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In the gambler's ruin problem of Section 4.5.1, suppose the gambler's fortune is presently \(i\), and suppose that we know that the gambler's fortune will eventually reach \(N\) (before it goes to 0 ). Given this information, show that the probability he wins the next gamble is $$ \begin{array}{ll} \frac{p\left[1-(q / p)^{i+1}\right]}{1-(q / p)^{i}}, & \text { if } p \neq \frac{1}{2} \\ \frac{i+1}{2 i}, & \text { if } p=\frac{1}{2} \end{array} $$

Short Answer

Expert verified
The probability of the gambler winning the next bet, given that they will eventually reach N dollars, is: $$ \begin{array}{ll} \frac{p\left[1-(q / p)^{i+1}\right]}{1-(q / p)^{i}}, & \text{ if } p \neq \frac{1}{2} \\ \frac{i+1}{2i}, & \text{ if } p = \frac{1}{2} \end{array} $$

Step by step solution

01

Expression for winning the next bet

If the gambler wins the next bet, they now have i+1 dollars. The probability of them reaching N dollars given that they win the next bet is the same as the probability of reaching N starting with i+1 dollars, which can be represented as P(N|i+1).
02

Expression for losing the next bet

If the gambler loses the next bet, they now have i-1 dollars. The probability of them reaching N dollars given that they lose the next bet is the same as the probability of reaching N starting with i-1 dollars, which can be represented as P(N|i-1). #Step 2: Use conditional probability formula to find the probability of winning the next bet given reaching N dollars#
03

Conditional probability formula

Let A be the event of winning the next bet, and B be the event of reaching N dollars. We want to find P(A|B). Using the conditional probability formula, we have: $$ P(A|B)=\frac{P(A \cap B)}{P(B)} $$ We can rewrite the numerator and denominator of this fraction using the expressions from Step 1. $$ P(A|B)=\frac{p \cdot P(N|i+1)}{p \cdot P(N|i+1) + q \cdot P(N|i-1)} $$
04

Plug in the formula for reaching N dollars

The formula for the probability of reaching N dollars starting with i dollars (found in Section 4.5.1) is: $$ P(N|i) = \begin{cases} \frac{1-(q/p)^i}{1-(q/p)^N}, & \text{ if } p \neq \frac{1}{2} \\ \frac{i}{N}, & \text { if } p = \frac{1}{2} \end{cases} $$ We need to plug this formula into the conditional probability expression from above. For \( p \neq \frac{1}{2} \): $$ P(A|B) = \frac{p\left[ \frac{1-(q/p)^{i+1}}{1-(q/p)^N}\right]}{p\left[\frac{1-(q/p)^{i+1}}{1-(q/p)^N}\right] + q\left[\frac{1-(q/p)^{i-1}}{1-(q/p)^N}\right]} $$
05

Simplify the expression

We simplify the expression by canceling common terms in the numerator and denominator to obtain the final probability expression for \( p \neq \frac{1}{2} \): $$ P(A|B)=\frac{p\left[1-(q / p)^{i+1}\right]}{1-(q / p)^{i}} $$ For \( p = \frac{1}{2} \), similar steps yield the final probability expression: $$ P(A|B)=\frac{i+1}{2 i} $$ Thus, the probability of the gambler winning the next bet given that they will eventually reach N dollars is: $$ \begin{array}{ll} \frac{p\left[1-(q / p)^{i+1}\right]}{1-(q / p)^{i}}, & \text{ if } p \neq \frac{1}{2} \\ \frac{i+1}{2i}, & \text{ if } p = \frac{1}{2} \end{array} $$

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in statistics that deals with assessing the likelihood of an event occurring given that another event has already happened. This principle is crucial for understanding complex probability scenarios, as it is often necessary to calculate probabilities in a chained sequence of events where prior outcomes influence subsequent ones.

