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Chapter 1: Problem 205

Fast Eddie needs to double his money; he can only do so by playing a certain win-lose game, in which the probability of winning is \(p\). However, he can play this game as many or as few times as he wishes, and in a particular game he can bet any desired fraction of his bankroll. The game pays even money (the odds are one-to-one). Assuming he follows an optimal strategy if one is available, what is the probability, as a function of \(p\), that Fast Eddie will succeed in doubling his money?

Short Answer

Expert verified
The probability that Fast Eddie will double his money is \( \frac{p}{1 - p} \).

Step by step solution

01

Understand the Problem

Fast Eddie wants to double his initial bankroll. He can bet on a game with a probability of winning equal to \( p \). The game pays even money, and he can choose how much to bet in each game. We need to find the probability that he successfully doubles his money.
02

Define the Variables

Let \( W(p) \) be the probability that Fast Eddie succeeds in doubling his money when the probability of winning a single game is \( p \).
03

Set Up the Recurrence Relation

We know that if Fast Eddie doubles his money, he wins all or some of his bets, but he will lose everything if he makes the wrong moves. The recurrence relation for the process can be stated as: \[ W(p) = p W(p) + (1 - p) W(p)^2 \] This equation simplifies to: \[ W(p) = p + (1 - p) W(p)^2 \].
04

Solve the Recurrence Relation

Using the quadratic formula to solve for \( W(p) \): Let \( W(p) = x \). The equation becomes: \[ x = p + (1 - p)x^2 \] Rearrange to: \[ (1 - p)x^2 - x + p = 0 \]
05

Apply the Quadratic Formula

Solve the quadratic equation using the formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 - p \), \( b = -1 \), and \( c = p \). Plugging in these values gives: \[ x = \frac{1 \pm \sqrt{1^2 - 4(1 - p)p}}{2(1 - p)} \] \[ x = \frac{1 \pm \sqrt{1 - 4p + 4p^2}}{2(1 - p)} \] \[ x = \frac{1 \pm \sqrt{(1 - 2p)^2}}{2(1 - p)} \] Simplifies to: \[ x = \frac{1 \pm |1 - 2p|}{2(1 - p)} \].
06

Interpret the Solution

Since the probability must be between 0 and 1, choose the solution that results in a value between 0 and 1. When \( p \leq 0.5 \): \[ W(p) = \frac{1 - (1 - 2p)}{2(1 - p)} = \frac{2p}{2(1 - p)} = \frac{p}{1 - p} \] When \( p > 0.5 \): \[ W(p) = \frac{1 + (2p - 1)}{2(1 - p)} = \frac{2p}{2(1 - p)} = \frac{p}{1 - p} \] Thus, the formula for the probability of doubling his money is the same for all values of \( p \), which is: \[ W(p) = \frac{p}{1 - p} \].

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Key Concepts

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

Win-Lose Games
Win-lose games refer to scenarios or gambling situations where the outcome is binary; you either win or lose. In our exercise, Fast Eddie faces a win-lose game where he has a probability of winning denoted as \( p \). This means every time he plays, he either wins with a probability \( p \) or loses with a probability \( 1 - p \). These kinds of games are fundamental to probability theory and help us understand larger concepts such as expected value and risk management. Understanding the nature of win-lose games is crucial since almost all strategic decisions in such games require a comprehension of the underlying probabilities.
Recurrence Relation
A recurrence relation is a kind of equation that defines a sequence where each term is related to one or more previous terms. For Fast Eddie, we established a recurrence relation to determine the probability of doubling his money. The exact relation was given by: \[ W(p) = p + (1 - p) W(p)^2 \] This equation states that Fast Eddie's probability of success \( W(p) \) depends on the probability of winning a single game \( p \), losing \( 1 - p \), and the probability \( W(p) \) squared. Solving this recurrence relation is essential for understanding how likely he will double his money based on the different probabilities \( p \).
Quadratic Formula
The quadratic formula is a solution technique for quadratic equations of the form \( ax^2 + bx + c = 0 \). In Fast Eddie's problem, we ended up with a quadratic equation derived from the recurrence relation: \[ (1 - p)x^2 - x + p = 0 \ \] Here, we used the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\ \] Substituting \( a = 1 - p \), \( b = -1 \), and \( c = p \), we get: \[ x = \frac{1 \pm |1 - 2p|}{2(1 - p)} \ \] The quadratic formula is powerful because it allows us to solve for \( x \) (which is \( W(p) \) in our recurrence relation) and find the probability of Eddie doubling his money correctly.
Optimal Betting Strategy
An optimal betting strategy refers to the plan or approach that maximizes a player's chance of achieving a desired outcome, in this case, doubling the money. For Fast Eddie, the optimal betting strategy is based on continuously adjusting how much he bets in each game according to the probability \( p \) and the formula derived: \[ W(p) = \frac{p}{1 - p} \ \] This formula provides a straightforward way to determine the probability of success, guiding Eddie on how much risk he should take per bet. The optimal strategy ensures that Eddie maximizes his chances of reaching the goal while managing his risk appropriately. It shows how he can leverage accurate mathematical models to decide the best course of action in win-lose scenarios.

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