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Chapter 4: Problem 20

A gambling book recommends the following "winning strategy" for the game of roulette. It recommends that a gambler bet \(\$ 1\) on red. If red appears (which has probability \(\left.\frac{18}{38}\right)\), then the gambler should take her \(\$ 1\) profit and quit. If the gambler loses this bet (which has probability \(\frac{20}{38}\) of occurring), she should make additional \(\$ 1\) bets on red on each of the next two spins of the roulette wheel and then quit. Let \(X\) denote the gambler's winnings when she quits. (a) Find \(P\\{X>0\\}\). (b) Are you convinced that the strategy is indeed a "winning" strategy? Explain your answer! (c) Find \(E[X]\)

Short Answer

Expert verified
\(P\\{X>0\\} \approx 0.6694\), which means the gambler has a 66.94% chance of winning. However, the expected winnings are \(E[X] \approx -0.0536\), indicating that the gambler is expected to lose money in the long run. Therefore, this strategy is not a winning strategy as it results in negative expected value.

Step by step solution

01

Probabilities and scenarios

First, let's list out the possible scenarios of the strategy along with their probabilities: 1. Gambler wins on the first spin: Probability: \(\frac{18}{38}\) 2. Gambler loses on the first spin, wins on the second spin: Probability: \(\frac{20}{38} \times \frac{18}{38}\) 3. Gambler loses on the first two spins, wins on the third spin: Probability: \(\frac{20}{38} \times \frac{20}{38} \times \frac{18}{38}\) 4. Gambler loses on all three spins: Probability: \(\frac{20}{38} \times \frac{20}{38} \times \frac{20}{38}\)
02

Determine P{X>0}

To find the probability that the gambler has positive winnings, we need to sum the probabilities of scenario 1, 2, and 3, since these are the scenarios in which she wins: \(P\\{X>0\\} = \frac{18}{38} + \frac{20}{38} \times \frac{18}{38} + \frac{20}{38} \times \frac{20}{38} \times \frac{18}{38}\) Now, we calculate this probability: \(P\\{X>0\\} = \frac{18}{38} + \frac{20 \times 18}{38^2} + \frac{20 \times 20 \times 18}{38^3} \approx 0.6694\)
03

Evaluate the strategy as "winning"

This strategy gives the gambler a 66.94% chance of winning. However, to determine if the strategy is indeed a "winning" strategy, we need to consider the expectation of the gambler's winnings.
04

Find E[X]

We need to find the expected winnings for each scenario: 1. Gambler wins on the first spin: Winnings: $1 Probability: \(\frac{18}{38}\) 2. Gambler loses on the first spin, wins on the second spin: Winnings: \(0 (since she loses \)1 and then wins $1) Probability: \(\frac{20}{38} \times \frac{18}{38}\) 3. Gambler loses on the first two spins, wins on the third spin: Winnings: \(-1 (since she loses \)2 and then wins $1) Probability: \(\frac{20}{38} \times \frac{20}{38} \times \frac{18}{38}\) 4. Gambler loses on all three spins: Winnings: $-3 Probability: \(\frac{20}{38} \times \frac{20}{38} \times \frac{20}{38}\) Now, we calculate the expected winnings: \(E[X] = (1) \times (\frac{18}{38}) + (0) \times (\frac{20}{38} \times \frac{18}{38}) + (-1) \times (\frac{20}{38} \times \frac{20}{38} \times \frac{18}{38}) + (-3) \times (\frac{20}{38} \times \frac{20}{38} \times \frac{20}{38}) \approx -0.0536\) The expected winnings following this strategy are approximately $-0.0536, which means the gambler is expected to lose money in the long run. This strategy is not a winning strategy as it results in negative expected value.

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Key Concepts

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

Expected Value in Gambling
When you engage in any gambling activity, like betting on a roulette spin, you're dealing with random events and uncertainty. The expected value is a concept that can provide insight into the long-term average outcome of these random events. It reflects what you can expect to win or lose on average if you were to repeat the bet many times.

