricts all cards to just one. This significantly lowers the number of possible combinations.
For example, with a straight flush starting at any card, there is precisely one possible combination per suit. Multiplying this by four suits results in just 40 possible straight flush hands from a total of over 2.5 million possible hands. Consequently, the probability of being dealt a straight flush is \( \frac{40}{2,598,960} \), approximately 0.0015%, making it an exceptional hand.
  • During poker, correctly judging the rarity of a straight flush can help you gauge betting strategy.
  • Understanding the dramatic difference in probability between a straight and a straight flush is essential for mastering the game.
Due to its rarity, a straight flush is one of the most treasured and strategically impactful hands in poker.

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Most popular questions from this chapter

Show that \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\left(\begin{array}{c}n \\ n-k\end{array}\right)\). Give an interpretation involving subsets.

A computer consulting firm presently has bids out on three projects. Let \(A_{i}=\\{\) awarded project \(i\\}\), for \(i=1,2,3\), and suppose that \(P\left(A_{1}\right)=.22, P\left(A_{2}\right)=.25, P\left(A_{3}\right)=.28\), \(P\left(A_{1} \cap A_{2}\right)=.11, P\left(A_{1} \cap A_{3}\right)=.05, P\left(A_{2} \cap A_{3}\right)=.07\), \(P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01\). Express in words each of the following events, and compute the probability of each event: a. \(A_{1} \cup A_{2}\) b. \(A_{1}^{\prime} \cap A_{2}^{\prime}\left[\right.\) Hint: \(\left.\left(A_{1} \cup A_{2}\right)^{\prime}=A_{1}^{\prime} \cap A_{2}^{\prime}\right]\) c. \(A_{1} \cup A_{2} \cup A_{3}\) d. \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}^{\prime}\) e. \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}\) f. \(\left(A_{1}^{\prime} \cap A_{2}^{\prime}\right) \cup A_{3}\)

a. Beethoven wrote 9 symphonies, and Mozart wrote 27 piano concertos. If a university radio station announcer wishes to play first a Beethoven symphony and then a Mozart concerto, in how many ways can this be done? b. The station manager decides that on each successive night (7 days per week), a Beethoven symphony will be played, followed by a Mozart piano concerto, followed by a Schubert string quartet (of which there are 15). For roughly how many years could this policy be continued before exactly the same program would have to be repeated?

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\\{\) the first board is green \(\\}\) and \(B=\\{\) the second board is green \(\\}\). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

As of April 2006, roughly 50 million .com web domain names were registered (e.g., yahoo.com). a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? [Note: A character length of three or more is now mandated.] b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.] c. Answer the questions posed in (b) for four-character sequences. d. As of April \(2006,97,786\) of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?
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