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Chapter 2: Problem 34

The second Earl of Yarborough is reported to have bet at odds of 1000 to 1 that a bridge hand of 13 cards would contain at least one card that is ten or higher. (By ten or higher we mean that a card is either a ten, a jack, a queen, a king, or an ace.) Nowadays, we call a hand that has no cards higher than 9 a Yarborough. What is the probability that a randomly selected bridge hand is a Yarborough?

Short Answer

Expert verified
The probability that a randomly selected bridge hand is a Yarborough is approximately 0.0014556% or 0.000014556. This is much lower than the odds offered by the Earl of Yarborough, making the bet favorable for him.

Step by step solution

01

Find the total number of bridge hands

Using the combination formula, we can determine the total number of bridge hands by choosing 13 cards from the 52-card deck. The combination formula is \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\], where n is the total number of items, k is the number of items to be chosen, and ! denotes the factorial. In our case, n = 52, and k = 13. \[\binom{52}{13} = \frac{52!}{13!(52-13)!} = \frac{52!}{13!39!}\] Calculating this expression, we get the total number of bridge hands: \[Total \ number \ of \ bridge \ hands = \binom{52}{13} = 635,013,559,600\]
02

Find the number of Yarborough hands

A Yarborough hand does not contain any card that is 10 or higher. There are 4 suits, with 9 cards in each suit below 10 (2, 3, 4, 5, 6, 7, 8, 9). Thus, there are 36 cards that can be part of a Yarborough hand. To find the number of Yarborough hands, we need to choose 13 cards from these 36 cards: \[\binom{36}{13} = \frac{36!}{13!(36-13)!} = \frac{36!}{13!23!}\] Calculating this expression, we get the number of Yarborough hands: \[Number \ of \ Yarborough \ hands = \binom{36}{13} = 9,236,080\]
03

Calculate the probability of a Yarborough hand

Now that we have both the total number of bridge hands and the number of Yarborough hands, we can calculate the probability of a randomly selected bridge hand being a Yarborough hand: \[Probability = \frac{Number \ of \ Yarborough \ hands}{Total \ number \ of \ bridge \ hands} = \frac{9,236,080}{635,013,559,600}\] Simplifying the fraction, we get the probability: \[Probability = 0.000014556\] So, the probability that a randomly selected bridge hand is a Yarborough is approximately 0.0014556%. This is significantly lower than the odds offered by the Earl of Yarborough, which suggests that his bet was a profitable one for him in the long run.

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Most popular questions from this chapter

Consider the following technique for shuffling a deck of \(n\) cards: For any initial ordering of the cards, go through the deck one card at a time and at each card, flip a fair coin. If the coin comes up heads, then leave the card where it is; if the coin comes up tails, then move that card to the end of the deck. After the coin has been flipped \(n\) times, say that one round has been completed. For instance, if \(n=4\) and the initial ordering is 1,2,3 \(4,\) then if the successive flips result in the outcome \(h, t, t, h,\) then the ordering at the end of the round is \(1,4,2,3 .\) Assuming that all possible outcomes of the sequence of \(n\) coin flips are equally likely, what is the probability that the ordering after one round is the same as the initial ordering?

An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color? (b) of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before the next selection. This is known as sampling with replacement.

Two dice are thrown \(n\) times in succession. Compute the probability that double 6 appears at least once. How large need \(n\) be to make this probability at least \(\frac{1}{2} ?\)

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Two cards are chosen at random from a deck of 52 playing cards. What is the probability that they (a) are both aces? (b) have the same value?
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