In gambling, the formula for expected value (\( E[X] \)) is given by the sum of all possible outcomes, each multiplied by their respective probability and the amount won or lost. It is calculated with the formula: \[ E[X] = \text{Sum of} (\text{probability of outcome}) \times (\text{winnings for outcome}) \]
Considering the roulette example from the exercise, the expected value was found to be approximately $-0.0536 per game. This implies that for every dollar bet, the gambler is statistically likely to lose about 5.36 cents. This concept is crucial as it reveals the inherent disadvantage faced by players in most gambling games, driven by the odds set by the gambling institution.
Probabilities of Independent Events
The concept of probabilities of independent events is vital in understanding games like roulette. In statistics, two events are independent if the occurrence of one event does not affect the occurrence of the other. Each spin of the roulette wheel is an independent event; the previous spins do not influence the outcome of the next spin.

In practice, calculating the combined probability of independent events involves multiplying the individual probabilities of each event. For instance, the probability that a gambler will lose the first spin but win the second (as proposed in the exercise's strategy) is calculated as: \[ \left(\frac{20}{38}\right) \times \left(\frac{18}{38}\right) \]
It's essential to recognize that, due to independence, the roulette wheel has no 'memory'. Regardless of past outcomes, the probability of winning or losing any given spin remains constant. This principle helps in evaluating the effectiveness of betting strategies, as it underlines the futility of trying to predict outcomes based on past results.
Roulette Betting Strategies
Roulette betting strategies often involve patterns of betting that aim to take advantage of the game's odds. The strategy discussed in the exercise鈥攂etting on red after losses鈥攊s known as a type of Martingale strategy. It operates under the assumption that after a loss, you are somehow 'due' for a win.

However, due to the independence of spins in roulette, this strategy does not increase the probability of winning. Each spin is a separate event with a fixed probability of winning. The reality is, no betting system can overcome the house edge inherent in the design of the roulette wheel.

Limitations of Betting Strategies

Some common misconceptions with roulette strategies include the belief that:
  • Certain numbers are 'hot' or 'cold' based on past spins
  • Increase in bet size can recoup past losses
  • A pattern or sequence gives insight into future results

A sound understanding of expected value and the independent nature of roulette spins can help dispel these myths. In the long run, the house edge will prevail, and the most effective strategy is often to manage your bankroll wisely, set a loss limit, and enjoy the game for its entertainment value rather than as a source of income.

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Most popular questions from this chapter

A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample.

In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess. correctly or if neither guesses correctly, then no money is exchanged. Consider a specified player and denote by \(X\) the amount of money he wins in a single game of two-finger Morra. (a) If each player acts independently of the other, and if each player makes his choice of the number of fingers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the 4 possibilities is equally likely, what are the possible values of \(X\) and what are their associated probabilities? (b) Suppose that each player acts independently of the other. If each player decides to hold up the same number of fingers that he guesses his opponent will hold up, and if each player is equally likely to hold up \(\mathrm{I}\) or 2 fingers, what are the possible values of \(X\) and their associated probabilities?

Five distinct numbers are randomly distributed to players numbered 1 through 5\. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2 compare their numbers; the winner then compares with player 3, and so on. Let \(X\) denote the number of times player 1 is a winner. Find \(P\\{X=i\\}, i=0,1,2,3,4\).

The number of times that a person contracts a cold in a given year is a Poisson random variable with parameter \(\lambda=5\). Suppose that a new wonder drug (based on large quantities of vitamin \(\mathrm{C}\) ) has just been marketed that reduces the Poisson parameter to \(\lambda=3\) for 75 percent of the population. For the other 25 percent of the population the drug has no appreciable effect on colds. If an individual tries the drug for a year and has 2 colds in that time, how likely is it that the drug is heneficial for him or her?

Two fair dice are rolled. Let \(X\) equal the product of the 2 dice. Compute \(P\\{X=i\\}\) for \(i=1,2, \ldots\)
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