In the context of the gambler's ruin problem, conditional probability allows us to evaluate the chances of the gambler winning the next gamble under the condition that they will eventually reach a certain wealth level, denoted as 'N'. This can be mathematically expressed using the formula for conditional probability:
\[\begin{equation}P(A|B) = \frac{P(A \cap B)}{P(B)}\end{equation}\]
Where A represents the event of winning the next bet and B represents the event of reaching the wealth level 'N'. By calculating the intersection of A and B, and dividing by the probability of B, we can quantify the gambler's chances in a much more precise way. Grasping this concept ensures a better understanding of how events relate to one another within the scope of probability theory.
Probability Formulas
Probability formulas are mathematical equations used to calculate the likelihood of varying events. In many cases, especially in games of chance and stochastic processes, these formulas are essential to predict outcomes and make strategic decisions. For the gambler's ruin problem, we use specific probability formulas to determine the chances of the gambler reaching a certain level of wealth.

The steps detailed in the solution involve applying probability formulas to account for different scenarios: winning and losing a bet. By plugging the results into the conditional probability equation, the gambler's next move can be assessed with greater accuracy. Here's the tailored probability formula for the gambler's scenario, distinguishing between cases based on the gambler's odds ( p ) of winning each individual bet:
\[\begin{equation}P(N|i) = \begin{cases}\frac{1-(q/p)^i}{1-(q/p)^N}, & \text{ if } p eq \frac{1}{2} \frac{i}{N}, & \text { if } p = \frac{1}{2}\end{cases}\end{equation}\]
This equation computes the probability of the gambler achieving a wealth level of 'N' starting with 'i' currency units. It's a pivotal tool contributing to the understanding of long-term prospects in gambling and similar stochastic situations.
Stochastic Processes
Stochastic processes are mathematical models used to describe systems or phenomena that evolve over time in a way that involves randomness or unpredictability. They play a vital role in fields such as physics, finance, and of course, games of chance like those found in casinos.

In studying the gambler's ruin problem, we're dealing with a discrete-time stochastic process. Each bet represents a step in time where the gambler's fortune changes in a random fashion—either increasing or decreasing—with defined probabilities. The process continues until a certain absorbing state is reached, in this case, the gambler's wealth hitting zero or reaching 'N' dollars.

This exercise examines the gambler's evolution based on a sequence of independent and identically distributed random variables, which is characteristic of many stochastic processes. Understanding these processes is key to quantifying risk and making informed decisions in uncertain environments, reflecting the inherent volatility present in systems incorporating random events.

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Most popular questions from this chapter

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(i, i=0,1,2,3\), if the first urn contains \(i\) white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left\\{X_{n}, n=0,1,2, \ldots\right\\}\) is a Markov chain and calculate its transition probability matrix.

Let \(\pi_{i}\) denote the long-run proportion of time a given irreducible Markov chain is in state \(i\). (a) Explain why \(\pi_{i}\) is also the proportion of transitions that are into state \(i\) as well as being the proportion of transitions that are from state \(i\). (b) \(\pi_{i} P_{i j}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{i j}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$

A professor continually gives exams to her students. She can give three possible types of exams, and her class is graded as either having done well or badly. Let \(p_{i}\) denote the probability that the class does well on a type \(i\) exam, and suppose that \(p_{1}=0.3, p_{2}=0.6\), and \(p_{3}=0.9 .\) If the class does well on an exam, then the next exam is equally likely to be any of the three types. If the class does badly, then the next exam is always type \(1 .\) What proportion of exams are type \(i, i=1,2,3 ?\)

Consider a branching process having \(\mu<1\). Show that if \(X_{0}=1\), then the expected number of individuals that ever exist in this population is given by \(1 /(1-\mu)\). What if \(X_{0}=n ?\)

Consider a process \(\left\\{X_{n}, n=0,1, \ldots\right\\}\), which takes on the values 0,1 , or 2 . Suppose $$ \begin{aligned} &P\left\\{X_{n+1}=j \mid X_{n}=i, X_{n-1}=i_{n-1}, \ldots, X_{0}=i_{0}\right\\} \\ &\quad=\left\\{\begin{array}{ll} P_{i j}^{\mathrm{I}}, & \text { when } n \text { is even } \\ P_{i i}^{\mathrm{II}}, & \text { when } n \text { is odd } \end{array}\right. \end{aligned} $$ where \(\sum_{j=0}^{2} P_{i j}^{\mathrm{I}}=\sum_{j=0}^{2} P_{i j}^{\mathrm{II}}=1, i=0,1,2 .\) Is \(\left\\{X_{n}, n \geqslant 0\right\\}\) a Markov chain? If not, then show how, by enlarging the state space, we may transform it into a Markov chain.

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