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Chapter 4: Problem 20

A gambling book recommends the following "winning strategy" for the game of roulette: Bet \(\$ 1\) on red. If red appears (which has probability \(\frac{18}{38}\) ), then take the \(\$ 1\) profit and quit. If red does not appear and you lose this bet (which has probability \(\frac{20}{38}\) of occurring , make additional \(\$ 1\) bets on red on each of the next two spins of the roulette wheel and then quit. Let \(X\) denote your winnings when you quit. (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
(a) The probability of winning (P{X > 0}) is given by: \(P\\{X>0\\} = \frac{18}{38} + \frac{20}{38} \times \frac{18}{38} + \frac{20}{38} \times \frac{20}{38} \times \frac{18}{38}\) (b) The strategy can be considered a "winning" strategy if the probability of winning (P{X > 0}) is higher than the probability of not winning (1 - P{X > 0}). (c) The expected value of winnings (E[X]) is given by: E[X] = \(\frac{18}{38} \times 1 + \frac{20}{38} \times \frac{18}{38} \times 0 + \frac{20}{38} \times \frac{20}{38} \times \frac{18}{38} \times (-1)\)

Step by step solution

01

Calculate the probabilities of each scenario

1. Winning on the first spin: The probability of winning on the first spin is equal to the probability of getting red, which is \(\frac{18}{38}\). 2. Losing on the first spin, but winning on the second spin: The probability for this scenario is the product of the probability of not getting red on the first spin and the probability of getting red on the second spin, which is \(\frac{20}{38} \times \frac{18}{38}\). 3. Losing on the first and second spins, but winning on the third spin: The probability for this scenario is the product of the probability of not getting red on the first spin, not getting red on the second spin, and then getting red on the third spin, which is \(\frac{20}{38} \times \frac{20}{38} \times \frac{18}{38}\).
02

Calculate the probability of winning (P{X > 0})

Add up the probabilities of the three scenarios to find the probability of winning: \(P\\{X>0\\} = \frac{18}{38} + \frac{20}{38} \times \frac{18}{38} + \frac{20}{38} \times \frac{20}{38} \times \frac{18}{38}\)
03

Determine if the strategy is a "winning" strategy

To determine if the strategy is a "winning" strategy, compare the probability of winning (P{X > 0}) with the probability of not winning, which is 1 - P{X > 0}. If the probability of winning is higher than the probability of not winning, then the strategy can be considered a "winning" strategy.
04

Calculate the expected value of winnings (E[X])

To calculate the expected value of winnings, multiply the probability of each scenario by the winnings in each scenario and then add up the results: E[X] = (Probability of winning on the first spin 脳 winnings in the first scenario) + (Probability of winning on the second spin 脳 winnings in the second scenario) + (Probability of winning on the third spin 脳 winnings in the third scenario) E[X] = \(\frac{18}{38} \times 1 + \frac{20}{38} \times \frac{18}{38} \times 0 + \frac{20}{38} \times \frac{20}{38} \times \frac{18}{38} \times (-1)\) Calculate E[X] to find the expected value of winnings.

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Key Concepts

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

Conditional Probability
Understanding the concept of conditional probability is essential when evaluating strategies in games of chance like roulette. To put it simply, conditional probability is the likelihood of an event or outcome occurring, given that another event has already occurred.

In the context of the roulette strategy mentioned in the exercise, looking at the probability of winning on the second or third spin is a classic example of conditional probability in action. After losing the first bet, the subsequent bet's outcome hinges on the condition that the previous spin did not turn up red. Therefore, to calculate the overall probability of winning across multiple spins, one must account for these conditional probabilities鈥攕uch as the likelihood of winning on the second try given the first attempt was a loss. By mastering conditional probability, students can better understand and critique betting strategies that rely on outcomes of previous events.
Expected Value Calculation
The expected value calculation is a cornerstone in probability theory, representing the average outcome one can expect from an event if it were repeated many times. For a roulette player using the recommended strategy, calculating the expected value (often denoted as E[X]) of their winnings allows them to evaluate the strategy's 'long-term' performance.

