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Chapter 11: Problem 13

Smith is in jail and has 1 dollar; he can get out on bail if he has 8 dollars. A guard agrees to make a series of bets with him. If Smith bets \(A\) dollars, he wins \(A\) dollars with probability .4 and loses \(A\) dollars with probability .6 . Find the probability that he wins 8 dollars before losing all of his money if (a) he bets 1 dollar each time (timid strategy). (b) he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy). (c) Which strategy gives Smith the better chance of getting out of jail?

Short Answer

Expert verified
(a) \(\frac{1}{7} \approx 0.143\), (b) 0.064, (c) Timid strategy is better.

Step by step solution

01

Understand Timid Strategy

Under the timid strategy, Smith bets 1 dollar each time. If he wins, his money increases by 1 dollar; if he loses, he loses the 1 dollar he bet. Thus, he either increases to the next integer or decreases to zero. We will calculate each intermediate probability starting at 1 and ending at 8.
02

Applying Timid Strategy Sequence

Define \(p_i\) to be the probability that Smith wins 8 dollars starting with \(i\) dollars. We know from boundary conditions: \(p_0 = 0\) and \(p_8 = 1\). The recursive formula for other \(i\) is \(p_i = 0.4p_{i+1} + 0.6p_{i-1}\). Solve the difference equation for \(p_1\).
03

Solve Timid Strategy Equation

Solving the recursive equation given boundary conditions, we assume \(p_i = A + Bi\). Substituting, find \(A\) and \(B\) using \(p_0 = 0\), giving \(p_i = \frac{1}{7}i\). Thus, \(p_1 = \frac{1}{7}\).
04

Understand Bold Strategy

In the bold strategy, Smith will bet as much as possible without exceeding the goal. Starting at 1 dollar, each winning move doubles his money until reaching 8 dollars. Calculate this probability at each doubling.
05

Applying Bold Strategy Sequence

Smith bets all his money each time: \(1 \to 2\), \(2 \to 4\), \(4 \to 8\). Each successful move has a probability of 0.4. Going all the way to 8 requires three successful moves in a row: \(P(success) = 0.4^3 = 0.064\).
06

Comparison of Strategies

Now compare probabilities: Timid strategy yields \(p_1 = \frac{1}{7} \approx 0.143\) and Bold strategy gives \(0.064\). The timid strategy is better.

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Key Concepts

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

Timid Strategy
Imagine Smith, who is in jail and desperately wants to win enough money to make bail. He has a timid strategy available. What does it mean to be timid in this context? Under the timid strategy, Smith is conservative in his approach. He bets only 1 dollar each time, irrespective of the outcome. This way, he's not risking much. If he wins, his dollar entices another dollar into his pile, leaving him with 2 dollars, and if he loses, he's down by a dollar. The benefit of this cautious approach is that it minimizes the risk of losing the entire fortune fast. The idea behind the timid strategy is built on recursive probability. This involves calculating probabilities using prior outcomes. By setting up an equation for the probability of reaching 8 dollars, Smith evaluates his likelihood to win progressively at each level. Each success or failure contributes to his understanding of a future win. As calculated, timidly betting gives him a probability of approximately 0.143 to reach the 8 dollars needed for bail.
Bold Strategy
Imagine taking bold, calculated risks. This is exactly what Smith does under the bold strategy. Instead of playing it safe with small bets, he risks it all by betting the maximum he can. This could mean the entire fortune he has at hand. Smith doubles his money with every successful bold bet. So, if Smith starts with 1 dollar, a win takes him to 2 dollars, another to 4 dollars, and one more successful bet brings him to the crucial 8 dollars. Now, why would someone choose such a strategy? The bold approach is often enticing when looking for quick gains. However, it comes with the high risk of losing everything. For Smith, this strategy only works out if he can make three successful bold moves in a row. The probability of this happening is calculated as 0.4^3, which equals 0.064. This probability shows that while being bold can lead to the desired outcome quickly, the chance of success is relatively low compared to the timid approach.
Recursive Probability
Recursive probability is like solving a puzzle step by step. Let's apply this to Smith's situation in jail. He uses past results to formulate future probabilities. In simple terms, it's about building probabilities using known boundaries and conditions. For instance, in the timid strategy, Smith knows he can calculate based on whether his position increases or decreases. Initially, losing everything means he stops, which sets the boundary conditions. He knows that at 0 dollars, there is no probability of winning, so the condition is straightforward: it is zero. At 8 dollars, he's assured victory, with a winning probability of 1. By considering intermediate steps and implementing recursive equations, he links each possibility together. For Smith, the use of recursive probability provides a structured framework to calculate outcomes when betting timidly, and eventually led him to see that a timid strategy might be more favorable in terms of probability.

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Most popular questions from this chapter

Assume that an ergodic Markov chain has states \(s_{1}, s_{2}, \ldots, s_{k} .\) Let \(S_{j}^{(n)}\) denote the number of times that the chain is in state \(s_{j}\) in the first \(n\) steps. Let \(\mathbf{w}\) denote the fixed probability row vector for this chain. Show that, regardless of the starting state, the expected value of \(S_{j}^{(n)}\), divided by \(n\), tends to \(w_{j}\) as \(n \rightarrow \infty .\) Hint : If the chain starts in state \(s_{i},\) then the expected value of \(S_{j}^{(n)}\) is given by the expression $$ \sum_{h=0}^{n} p_{i j}^{(h)} $$

Define \(f(r)\) to be the smallest integer \(n\) such that for all regular Markov chains with \(r\) states, the \(n\) th power of the transition matrix has all entries positive. It has been shown, \(^{14}\) that \(f(r)=r^{2}-2 r+2\). (a) Define the transition matrix of an \(r\) -state Markov chain as follows: For states \(s_{i},\) with \(i=1,2, \ldots, r-2, \mathbf{P}(i, i+1)=1, \mathbf{P}(r-1, r)=\mathbf{P}(r-1,1)=\) \(1 / 2,\) and \(\mathbf{P}(r, 1)=1\). Show that this is a regular Markov chain. (b) For \(r=3\), verify that the fifth power is the first power that has no zeros. (c) Show that, for general \(r,\) the smallest \(n\) such that \(\mathbf{P}^{n}\) has all entries positive is \(n=f(r)\)

Let \(\mathbf{P}\) be the transition matrix of an \(r\) -state ergodic chain. Prove that, if the diagonal entries \(p_{i i}\) are positive, then the chain is regular.

Two players match pennies and have between them a total of 5 pennies. If at any time one player has all of the pennies, to keep the game going, he gives one back to the other player and the game will continue. Show that this game can be formulated as an ergodic chain. Study this chain using the program ErgodicChain.

Consider the following process. We have two coins, one of which is fair, and the other of which has heads on both sides. We give these two coins to our friend, who chooses one of them at random (each with probability \(1 / 2\) ). During the rest of the process, she uses only the coin that she chose. She now proceeds to toss the coin many times, reporting the results. We consider this process to consist solely of what she reports to us. (a) Given that she reports a head on the \(n\) th toss, what is the probability that a head is thrown on the \((n+1)\) st toss? (b) Consider this process as having two states, heads and tails. By computing the other three transition probabilities analogous to the one in part (a), write down a "transition matrix" for this process. (c) Now assume that the process is in state "heads" on both the \((n-1)\) st and the \(n\) th toss. Find the probability that a head comes up on the \((n+1)\) st toss. (d) Is this process a Markov chain?
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