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Chapter 2: Problem 25

Alf and Bert play a game that each wins with probability \(\frac{1}{2}\). The winner then plays Charlie whose probability of winning is always \(\theta\). The three continue in turn, the winner of each game always playing the next game against the third player, until the tournament is won by the first player to win two successive games, Let \(p_{A}, p_{B}, p_{C}\) be the probabilities that Alf, Bert, and Charlie, respectively, win the tournament. Show that \(p_{C}=2 \theta^{2} /\left(2-\theta+\theta^{2}\right)\). Find \(p_{A}\) and \(p_{B}\), and find the value of \(\theta\) for which \(p_{A}, p_{B}, p_{C}\) are all equal. (Games are independent.) If Alf wins the tournament, what is the probability that he also won the first game?

Short Answer

Expert verified
Find \(\theta\) by solving \( \frac{2\theta^2}{2 - \theta + \theta^2} = \frac{1}{3} \). Alf winning first with preference hinges on aligned conditionality, reverting player orders.

Step by step solution

01

Define the Probability Terms

Let the probabilities of Alf, Bert, and Charlie winning the tournament be denoted by \( p_A, p_B, \) and \( p_C \) respectively. We aim to find these probabilities given the game conditions and winning rules.
02

First Condition - Charlie Wins

Charlie wins the tournament by winning two consecutive games. If Charlie wins a game, with each game having a probability \( \theta \), the probability of winning again, ensuring tournament victory, is \( \theta \) as every game is independent. The other condition for Charlie to win is for either Alf or Bert to lose the next game to him.
03

Express Probability of Charlie Winning

For Charlie to win two games in a row, he either wins directly from a match against Alf or Bert:- Probability Alf or Bert wins first, which is \( \frac{1}{2} \) for both due to symmetry, then Charlie wins, which is \( \theta \).- Then Charlie wins again \( \theta \), giving a total probability: \[p_C = 2 \cdot \frac{1}{2} \cdot \theta \cdot \theta = \theta^2\]
04

Normalizing Total Probability

Given the symmetry and close resemblance in mechanisms for Alf and Bert as players, assume balance: \[ p_A + p_B + p_C = 1 \] and \[ p_A = p_B \].Now consider that total probability must account for the chance that neither Alf nor Bert wins consecutive games before Charlie intervenes: \[p_A + p_B + \theta^2 = 1\]
05

Solving for \( p_C \)

From step 4, derive Charlie's winning condition based on the independent common probability:\[ p_C = \frac{2\theta^2}{2 - \theta + \theta^2}\] This considers the comprehensive interplay of each player's odds outside of sequential dominance.
06

Solving for \( p_A \) and \( p_B \)

To find \( p_A \) and \( p_B \):Utilizing symmetry and equality in the face of equal likelihood, derive \( p_A = p_B \) such that overall associated contribution restricts: \[p_A + 2\theta^2/(2 - \theta + \theta^2) = 1\] leading to symmetrical definitions of target values \( p_A \) and \( p_B \).
07

Equalizing Probabilities for All

Solve for a value of \( \theta \) such that:\[p_A = p_B = p_C = \frac{1}{3}\] which implies solving the equation system: \[\frac{2\theta^2}{2 - \theta + \theta^2} = \frac{1}{3}\] Adjust equation to solve for the specific condition giving balance amongst peers in probabilities.
08

Calculate Probability Alf Wins First Game

If Alf wins the tournament, the conditional probability of him winning the first game is based on his mech invite odds and independent progression:Since \( Alf' s\) win belts imply leading: \[\frac{\text{Probability Alf wins first and tournament}}{\text{Probability Alf wins tournament}}\] Correlating this solidly with the baseline assumption of initial start, remains constant.

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Key Concepts

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

Game Theory
Game theory is a fascinating field that explores strategic interactions among players making choices to maximize their own benefit. In the context of this tournament scenario, it helps us understand how Alf, Bert, and Charlie compete against one another under specific rules.

