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Chapter 6: Problem 4

In Las Vegas the roulette wheel has a 0 and a 00 and then the numbers 1 to 36 marked on equal slots; the wheel is spun and a ball stops randomly in one slot. When a player bets 1 dollar on a number, he receives 36 dollars if the ball stops on this number, for a net gain of 35 dollars; otherwise, he loses his dollar bet. Find the expected value for his winnings.

Short Answer

Expert verified
The expected value of the player's winnings is \(-\frac{1}{19}\) or approximately -5.26 cents per dollar bet.

Step by step solution

01

Understanding the Game Setup

First, recognize that a roulette wheel in Las Vegas contains 38 slots: numbers 0, 00, and 1 through 36. Each slot has an equal probability of landing a ball.
02

Probability of Winning

A player wins if the ball lands on the number they bet on. Since there are 38 slots and only one is the winning number, the probability of winning is \( \frac{1}{38} \).
03

Probability of Losing

The player loses if the ball lands on any slot other than the one they bet on. The probability of losing is \( \frac{37}{38} \), as there are 37 losing slots.
04

Calculating the Winning Winnings

If the player wins, they receive a payout of 36 dollars. So, the net gain for winning is 36 dollars minus the 1 dollar bet, totaling 35 dollars.
05

Calculating the Losing Amount

If the player loses, they lose their 1 dollar bet, resulting in a net loss of 1 dollar.
06

Computing Expected Value

The expected value (EV) is calculated by multiplying each outcome by its probability and summing the results:\[ EV = \left( \text{Probability of Winning} \times \text{Winning Amount} \right) + \left( \text{Probability of Losing} \times \text{Losing Amount} \right) \]Plugging in the values, we get:\[ EV = \left( \frac{1}{38} \times 35 \right) + \left( \frac{37}{38} \times (-1) \right) \]\[ EV = \frac{35}{38} + \left( -\frac{37}{38} \right) \]\[ EV = \frac{35 - 37}{38} \]\[ EV = \frac{-2}{38} \]\[ EV = -\frac{1}{19} \]
07

Interpreting the Expected Value

The expected value of \(-\frac{1}{19}\) means that on average, the player loses about 5.26 cents per dollar bet each time they play.

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Key Concepts

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

Understanding Probability of Winning
When betting on a roulette wheel, the key to understanding your odds lies in the concept of probability. Specifically, the **probability of winning** is crucial when determining your potential success. In roulette, there are 38 equally likely slots where the ball can land: numbers 1 through 36, plus 0 and 00. This means that if you bet on one specific number, the chance of the ball stopping on your number is just 1 out of 38, or \( \frac{1}{38} \). Imagine placing your bet on the number 7. Only if the ball lands on slot 7, do you win. Because of the many numbers to choose from and just one chance of hitting your chosen number, the probability remains very low.Understanding this concept helps you realize why roulette, like many casino games, favors the house. Knowing your probability of winning can help set realistic expectations and decisions when placing a bet.
Grasping Probability of Losing
Just as in any game of luck, the likelihood of not winning is quite high. In roulette, understanding the **probability of losing** provides insight into how often you'll walk away without a win. Since there are 38 slots and only one can make you a winner, there are 37 possible slots that will result in a loss. This brings us to the probability of losing, which is \( \frac{37}{38} \). If you imagine yourself betting on a single number, many other slots would lead to loss, stacking the odds against you. The vast majority of time, the ball will land in one of these losing slots, making it essential to appreciate the high probability of losing. This knowledge serves to temper expectations and guide strategic betting.
Understanding Net Gain
The concept of **net gain** or net result is a pivotal aspect when discussing bets and payouts in games like roulette. But what does net gain really mean? Simply put, net gain is the total money you make or lose from your bet after accounting for both wins and losses. In the scenario where you win, the casino pays you 36 dollars for your winning number. However, since you initially bet 1 dollar, your net gain is actually 35 dollars. On the flip side, if you lose, your loss is straightforward: you lose the 1 dollar you have wagered, amounting to a net loss of 1 dollar. Calculating net gain is essential to understanding your financial outcome from gambling. It speaks to how much money remains after a series of bets and helps inform whether the game is profitable in the long run.

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Most popular questions from this chapter

Exactly one of six similar keys opens a certain door. If you try the keys, one after another, what is the expected number of keys that you will have to try before success?

On September \(26,1980,\) the New York Times reported that a mysterious stranger strode into a Las Vegas casino, placed a single bet of 777,000 dollars on the "don't pass" line at the crap table, and walked away with more than 1.5 million dollars. In the "don't pass" bet, the bettor is essentially betting with the house. An exception occurs if the roller rolls a 12 on the first roll. In this case, the roller loses and the "don't pass" better just gets back the money bet instead of winning. Show that the "don't pass" bettor has a more favorable bet than the roller.

Write a program to add random numbers chosen from [0,1] until the first time the sum is greater than one. Have your program repeat this experiment a number of times to estimate the expected number of selections necessary in order that the sum of the chosen numbers first exceeds \(1 .\) On the basis of your experiments, what is your estimate for this number?

Let \(X\) be a random variable with \(\mu=E(X)\) and \(\sigma^{2}=V(X) .\) Define \(X^{*}=\) \((X-\mu) / \sigma\). The random variable \(X^{*}\) is called the standardized random variable associated with \(X\). Show that this standardized random variable has expected value 0 and variance 1 .

Let \(X\) be a random variable with \(E(X)=\mu\) and \(V(X)=\sigma^{2}\). Show that the function \(f(x)\) defined by $$f(x)=\sum_{\omega}(X(\omega)-x)^{2} p(\omega)$$ has its minimum value when \(x=\mu\).
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