Similar to the step-by-step solution, the expected value is found by multiplying each potential payout by the probability of its corresponding scenario and then summing those products. The final figure can either be positive or negative, which denotes whether the strategy can be considered profitable or not. If the expected value of the winnings (E[X]) is greater than zero, the strategy might be considered 'winning' over time. However, it's important to remember that a positive expected value doesn't guarantee profit in every session, as outcomes can still vary widely in the short term.
Probability Theory
Probability theory is the branch of mathematics that deals with predicting the likelihood of various outcomes. It allows us to make informed guesses about future events based on mathematical principles. When applied to games such as roulette, probability theory can be used to analyze different betting strategies and determine their effectiveness.

In the problem from the textbook, probability theory is employed to calculate the chances of winning at different stages of the game and evaluate the overall success rate of the proposed strategy. This theory is fundamental in understanding random processes, and its principles are utilized in calculating both the cumulative probability of victory (P{X>0}) and the expected monetary gain or loss from following such a strategy (E[X]). A comprehensive understanding of probability theory is crucial for students who wish to delve into the complexities of stochastic events and analyze the risk and reward in decision-making scenarios.

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Most popular questions from this chapter

An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly \(n\) selections?

There are two possible causes for a breakdown of a machine. To check the first possibility would cost \(C_{1}\) dollars, and, if that were the cause of the breakdown, the trouble could be repaired at a cost of \(R_{1}\) dollars. Similarly, there are costs \(C_{2}\) and \(R_{2}\) associated with the second possibility. Let \(p\) and \(1-\) \(p\) denote, respectively, the probabilities that the breakdown is caused by the first and second possibilities. Under what conditions on \(p, C_{i}, R_{i}, i=1,2\) should we check the first possible cause of breakdown and then the second, as opposed to reversing the checking order, so as to minimize the expected cost involved in returning the machine to working order? Note: If the first check is negative, we must still check the other possibility.

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers 1 through \(80 .\) A player can select from 1 to 15 numbers; a win occurs if some fraction of the player's chosen subset matches any of the 20 numbers drawn by the house. The payoff is a function of the number of elements in the player's selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of \(20,\) and the payoff is \(\$ 2.2\) won for every dollar bet. (As the player's probability of winning in this case is \(\frac{1}{4},\) it is clear that the "fair" payoff should be \(\$ 3\) won for every \(\$ 1\) bet.) When the player selects 2 numbers, a payoff (of odds) of \(\$ 12\) won for every \(\$ 1\) bet is made when both numbers are among the 20 (a) What would be the fair payoff in this case? Let \(P_{n, k}\) denote the probability that exactly \(k\) of the \(n\) numbers chosen by the player are among the 20 selected by the house. (b) Compute \(P_{n, k}\) (c) The most typical wager at Keno consists of selecting 10 numbers. For such a bet the casino pays off as shown in the following table. Compute the expected payoff: $$\begin{array}{cc} \hline \multicolumn{2}{c} {\text { Keno Payoffs in 10 Number Bets }} \\ \hline \text { Number of matches } & \text { Dollars won for each \$1 bet } \\\ \hline 0-4 & -1 \\ 5 & 1 \\ 6 & 17 \\ 7 & 179 \\ 8 & 1,299 \\ 9 & 2,599 \\ 10 & 24,999 \\ \hline \end{array}$$

A salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to a sale with probability \(.3,\) and his second will lead independently to a sale with probability .6. Any sale made is equally likely to be either for the deluxe model, which costs \(\$ 1000\), or the standard model, which costs \(\$ 500 .\) Determine the probability mass function of \(X,\) the total dollar value of all sales.

Suppose that it takes at least 9 votes from a 12 member jury to convict a defendant. Suppose also that the probability that a juror votes a guilty person innocent is \(.2,\) whereas the probability that the juror votes an innocent person guilty is .1. If each juror acts independently and if 65 percent of the defendants are guilty, find the probability that the jury renders a correct decision. What percentage of defendants is convicted?
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