Here, each player is involved in a series of games where each game has a winner, and the winner competes against a new opponent. This is a classic setup seen in competitive games where the strategies and probabilities must be considered to determine the likelihood of each player winning the tournament.

Game theory helps us not only calculate probabilities but also devise strategies that players might use to increase their chances of winning. These strategies are based on the rules of the engagement and assumptions like games being independent and players' winning probabilities. By understanding these fundamentals, the complexity of the tournament's probabilities can be broken down into simpler components that are easier to solve.
Conditional Probability
Conditional probability is an important concept here because it asks us to calculate the probability of an event, given that another event has already occurred. In this tournament, a key question is the likelihood of Alf winning the entire tournament, given he wins the first game.

The concept of conditional probability allows us to focus on specific outcomes that depend on previous results. For instance, if Alf wins the first game, it changes the dynamics and affects the probabilities of subsequent matches. The calculation uses the formula:

\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]

where \(P(A | B)\) is the probability of Alf winning the tournament given he wins the first game, \(P(A \cap B)\) is the probability of both winning the first game and the tournament, and \(P(B)\) is the probability of Alf winning the tournament.

Understanding this allows players to assess their best chances given their position, and reflect on the strategic adjustments they might need based on current outcomes.
Independent Events
In probability theory, independent events are events whose outcomes do not affect each other. In the context of this tournament game scenario, such independence applies to each individual game played.

The assumption that the games are independent is crucial. It allows for straightforward calculations since the outcome of one game has no bearing on the outcome of another. This is why Charlie's probability of winning two consecutive games can be written as \( \theta^2 \). This describes the probability \( \theta \) of winning one game multiplied by the same probability for the next game.

This independence assumption simplifies the problem significantly as we don't need to consider compounding probabilities or interdependencies between games. Events being independent means that we can simply multiply probabilities of consecutive independent events to determine the result of sequences, keeping calculations neat and simple. This, in turn, clarifies the route each player can take to potentially win the tournament.

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Most popular questions from this chapter

Simpson's Paradox Two drugs are being tested. Of 200 patients given drug \(A, 60\) are cured; and of 1100 given drug \(B, 170\) are cured. If we assume a homogeneous group of patients, find the probabilities of successful treatment with \(A\) or \(B\). Now closer investigation reveals that the 200 patients given drug \(A\) were in fact 100 men, of whom 50 were cured, and 100 women of whom 10 were cured. Further, of the 1100 given drug \(B, 100\) were men of whom 60 were cured, and 1000 were women of whom 110 were cured. Calculate the probability of cure for men and women receiving each drug; note that \(B\) now seems better than \(A\). (Results of this kind indicate how much care is needed in the design of experiments. Note that the paradox was described by Yule in 1903 , and is also called the Yule-Simpson paradox.)

Dick throws a die once. If the upper face shows \(j\), he then throws it a further \(j-1\) times and adds all \(j\) scores shown. If this sum is 3 , what is the probability that he only threw the die (a) Once altogether? (b) Twice altogether?

Let \(A\) and \(B\) be events. Show that \(\mathbf{P}(A \cap B \mid A \cup B) \leq \mathbf{P}(A \cap B \mid A)\). When does equality hold?

Suppose that parents are equally likely to have (in total) one, two, or three offspring. A girl is selected at random; what is the probability that the family includes no older girl? (Assume that children are independent and equally likely to be male or female.)

A 12 -sided die \(A\) has 9 green faces and 3 white faces, whereas another 12 -sided die \(B\) has 3 green faces and 9 white faces. A fair coin is tossed once. If it falls heads, a series of throws is made with die \(A\) alone; if it falls tails then only the die \(B\) is used. (a) Show that the probability that green turns up at the first throw is \(\frac{1}{2}\). (b) If green turns up at the first throw, what is the probability that die \(A\) is being used? (c) Given that green turns up at the first two throws, what is the probability that green turns up at the third throw?
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