91Ó°ÊÓ

Allan and Beth currently have \(\$ 2\) and \(\$ 3\), respectively. A fair coin is tossed. If the result of the toss is \(\mathrm{H}\), Allan wins \(\$ 1\) from Beth, whereas if the coin toss results in \(\mathrm{T}\), then Beth wins \(\$ 1\) from Allan. This process is then repeated, with a coin toss followed by the exchange of \(\$ 1\), until one of the two players goes broke (one of the two gamblers is ruined). We wish to determine \(a_{2}=P\) (Allan is the winner \(\mid\) he starts with \(\$ 2\) ) To do so, let's also consider \(a_{i}=P\) (Allan wins | he starts with \(\$ i\) ) for \(i=0,1,3,4\), and 5 . a. What are the values of \(a_{0}\) and \(a_{5}\) ? b. Use the law of total probability to obtain an equation relating \(a_{2}\) to \(a_{1}\) and \(a_{3}\). [Hint: Condition on the result of the first coin toss, realizing that if it is a \(\mathrm{H}\), then from that point Allan starts with \$3.] c. Using the logic described in (b), develop a system of equations relating \(a_{i}(i=1,2,3,4)\) to \(a_{i-1}\) and \(a_{i+1}\). Then solve these equations. [Hint: Write each equation so that \(a_{i}-a_{i-1}\) is on the left hand side. Then use the result of the first equation to express each other \(a_{i}-a_{i-1}\) as a function of \(a_{1}\), and add together all four of these expressions \((i=2,3,4,5)\).] d. Generalize the result to the situation in which Allan's initial fortune is \(\$ a\) and Beth's is \(\$ b\). Note: The solution is a bit more complicated if \(p=P(\) Allan wins \(\$ 1) \neq .5 .\)

Step by step solution

01

Determine Outcomes for Boundary Conditions

First, we determine the probabilities for the boundary conditions: when Allan has $0 or $5. If Allan has $0 (he is broke), he cannot win. So, the probability that Allan wins, given he starts with $0, is $a_0 = 0. If Allan has all the money (i.e., $5), he is already declared the winner. Thus, $a_5 = 1.

02

Use Law of Total Probability for $a_2$

We'll use the law of total probability, considering the first coin toss to derive an equation for $a_2$. If the first coin toss results in H (P=0.5), Allan wins $1 and then has $3, so his probability of ultimately winning becomes $a_3. If the coin toss results in T (P=0.5), Allan loses $1 and then has $1, and his probability of winning becomes $a_1. The equation is: a_2 = 0.5 imes a_3 + 0.5 imes a_1.

03

Create System of Equations for All Initial Values

Apply a similar logic to all values of $a_i$, where i = 1, 2, 3, 4: For $a_1$: a_1 = 0.5 imes a_2 + 0.5 imes a_0 a_0 = 0 (from Step 1), thus a_1 = 0.5 imes a_2. For $a_3$: a_3 = 0.5 imes a_4 + 0.5 imes a_2. For $a_4$: a_4 = 0.5 imes a_5 + 0.5 imes a_3. a_5 = 1 (from Step 1), thus a_4 = 0.5 imes 1 + 0.5 imes a_3.

04

Solve the System of Equations

Substitute the expressions obtained in Step 3: From the equation for $a_1$: a_1 = 0.5 imes (0.5 imes a_3 + 0.5 imes a_1). From the equation for $a_3$ and $a_4$: a_4 = 0.5 + 0.5 imes (0.5 imes a_4 + 0.5 imes a_2). We are solving a linear system. Solving yields: $a_2 = 0.4, $a_1 = 0.2, $a_3 = 0.6, and $a_4 = 0.8.

05

Generalize to $a$ Dollars for Allan and $b$ for Beth

For generalized situations where Allan begins with \(a\) dollars, and the game ends when one reaches \(a+b\), the probability of Allan winning follows:If the probability of Allan winning a single coin toss is \(p (0.5 in this case), use the binomial probabilities:Allan's winning probability starting with a capital \)a, given total \(a+b is:\[P(a, a+b) = \frac{(1-p)^b - p^b}{(1-p)^{a+b} - p^{a+b}} \]Given \)p = 0.5, the expression simplifies to the linear interpolation between 0 and 1, directly solved above.

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Key Concepts

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

Markov Chains

A Markov Chain is a type of stochastic process that undergoes transitions from one state to another within a finite or countable state space. It relies primarily on the principle of memorylessness, meaning the next state depends only on the current state and not on the sequence of states that preceded it.
In our exercise involving Allan and Beth, the states represent their potential financial conditions that change based on the outcome of a fair coin toss. The transitions between these states are probabilistically determined by the coin toss, and this defines a Markov Chain:

• The possible states are defined by the total amount of money bet between Allan and Beth, which in this case ranges from 0 to 5 dollars.
• The transition probabilities are determined by the outcomes of the coin toss—50% chance for heads and 50% for tails.
• Each state transition should follow the Markovian property, i.e., the transition depends only on the current state.

Understanding Markov Chains helps elucidate how Allan’s winning probabilities evolve as the game progresses.

Law of Total Probability

The Law of Total Probability is a fundamental rule connecting marginal probabilities to conditional probabilities. In simpler terms, it helps break down complex probability problems by partitioning them into manageable parts.
In our scenario, this law is applied by considering the potential outcomes of the first coin toss.

• If the first toss is heads ( P=0.5), Allan receives an additional dollar, transitioning from a state of having $2 to $3.
• If it lands tails ( P=0.5), Allan loses a dollar and shifts from $2 to $1.

Thus, we can express Allan's probability of ultimately winning when starting with $2 as the weighted sum of the probabilities after the first toss: a_2 = 0.5 imes a_3 + 0.5 imes a_1.
Using this breakdown allows us to compute each probability involving sequential events or stages effectively.

Stochastic Processes

Stochastic processes are mathematical objects used to model systems that evolve over time in a random manner. They are crucial in understanding real-world processes that are inherently unpredictable, like stock price movements or, in this case, the coin toss game between Allan and Beth.
Stochastic processes can be used in various scenarios:

• Discrete-time stochastic processes, like our coin toss scenario, are events occurring at fixed time intervals (each coin flip).
• Understanding stochastic processes provides insights into long-term behavior where analysis of probabilities (like Allan's winning chance) is extended over multiple iterations.

Our coin toss game is a practical example of a stochastic process, where outcomes rely on probabilities set by the coin's fairness. Computational models use these frameworks to predict results in seemingly random environments.

Gambler's Ruin Problem

The Gambler's Ruin Problem is a classic example of Markov Chains and stochastic processes. It explores the probability of a gambler's eventual bankruptcy when engaging in a fair or unfair betting game.
This problem posits situations where a gambler bets repetitively, with winning or losing states, until they either go broke (ruin) or reach a winning goal.
The scenario of Allan and Beth centers around a similar setup:

• Allan starts with $2, and his ruin will occur if he loses all his money (reaches $0), or he wins $5 by taking all of Beth's money.
• The game concludes when either of these events occurs, offering a finite state space within the transition model.

Understanding this problem helps in computing and anticipating the outcome probabilities and gaining insight into strategic moves based on conditional winning probabilities in betting games.