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Chapter 4: Problem 61

Suppose in the gambler's ruin problem that the probability of winning a bet depends on the gambler's present fortune. Specifically, suppose that \(\alpha_{i}\) is the probability that the gambler wins a bet when his or her fortune is \(i\). Given that the gambler's initial fortune is \(i\), let \(P(i)\) denote the probability that the gambler's fortune reaches \(N\) before 0 . (a) Derive a formula that relates \(P(i)\) to \(P(i-1)\) and \(P(i+1)\). (b) Using the same approach as in the gambler's ruin problem, solve the equation of part (a) for \(P(i)\) (c) Suppose that \(i\) balls are initially in urn 1 and \(N-i\) are in urn 2, and suppose that at each stage one of the \(N\) balls is randomly chosen, taken from whichever urn it is in, and placed in the other urn. Find the probability that the first urn becomes empty before the second.

Short Answer

Expert verified
To find the probability that urn 1 becomes empty before urn 2, we relate \(P(i)\) to \(P(i-1)\) and \(P(i+1)\) using the equation: \[P(i)=\alpha_{i}P(i+1)+(1-\alpha_{i})P(i-1)\] where \(\alpha_{i}=\frac{i}{N}\), and boundary conditions \(P(0)=0\) and \(P(N)=1\). We can then iteratively solve for \(P(i)\) using: \[P(i)=\frac{\frac{i}{N}P(i+1)+(1-\frac{i}{N})P(i-1)}{1+\frac{i}{N}-\frac{i}{N}}\] Finally, iterating from \(i=1\) to \(i=N-1\), we obtain the probability that urn 1 becomes empty before urn 2.

Step by step solution

01

Relating \(P(i)\) to \(P(i-1)\) and \(P(i+1)\)

Let's analyze the situation when the gambler has \(i\) fortune. There are two possible outcomes: 1. The gambler wins the bet with probability \(\alpha_{i}\): In this case, the fortune increases to \(i+1\), and the probability of reaching \(N\) before 0 is \(P(i+1)\). 2. The gambler loses the bet with probability \(1-\alpha_{i}\): In this case, the fortune decreases to \(i-1\), and the probability of reaching \(N\) before 0 is \(P(i-1)\). The probability of reaching \(N\) before 0 equals the sum of the probabilities of these two possibilities. So, we have the following equation: \[P(i)=\alpha_{i}P(i+1)+(1-\alpha_{i})P(i-1)\]
02

Solving the equation for \(P(i)\)

To solve the equation for \(P(i)\), we can rewrite it as: \[(1-\alpha_{i})P(i-1)+\alpha_{i}P(i+1)-P(i)=0\] Now, we can apply the same approach as in the gambler's ruin problem by dividing the range from \(0\) to \(N\) into intervals and solving the equation for each interval. We have the boundary conditions: \(P(0)=0\) since the gambler has no chance of reaching \(N\) if they start with 0 fortune, and \(P(N)=1\) since if the gambler starts with \(N\), they have already reached \(N\). We can solve the equation for \(P(i)\) iteratively by using the boundary conditions and starting from \(i=1\) until \(i=N-1\): \[P(i)=\frac{\alpha_{i}P(i+1)+(1-\alpha_{i})P(i-1)}{1+\alpha_{i}-\alpha_{i}}\]
03

Applying the formula to urn problem

In the urn problem, we have \(i\) balls initially in urn 1 and \(N-i\) balls in urn 2. We need to find the probability that urn 1 becomes empty before urn 2. At each stage, one of the \(N\) balls is randomly chosen and taken from whichever urn it is in and placed in the other urn. The probability of choosing a ball from urn 1 is \(\frac{i}{N}\), and from urn 2 is \(\frac{N-i}{N}\). Thus, \(\alpha_{i}=\frac{i}{N}\), then we can solve for \(P(i)\) using the formula in Step 2: \[P(i)=\frac{\frac{i}{N}P(i+1)+(1-\frac{i}{N})P(i-1)}{1+\frac{i}{N}-\frac{i}{N}}\] Finally, by iterating from \(i=1\) to \(i=N-1\), we can obtain the probability that urn 1 becomes empty before urn 2.

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Key Concepts

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

Probability Models
Probability models are a fundamental tool in understanding random phenomena. They provide a structured framework to quantify the likelihood of various outcomes in a random process. In the context of the gambler's ruin problem, a probability model allows us to analyze and compute the chances of a gambler eventually winning or going broke.

For instance, when a gambler is betting, the outcome of each bet can be modeled using probability distributions, which is essentially a mathematical function that tells us how the probabilities are distributed among different possible outcomes. When these bets are placed successively, the random process is best described using a probability model known as a Markov chain, where the next state depends only on the current state and not on the sequence of events that preceded it.

