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Chapter 1: Problem 26

Two cards are drawn successively from a deck of 52 cards. Find the probability that the second card is higher in rank than the first card. Hint: Show that \(1=\) \(P\) (higher) \(+P\) (lower) \(+P(\) same \()\) and use the fact that \(P(\) higher \()=P\) (lower).

Short Answer

Expert verified
The probability that the second card is higher than the first is \( \frac{8}{17} \).

Step by step solution

01

Understand the Problem

We are tasked with finding the probability that the second card drawn from a deck of 52 cards is of a higher rank than the first card. The hint suggests using symmetry in relations to determine these probabilities.
02

Define Outcomes

When two cards are drawn, the outcomes for the second card can either be higher in rank than the first card, lower in rank, or of the same rank. Thus we can write the equation as: \( P(\text{higher}) + P(\text{lower}) + P(\text{same}) = 1 \).
03

Use Symmetry

According to the symmetry in card drawing, \( P(\text{higher}) = P(\text{lower}) \). This is because there is no initial bias in either direction; every card higher than a given card has a corresponding card lower in rank.
04

Calculate Equal Ranks Probability

The probability that both cards are of the same rank is given by the presence of 4 cards for any particular rank in a deck. Thus for two card draws, \( P(\text{same}) = \frac{13}{52} \times \frac{3}{51} = \frac{1}{17} \).
05

Solve for Higher Probability

Substitute \( P(\text{same}) \) into the equation: \( P(\text{higher}) + P(\text{lower}) + \frac{1}{17} = 1 \). Using \( P(\text{higher}) = P(\text{lower}) \), we set \( 2P(\text{higher}) = 1 - \frac{1}{17} = \frac{16}{17} \).
06

Final Calculation

Solve for \( P(\text{higher}) \) by dividing both sides by 2: \( P(\text{higher}) = \frac{16}{17} \times \frac{1}{2} = \frac{8}{17} \).

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Key Concepts

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

Symmetry in Probability
Symmetry is a powerful concept in probability that makes complex calculations simpler. In the context of card drawing, symmetry refers to the balanced nature of possibilities. When you draw a card, there's an equal chance that another card drawn can be higher or lower. This happens because neither direction has an inherent bias. In simpler terms, nature of card ranks ensures that the number of cards higher than any given card equates to the number of cards lower than it. This balance is the symmetry in probability.

By relying on this symmetry, we establish the equation:
  • For any card drawn, the probability that the second card is higher equals the probability that it is lower.
  • This leads to the relationship: \( P(\text{higher}) = P(\text{lower}) \).
Understanding symmetry simplifies our approach and helps in solving probability problems efficiently.
Card Drawing
Drawing cards from a deck is a classic example of probability in action. With a full deck of 52 cards, each card is unique in terms of rank and suit.
  • The potential draws and their outcomes form the basis of many probability calculations, including higher or lower comparisons.
  • In this scenario, we focus on rank, disregarding the suits.

Every card draw reduces the number of available cards, influencing the subsequent probability of the next draw. This dependency makes card probability calculations slightly more intricate than independent events, as each draw impacts the possible outcomes of the next.
Rank Comparison
Comparing ranks is essential to solve our card problem. Knowing that each suit (hearts, diamonds, clubs, spades) has cards ranked from ace to king helps structure this comparison.

  • When comparing ranks: If the first card drawn is a king, no other card can be higher.
  • If it's an ace, every other card is higher.

These comparisons are vital as they directly impact each draw's probability. They help break down what needs to be calculated. Rank comparison uses symmetry, as the problem indirectly hints, because for every card of a certain rank, a mirror card exists on the opposite side of the spectrum.
Probability Calculation
Probability calculations involve understanding the likelihood of certain outcomes. In our card drawing scenario, we have three outcomes: higher, lower, and same rank. The sum of these probabilities is always 1.

  • A key point is calculating the probability of same rank, which involves both cards having the same rank from the four available cards in the deck, giving us \( P(\text{same}) = \frac{1}{17} \).
  • For the remaining outcomes (higher or lower), the symmetry simplifies the calculation as \( 2P(\text{higher}) = \frac{16}{17} \).

Finally, solving for our desired probability is straightforward: \( P(\text{higher}) = \frac{8}{17} \). Utilizing symmetry simplifies the need for extensive calculations. Focus on clear logical steps, and you鈥檒l arrive at accurate results in your probability tasks.

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Most popular questions from this chapter

In an upcoming national election for the President of the United States, a pollster plans to predict the winner of the popular vote by taking a random sample of 1000 voters and declaring that the winner will be the one obtaining the most votes in his sample. Suppose that 48 percent of the voters plan to vote for the Republican candidate and 52 percent plan to vote for the Democratic candidate. To get some idea of how reasonable the pollster's plan is, write a program to make this prediction by simulation. Repeat the simulation 100 times and see how many times the pollster's prediction would come true. Repeat your experiment, assuming now that 49 percent of the population plan to vote for the Republican candidate; first with a sample of 1000 and then with a sample of 3000 . (The Gallup Poll uses about 3000.) (This idea is discussed further in Chapter 9 , Section 9.1.)

Estimate, by simulation, the average number of children there would be in a family if all people had children until they had a boy. Do the same if all people had children until they had at least one boy and at least one girl. How many more children would you expect to find under the second scheme than under the first in 100,000 families? (Assume that boys and girls are equally likely.)

The Labouchere system for roulette is played as follows. Write down a list of numbers, usually \(1,2,3,4 .\) Bet the sum of the first and last, \(1+4=5,\) on red. If you win, delete the first and last numbers from your list. If you lose, add the amount that you last bet to the end of your list. Then use the new list and bet the sum of the first and last numbers (if there is only one number, bet that amount). Continue until your list becomes empty. Show that, if this happens, you win the sum, \(1+2+3+4=10\), of your original list. Simulate this system and see if you do always stop and, hence, always win. If so, why is this not a foolproof gambling system?

You are offered the following game. A fair coin will be tossed until the first time it comes up heads. If this occurs on the \(j\) th toss you are paid \(2^{j}\) dollars. You are sure to win at least 2 dollars so you should be willing to pay to play this game - but how much? Few people would pay as much as 10 dollars to play this game. See if you can decide, by simulation, a reasonable amount that you would be willing to pay, per game, if you will be allowed to make a large number of plays of the game. Does the amount that you would be willing to pay per game depend upon the number of plays that you will be allowed?

Modify the program CoinTosses so that it tosses a coin \(n\) times and records whether or not the proportion of heads is within .1 of .5 (i.e., between .4 and .6 ). Have your program repeat this experiment 100 times. About how large must \(n\) be so that approximately 95 out of 100 times the proportion of heads is between .4 and \(.6 ?\)
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