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Chapter 3: Problem 13

Suppose that you are gambling against an infinitely rich adversary and at each stage you either win or lose 1 unit with respective probabilities \(p\) and \(1-p .\) Show that the probability that you eventually go broke is $$ \begin{array}{cl} 1 & \text { if } p \leq \frac{1}{2} \\ (q / p)^{i} & \text { if } p>\frac{1}{2} \end{array} $$ where \(q=1-p\) and where \(i\) is your initial fortune.

Short Answer

Expert verified
To summarize, the probability of eventually going broke, given your initial fortune i and winning probability p, can be found using the following formula: \[ \begin{array}{cl} 1 & \text { if } p \leq \frac{1}{2} \\ \left(\frac{1-p}{p}\right)^{i} & \text { if } p>\frac{1}{2} \end{array} \]

Step by step solution

01

Consider the two cases separately

There are two cases we need to consider: when our winning probability p is less than or equal to 0.5, and when p is greater than 0.5. In the first case, the probability of going broke is assumed to be 1. In the second case, we need to find the probability of going broke as a function of i.
02

Set up a recursive equation for the probabilities

Let's denote the probability of going broke for an initial fortune i as P(i). Then, we can set up a recursive equation for the probabilities of going broke: \[P(i) = pP(i+1) + qP(i-1)\] This equation represents the fact that, at each step, we have a probability of p to win 1 unit (and have an initial fortune i+1) and a probability of q to lose 1 unit (and have an initial fortune i-1).
03

Consider the boundary conditions

We have two boundary conditions for this problem: 1. If we have no money (i = 0), we are already broke (P(0) = 1). 2. If we have infinite money (i 鈫� 鈭�), the probability of going broke is 0 (P(鈭�) = 0).
04

Solve the recursive equation for the case when p 鈮� 0.5

If p 鈮� 0.5, then the probability of going broke is 1. Hence, we don't need to solve the recursive equation in this case.
05

Solve the recursive equation for the case when p > 0.5

For the case when p > 0.5, we can rewrite the recursive equation as: \[P(i+1) = \frac{1}{p}\left[P(i) - qP(i-1)\right]\] To find a pattern, let's compute \(P(i+2)\): \[P(i+2) = \frac{1}{p}\left[P(i+1) - qP(i)\right]\] Now, substitute the expression for \(P(i+1)\) from the previous equation: \[P(i+2) = \frac{1}{p}\left[\frac{1}{p}\left[P(i) - qP(i-1)\right] - qP(i)\right]\] Simplifying the equation and isolating the term with P(i): \[P(i+2) - \frac{q}{p}P(i+1) = \left(\frac{q}{p}\right)^2 P(i) - \frac{q}{p}P(i-1)\] It appears that we have a pattern: \[P(n) = \left(\frac{q}{p}\right)^n P(1) - \left(\frac{q}{p}\right)^{n-1}P(0)\] Now apply the boundary conditions to solve for P(1): \[P(0) = 1\] \[P(鈭�) = 0 = \lim_{n \to \infty} \left[\left(\frac{q}{p}\right)^n P(1) - \left(\frac{q}{p}\right)^{n-1}\right]\] Since p > 0.5, q < p, and \(\left(\frac{q}{p}\right)^n \to 0\) as n 鈫� 鈭�. Therefore, we have: \[P(1) = \frac{q}{p}\] Now, we can find the general formula for P(i): \[P(i) = \left(\frac{q}{p}\right)^i P(1) - \left(\frac{q}{p}\right)^{i-1}P(0) = \left(\frac{q}{p}\right)^i\] Thus, the probability of going broke when p > 0.5 is \(\left(\frac{q}{p}\right)^i\).

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Key Concepts

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

Gambling Probability
Understanding the concept of gambling probability is crucial when delving into games of chance where outcomes are influenced by random variables. This involves analyzing the likelihood of different outcomes and how they affect your chances of winning or losing. In the context of the given exercise, the probability revolves around the chance of going broke while gambling against an opponent with an endless bankroll.

When participating in such a gamble, each round presents two outcomes: winning or losing a unit of fortune. The probabilities of these outcomes are represented by the variables \( p \) for winning and \( q = 1-p \) for losing. We assume that these events are independent and identically distributed, meaning the outcome of one gamble does not influence the next. If you start with an initial fortune \( i \) and gamble repeatedly, the probability that you will eventually go broke is dependent on the value of \( p \).

When \( p \) is less than or equal to 0.5, it is generally understood that the game is unfavorable, or 'fair' at best, providing no edge to the gambler. Hence, the chance of eventually losing all your money is certain, that is, a probability of 1. On the other hand, if \( p > 0.5 \), you have a greater chance of winning than losing each round, and the probability of going broke becomes a function of both \( p \) and your initial fortune \( i \).
Recursive Equations
Recursive equations are mathematical tools used to define sequences of numbers, functions or other entities, where each term is derived from the preceding terms according to a fixed rule. They are particularly valuable in situations like our gambling scenario, where the future state depends on the current state.

To analyze the probability of going broke, we use recursive equations to express the chance of losing everything as function of our current fortune \( i \). The recursive formula established in the solution is given by:
\[ P(i) = pP(i+1) + qP(i-1) \]
With each round, this equation accounts for the probability of moving up to a higher fortune \( i+1 \) by winning, or down to a lower fortune \( i-1 \) by losing. It essentially builds a relationship between the probabilities of subsequent states of fortune. By solving this recursive equation, we can determine the explicit form of \( P(i) \) under different conditions of \( p \), essentially revealing our long-term prospects of financial ruin.
Boundary Conditions
Boundary conditions are crucial in the world of mathematics as they provide constraints that allow for the solution of equations, often used with differential or recursive equations. Think of them as the 'edges' of a problem where the solution must satisfy certain criteria.

In our gambling problem, two clear boundary conditions are identified:
  • When we're out of money \( (i=0) \) we are already broke, hence \( P(0) = 1 \).
  • With an infinitely large fortune, or as \( i \) approaches infinity, the probability of going broke tends toward zero, thus \( P(鈭�) = 0 \).
These provide the starting and ending points needed to solve the recursive equation. For \( p > 0.5 \), we use the boundary conditions to determine the constants within the general solution and to ensure that our probability formula makes sense in the context of gambling. For instance, no matter how favorable the game, if you start with nothing \( (P(0)) \), you're already broke. Conversely, with endless resources, the risk of depletion is non-existent.

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Most popular questions from this chapter

Barbara and Dianne go target shooting. Suppose that each of Barbara's shots hits the wooden duck target with probability \(p_{1}\), while each shot of Dianne's hits it with probability \(p_{2}\). Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that (a) both shots hit the duck; (b) Barbara's shot hit the duck? What independence assumptions have you made?

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer this question, pointing out that if \(A\) knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\) because he would then be one of two prisoners. What do you think of the jailer's reasoning?

In a game of bridge, West has no aces. What is the probability of his partner's having (a) no aces and (b) 2 or more aces? (c) What would the probabilities be if West had exactly 1 ace?

A ball is in any one of \(n\) boxes. It is in the \(i\) th box with probability \(P_{i}\). If the ball is in box \(i\), a search of that box will uncover it with probability \(\alpha_{i}\). Show that the conditional probability that the ball is in box \(j\), given that a search of box \(i\) did not uncover it, is $$ \begin{array}{cl} \frac{P_{j}}{1-\alpha_{i} P_{i}} & \text { if } j \neq i \\ \frac{\left(1-\alpha_{i}\right) P_{i}}{1-\alpha_{i} P_{i}} & \text { if } j=i \end{array} $$

A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.
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