In the gambler's ruin example, the probability model takes into account the changing probabilities of winning a bet as the gambler's fortune changes, denoted by \(\alpha_{i}\). By setting up a probability model with the appropriate conditions and recursive relationships, one can analyze the gambler's chances of reaching a fortune of \(N\) before reaching a fortune of 0.
Stochastic Processes
At the heart of the gambler's ruin problem lies a stochastic process—a random process that evolves over time according to probability rules. Stochastic processes are used to model situations where outcomes are uncertain, which is the case every time our gambler places a bet.

The gambler's fortune over time can be described as a stochastic process. This process is discrete since the gambler makes a bet at distinct points in time. In the simplified model, after each bet, the gambler's fortune either goes up by 1 with probability \(\alpha_{i}\), or goes down by 1 with the complementary probability \(1 - \alpha_{i}\). The stochastic process here is known as a 'random walk' where each step is determined by the flip of a weighted coin, representing the probability of winning or losing the bet.

To determine the eventual outcome of the gambling sequence (ruin or reaching \(N\)), we need to look at the stochastic process over the long term. This involves analyzing the transition probabilities and applying them recursively, which we see with the equation that links \(P(i)\) to \(P(i-1)\) and \(P(i+1)\).
Random Variables
Random variables play a crucial role in probabilistic analysis and are deeply ingrained in problems like the gambler's ruin. A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In our case, the gambler's current fortune is represented as a random variable that changes with each bet placed.

Each possible state of the gambler's fortune is an instance of the random variable. The gambler's fortune after the next bet has a probability distribution that depends on their current fortune, hence the use of \(\alpha_{i}\), which represents the probability of a win given the gambler's current fortune \(i\). The Markov chain model indicates that the future state of this random variable depends only on the current state and not on the path taken to reach that state.

Understanding random variables is essential for computing probabilities of different outcomes in complex problems. In this problem, \(P(i)\), the probability of reaching a fortune of \(N\) before going broke, is itself a random variable that depends on the gambler's current amount of wealth. By connecting these random variables using recursive relationships, we can solve for the probability of the gambler's success.

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Most popular questions from this chapter

Let \(\left\\{X_{n}, n \geqslant 0\right\\}\) denote an ergodic Markov chain with limiting probabilities \(\pi_{i} .\) Define the process \(\left\\{Y_{n}, n \geqslant 1\right\\}\) by \(Y_{n}=\left(X_{n-1}, X_{n}\right)\). That is, \(Y_{n}\) keeps track of the last two states of the original chain. Is \(\left\\{Y_{n}, n \geqslant 1\right\\}\) a Markov chain? If so, determine its transition probabilities and find $$ \lim _{n \rightarrow \infty} P\left\\{Y_{n}=(i, j)\right\\} $$

Show that if state \(i\) is recurrent and state \(i\) does not communicate with state \(j\), then \(P_{i j}=0 .\) This implies that once a process enters a recurrent class of states it can never leave that class. For this reason, a recurrent class is often referred to as a closed class.

Suppose that a population consists of a fixed number, say, \(m\), of genes in any generation. Each gene is one of two possible genetic types. If exactly \(i\) (of the \(m\) ) genes of any generation are of type 1 , then the next generation will have \(j\) type 1 (and \(m-j\) type 2 ) genes with probability $$ \left(\begin{array}{c} m \\ j \end{array}\right)\left(\frac{i}{m}\right)^{j}\left(\frac{m-i}{m}\right)^{m-j}, \quad j=0,1, \ldots, m $$ Let \(X_{n}\) denote the number of type 1 genes in the \(n\) th generation, and assume that \(X_{0}=i\) (a) Find \(E\left[X_{n}\right]\). (b) What is the probability that eventually all the genes will be type \(1 ?\)

For the random walk of Example \(4.18\) use the strong law of large numbers to give another proof that the Markov chain is transient when \(p \neq \frac{1}{2}\). Hint: Note that the state at time \(n\) can be written as \(\sum_{i=1}^{n} Y_{i}\) where the \(Y_{i}\) s are independent and \(P\left\\{Y_{i}=1\right\\}=p=1-P\left\\{Y_{i}=-1\right\\}\). Argue that if \(p>\frac{1}{2}\), then, by the strong law of large numbers, \(\sum_{1}^{n} Y_{i} \rightarrow \infty\) as \(n \rightarrow \infty\) and hence the initial state 0 can be visited only finitely often, and hence must be transient. A similar argument holds when \(p<\frac{1}{2}\).

Consider the Ehrenfest urn model in which \(M\) molecules are distributed between two urns, and at each time point one of the molecules is chosen at random and is then removed from its urn and placed in the other one. Let \(X_{n}\) denote the number of molecules in urn 1 after the \(n\) th switch and let \(\mu_{n}=E\left[X_{n}\right]\). Show that (a) \(\mu_{n+1}=1+(1-2 / M) \mu_{n}\). (b) Use (a) to prove that $$ \mu_{n}=\frac{M}{2}+\left(\frac{M-2}{M}\right)^{n}\left(E\left[X_{0}\right]-\frac{M}{2}\right) $$
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