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Consider a gambler who, at each gamble, either wins or loses her bet with respective probabilities \(p\) and \(1-p\) A popular gambling system known as the Kelley strategy is to always bet the fraction \(2 p-1\) of your current fortune when \(p>\frac{1}{2} .\) Compute the expected fortune after \(n\) gambles of a gambler who starts with \(x\) units and employs the Kelley strategy.

Short Answer

Expert verified
The expected fortune after \(n\) gambles of a gambler who starts with \(x\) units and employs the Kelley strategy is \(2xn\).

Step by step solution

01

Determine the bet fraction using the Kelley strategy

The Kelley strategy is to bet the fraction \(2p-1\) of the current fortune when \(p>\frac{1}{2}\). We can write this expression as: \(f = 2p - 1\) Keep this fraction in mind, as we'll use it in the next steps.
02

Determine the outcome for each gamble

In each gamble, there are two possible outcomes: winning or losing. We are given their probabilities as \(p\) (winning) and \(1-p\) (losing). Let's denote the winnings as \(G_1\) and \(G_2\) for winning and losing, respectively. If the gambler wins, their fortune will increase by the bet fraction of the current fortune: \(G_1 = x + f * x\) If the gambler loses, their fortune will decrease by the bet fraction of the current fortune: \(G_2 = x - f * x\)
03

Compute the expected fortune after a single gamble

Now we calculate the expected fortune after a single gamble by multiplying each outcome by its probability: Expected fortune (single gamble) = \(p * G_1 + (1-p) * G_2\) Substitute the expressions for \(G_1\) and \(G_2\) from Step 2 and the fraction \(f\) from Step 1: Expected fortune (single gamble) = \(p * (x + (2p-1) * x) + (1-p) * (x - (2p-1) * x)\)
04

Simplify the expected fortune after a single gamble

We will simplify the expression from Step 3: Expected fortune (single gamble) = \(p * (x + 2px - x) + (1-p) * (x - 2px + x)\) Expected fortune (single gamble) = \(p * (2px) + (1-p) * (2x - 2px)\) Expected fortune (single gamble) = \(2px^2 + 2x - 2px^2\) Expected fortune (single gamble) = \(2x\) This means that after a single gamble, the expected fortune doubles regardless of the value of \(p\).
05

Compute the expected fortune after n gambles

Now that we have determined the expected fortune after a single gamble, we can calculate the expected fortune after \(n\) gambles. Expected fortune (n gambles) = Expected fortune (single gamble) \(*n\) Since we found out that the expected fortune after a single gamble is \(2x\), we simply multiply it by \(n\): Expected fortune (n gambles) = \(2x * n = 2xn\) So, the expected fortune after \(n\) gambles of a gambler who starts with \(x\) units and employs the Kelley strategy is \(2xn\).

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Key Concepts

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

Expected Fortune
Expected fortune is a key idea in understanding how much money a gambler can anticipate from their betting strategy over time. In this scenario, using the Kelley strategy, we want to figure out how much the gambler's wealth can grow after a series of bets. The expected fortune is the average outcome, calculated using the probabilities of winning and losing.

In the Kelley strategy, we start with an initial amount, say \(x\) units. After a single gamble using this strategy, we find that the expected fortune becomes double what it was because the formula simplifies to \(2x\). Therefore, after \(n\) gambles, the expected fortune will be \(2xn\), as the result grows linearly with the number of bets.

This calculation is crucial because it reflects consistency. Even though each individual bet contains an element of risk, the expected fortune after multiple gambles provides insight into the long-term success of the gambling strategy.
Probability
Probability is a fundamental concept in predicting the outcomes of gambling scenarios. It represents the chance that a given event, like winning or losing a bet, will occur. In this context, the probability of winning is given by \(p\), and the probability of losing is \(1-p\).

Understanding these probabilities is vital for the application of strategies, like the Kelley strategy. The strategy suggests a calculated fraction of the current fortune to bet, based on the likelihood of winning being greater than 50%, or \(p > \frac{1}{2}\).

This is because the probability plays a central role in managing risk. When \(p\) is greater than \(\frac{1}{2}\), the odds are in the bettor's favor, and the Kelley strategy maximizes potential gains while minimizing potential losses. By assessing these probabilities accurately, gamblers can make more informed decisions that increase their chances of success.
Gambling Systems
Gambling systems are strategies employed by bettors to maximize their chances of winning and manage their capital efficiently. The Kelley strategy is one such approach that uses mathematical calculations to decide how much of a gambler's fortune should be wagered.

This system is based on the premise of placing bets when the odds are in the bettor's favor, specifically when the probability of winning \(p\) is greater than 50%. It involves betting a precise fraction, \(2p - 1\) of the current fortune, to ensure growth over time while limiting losses.

Here’s why the Kelley strategy is popular:
  • It balances risk and reward effectively, by betting proportionally to the perceived odds.
  • It's grounded in robust mathematical principles, unlike other more arbitrary systems.
  • It promotes disciplined betting, preventing gamblers from wagering too much during favorable conditions or too little during unfavorable ones.
This calculated approach makes the Kelley strategy a favored system, particularly for serious gamblers seeking to optimize their fortune over the long haul.

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Most popular questions from this chapter

The moment generating function of \(X\) is given by \(M_{X}(t)=\exp \left\\{2 e^{t}-2\right\\}\) and that of \(Y\) by \(M_{Y}(t)=\) \(\left(\frac{3}{4} e^{t}+\frac{1}{4}\right)^{10} .\) If \(X\) and \(Y\) are independent, what are (a) \(P\\{X+Y=2\\} ?\) (b) \(P\\{X Y=0\\} ?\) (c) \(E[X Y] ?\)

Consider the following dice game, as played at a certain gambling casino: Players 1 and 2 roll a pair of dice in turn. The bank then rolls the dice to determine the outcome according to the following rule: Player \(i, i=1,2,\) wins if his roll is strictly greater than the bank's. For \(i=\) \(1,2,\) let $$I_{i}=\left\\{\begin{array}{ll}1 & \text { if } i \text { wins } \\\0 & \text { otherwise }\end{array}\right.$$ and show that \(I_{1}\) and \(I_{2}\) are positively correlated. Explain why this result was to be expected.

Urn 1 contains 5 white and 6 black balls, while urn 2 contains 8 white and 10 black balls. Two balls are randomly selected from urn 1 and are put into urn \(2 .\) If 3 balls are then randomly selected from urn \(2,\) compute the expected number of white balls in the trio. Hint: Let \(X_{i}=1\) if the \(i\) th white ball initially in urn 1 is one of the three selected, and let \(X_{i}=0\) otherwise. Similarly, let \(Y_{i}=1\) if the \(i\) th white ball from urn 2 is one of the three selected, and let \(Y_{i}=0\) otherwise. The number of white balls in the trio can now be written as \(\sum_{1}^{5} X_{i}+\sum_{1}^{8} Y_{i}\)

If 101 items are distributed among 10 boxes, then at least one of the boxes must contain more than 10 items. Use the probabilistic method to prove this result.

The county hospital is located at the center of a square whose sides are 3 miles wide. If an accident occurs within this square, then the hospital sends out an ambulance. The road network is rectangular, so the travel distance from the hospital, whose coordinates are \((0,0),\) to the point \((x, y)\) is \(|x|+|y| .\) If an accident occurs at a point that is uniformly distributed in the square, find the expected travel distance of the ambulance.

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