Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 18: Problem 10

Continuing from Exercise 9, suppose that player \(A\) is twice as strong as player \(B\). Suppose that the score is \(30: 30\). Determine the probability of player \(A\) winning. What is the probability of \(A\) winning if the score is \(15: 30\) ?

Short Answer

Expert verified
Answer: The probability of player A winning in both scenarios is \(\frac{4}{9}\).

Step by step solution

01

Scenario 1: Score 30:30 (Deuce)

In this scenario, we know that both players have scored 30 points each. To win, a player must score two consecutive points and therefore we can represent the winning probabilities as follows: - Probability of A winning first point: \(P(A1) = \frac{2}{3}\) (since A is twice as strong) - Probability of B winning first point: \(P(B1) = \frac{1}{3}\) - If A wins the first point, the probability of A winning the second point and the game: \(P(A2 | A1) = \frac{2}{3}\) Now let's calculate the probability of A winning the game using the conditional probabilities: \(P(A \ wins) = P(A1) * P(A2 | A1) = \frac{2}{3} * \frac{2}{3} = \frac{4}{9}\) Therefore, the probability of player A winning with a score of 30:30 is \(\boxed{\frac{4}{9}}\).
02

Scenario 2: Score 15:30

In this scenario, player A has 15 points and player B has 30 points. Since B already won the first game, we only need to determine the probability of A winning two consecutive games to tie the score and eventually win. - Probability of A winning next game: \(P(A1) = \frac{2}{3}\) - Probability of A winning the game after that: \(P(A2) = \frac{2}{3}\) Now, let’s calculate the probability of A winning the match: \(P(A \ wins) = P(A1) * P(A2) = \frac{2}{3} * \frac{2}{3} = \frac{4}{9}\) Therefore, the probability of player A winning with a score of 15:30 is \(\boxed{\frac{4}{9}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Conditional Probability
Understanding conditional probability is essential when the outcome of an event is influenced by the occurrence of a previous event. In the given exercise scenario, when player A is at a score of 30:30, they must win two consecutive points to win the game.

The conditional probability, written as \(P(A2 | A1)\), denotes the probability of player A winning the second point given that they have already won the first point. This concept relies heavily on the dependency between the two events.

The formula for calculating conditional probability is:
\[ P(A2 | A1) = \frac{P(A1 \cap A2)}{P(A1)} \]
For player A, since winning the first and second points are independent events, the \(P(A1 \cap A2)\) (the probability that both events happen) is simply the product of the probabilities of each event occurring separately.
Probability Theory
Probability theory is the branch of mathematics concerned with analysis of random phenomena. The exercise presents a practical aspect of probability theory—determining the likelihood of a player winning a game.

In theory, probability is always between 0 and 1, with 0 indicating impossibility and 1 indicating certainty. For player A to win the game, two independent events must occur: winning the first point and then the second point. The multiplication rule for independent events states that:
\[ P(A \text{ wins}) = P(A1) \times P(A2) \]
Thus, probability theory provides the framework for calculating the odds of player A's victory in different score scenarios by considering the events' independence or dependence.
Mathematical Statistics
Mathematical statistics involves collecting, analyzing, interpreting, and presenting data. In the context of our exercise, we can relate it to making predictions based on observed data—such as player A's strength in comparison to player B.

The computation of player A's probability of winning from different scores can be used to infer patterns and make generalizations. For instance, despite different initial scores (30:30 versus 15:30), player A's probability of winning remained the same due to their consistent strength in the game.

This consistency aligns with the concepts of mathematical expectation and variance, which are fundamental to determining the predictability and variability of an event in statistics. The exercise demonstrates how a seemingly abstract concept plays a crucial role in predicting outcomes in real-world scenarios, such as games or sports competitions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In his Letter to a Friend on Sets in Court Tennis, written in 1687 but not published until 1713, Jakob Bernoulli analyzed the probabilities at any point in a game or set of court tennis, whose scoring rules are virtually identical with those of tennis today. He determined the odds both when the players were evenly matched and when one player was stronger than the other. If two players \(A\) and \(B\) are evenly matched in a tennis game with the score \(15: 30\), determine the probability of player \(A\) winning. (Remember that one must win by two points.)

The so-called St. Petersburg Paradox was a topic of debate among those mathematicians involved in probability theory in the eighteenth century. The paradox involves the following game between two players. Player \(A\) flips a coin until a tail appears. If it appears on his first flip, player \(B\) pays him 1 ruble. If it appears on the second flip, \(B\) pays 2 rubles, on the third, 4 rubles, \(\ldots\), on the \(n\)th flip, \(2^{n-1}\) rubles. What amount should \(A\) be willing to pay \(B\) for the privilege of playing? Show first that \(A\) 's expectation, namely, the sum of the probabilities for each possible outcome of the game multiplied by the payoff for each outcome, is $$ \sum_{i=0}^{\infty} \frac{1}{2^{i}} 2^{i-1} $$ and then that this sum is infinite. Next, play the game 10 times and calculate the average payoff. What would you be willing to pay to play? Why does the concept of expectation seem to break down in this instance?

Use Bernoulli's formula to show that if greater certainty is wanted in the problem of Exercise 7 , say, \(c=10,000\), then the number of trials necessary is \(N=31,258\).

Suppose that the probability of success in an experiment is \(1 / 10\). How many trials of the experiment are necessary to ensure even odds on it happening at least once? Calculate this both by De Moivre's exact method and his approximation.

Calculate the Bernoulli numbers \(B_{8}, B_{10}\), and \(B_{12} .\) The sequence of Bernoulli numbers is usually completed by setting \(B_{0}=1, B_{1}=-\frac{1}{2}\), and \(B_{k}=0\) for \(k\) odd and greater than 1 .
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free