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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鈥檚 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 \(\boxed{\frac{4}{9}}\).

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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鈥攄etermining 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鈥攕uch 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.

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Most popular questions from this chapter

Calculate \(P(r

Derive De Moivre's result \(\log \left(\frac{Q}{M}\right) \approx-\frac{2 t^{2}}{n} \quad\) or equivalently \(\quad \log \left(\frac{M}{Q}\right) \approx \frac{2 t^{2}}{n}\) (Hint: Divide the arguments of the first two logarithm terms in the expression in the text by \(m\). Then simplify and replace the remaining logarithm terms by the first two terms of their respective power series.)

Show that if one defines the Bernoulli numbers \(B_{i}\) by setting $$ \frac{x}{e^{x}-1}=\sum_{i=0}^{\infty} \frac{B_{i}}{i !} x^{i} $$ then the values of \(B_{i}\) for \(i=2,4,6,8,10,12\) are the same as those calculated in the text and in Exercise 1 .

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.)

Add the highest-degree terms of the columns from Exercise 15 to get $$ s\left(\frac{s}{m}+\frac{1}{2 \cdot 3} \frac{s^{3}}{m^{3}}+\frac{1}{3 \cdot 5} \frac{s^{5}}{m^{5}}+\frac{1}{4 \cdot 7} \frac{s^{7}}{m^{7}}+\cdots\right) $$ which, setting \(x=s / m\), is equal to $$ s\left(\frac{2 x}{1 \cdot 2}+\frac{2 x^{3}}{3 \cdot 4}+\frac{2 x^{5}}{5 \cdot 6}+\frac{2 x^{7}}{7 \cdot 8}+\cdots\right) $$ Show that the series in the parenthesis can be expressed in finite terms as $$ \log \left(\frac{1+x}{1-x}\right)+\frac{1}{x} \log \left(1-x^{2}\right) $$ and therefore that the original series is $$ m x \log \left(\frac{1+x}{1-x}\right)+m \log \left(1-x^{2}\right) $$ Since \(s=m-1\) (or \(m x=m-1\) ), show therefore that the sum of the highest- degree terms of the columns of Exercise 15 is equal to $$ \begin{aligned} &(m-1) \log \left(\frac{1+\frac{m-1}{m}}{1-\frac{m-1}{m}}\right) \\ &\quad+m \log \left[\left(1+\frac{m-1}{m}\right)\left(1-\frac{m-1}{m}\right)\right] \end{aligned} $$ which in turn is equal to \((2 m-1) \log (2 m-1)-2 m \log m\).
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