1) Introduction to Expected Value ๐ฐ๐ฒ
Welcome back to Advanced Probability & Combinatorics! In Part 3, we explored conditional probability and independent events, learning how probabilities change with new information and how events relate to each other. Now, in Part 4, we will introduce a powerful concept called Expected Value.
Expected Value is a way to calculate the average outcome you can expect from a probabilistic situation over the long run. It is particularly useful in decision-making under uncertainty, such as in games of chance, investments, and business decisions. Expected value helps us to determine what is "fair" in a game, or to compare different options in terms of their average payoff.
Expected Value (Expectation) of a random variable is the average outcome of an experiment you would expect to see over many repetitions. It is calculated by summing up the product of each possible outcome and its probability.
In this topic, we will cover:
- Understanding the concept of Expected Value and its significance.
- Learning how to calculate the expected value for various scenarios.
- Interpreting expected value in the context of games of chance and decision-making.
- Exploring examples and applications of expected value in real-world situations.
Let's dive into expected value and learn how to make informed decisions when faced with uncertainty! ๐
2) Calculating Expected Value ๐งฎ
2.1 The Formula for Expected Value
To calculate the expected value of a random variable, we need to know all possible outcomes and their associated probabilities. If we let \(X\) be a random variable representing the outcomes of an experiment, and \(x_1, x_2, \ldots, x_n\) be the possible values of \(X\) with corresponding probabilities \(P(X=x_1), P(X=x_2), \ldots, P(X=x_n)\), then the expected value of \(X\), denoted as \(E(X)\) or \( \mu \), is given by:
\( E(X) = x_1 \cdot P(X=x_1) + x_2 \cdot P(X=x_2) + \ldots + x_n \cdot P(X=x_n) \)
In more compact summation notation, this is:
\( E(X) = \sum_{i=1}^{n} x_i \cdot P(X=x_i) \)
Where:
- \(E(X)\) is the expected value of the random variable \(X\).
- \(x_1, x_2, \ldots, x_n\) are the possible outcomes.
- \(P(X=x_i)\) is the probability of each outcome \(x_i\).
- \(n\) is the total number of possible outcomes.
The expected value is a weighted average of the possible outcomes, where each outcome is weighted by its probability.
Example 1: Expected Value of a Die Roll
What is the expected value when you roll a fair six-sided die?
**Solution:**
- Possible outcomes when rolling a die: \( \{1, 2, 3, 4, 5, 6\} \).
- Since the die is fair, each outcome has a probability of \( \frac{1}{6} \).
Using the expected value formula: \( E(X) = \sum_{i=1}^{n} x_i \cdot P(X=x_i) \)
\( E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) + 4 \cdot P(X=4) + 5 \cdot P(X=5) + 6 \cdot P(X=6) \)
\( E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \)
\( E(X) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \)
The expected value of rolling a fair six-sided die is 3.5. Note that 3.5 is not a possible outcome itself, but it represents the average value you would expect to get over many rolls.
Example 2: Expected Value in a Simple Lottery
Consider a lottery where you pick one number from 1 to 10. If your number matches the winning number, you win \$10. If it doesn't, you lose \$1. What is the expected value of playing this lottery once?
**Solution:**
- Possible outcomes and payoffs:
- Win (number matches): Payoff = +\$10. Probability of winning = \( \frac{1}{10} \).
- Lose (number doesn't match): Payoff = -\$1 (cost of playing). Probability of losing = \( \frac{9}{10} \).
Using the expected value formula, with payoffs as outcomes: \( E(X) = \sum_{i=1}^{n} x_i \cdot P(X=x_i) \)
\( E(\text{Lottery}) = (\text{Payoff for Win}) \cdot P(\text{Win}) + (\text{Payoff for Lose}) \cdot P(\text{Lose}) \)
\( E(\text{Lottery}) = (\$10) \cdot \frac{1}{10} + (-\$1) \cdot \frac{9}{10} \)
\( E(\text{Lottery}) = \$1 - \$0.9 = \$0.1 \)
The expected value of playing this lottery once is \$0.10 or 10 cents. This means that on average, for each ticket you buy, you can expect to gain 10 cents in the long run. A positive expected value suggests the game is favorable to the player in the long term.
3) Interpreting Expected Value: Fair Games and Decision Making โ๏ธ
3.1 Fair Games
In the context of games of chance, the concept of expected value is closely tied to the idea of a fair game.
A Fair Game is a game where the expected value is zero. In a fair game, in the long run, neither the player nor the game organizer is expected to profit or lose on average.
If the expected value of a game is positive for a player, it means the game is favorable to the player in the long run (on average, the player gains). If the expected value is negative, the game is unfavorable to the player (on average, the player loses), and favorable to the game organizer.
Example 3: Is the Lottery a Fair Game?
In Example 2, the expected value of the lottery was \$0.10 (10 cents). Is this lottery a fair game?
**Solution:**
Since the expected value is \$0.10, which is greater than zero, the expected value is not zero. Therefore, this lottery is not a fair game. It is slightly favorable to the player in the long run (though the profit is small, just 10 cents per play on average). However, typically, lotteries are designed to have a negative expected value for the player, making them profitable for the organizers.
3.2 Expected Value in Decision Making
Expected value is not just for games; itโs a powerful tool for making decisions in any situation involving uncertainty. When faced with choices that have probabilistic outcomes and associated values (like costs or benefits), we can use expected value to compare options and make more rational decisions.
Example 4: Decision Making - Investment Opportunity
Suppose you have an investment opportunity. There is a 60% chance it will yield a profit of \$5,000, and a 40% chance it will result in a loss of \$2,000. What is the expected value of this investment? Should you invest based solely on expected value?
**Solution:**
- Possible outcomes and values:
- Profit: Value = +\$5,000. Probability = 0.60.
- Loss: Value = -\$2,000. Probability = 0.40.
Using the expected value formula: \( E(X) = \sum_{i=1}^{n} x_i \cdot P(X=x_i) \)
\( E(\text{Investment}) = (\$5,000) \cdot 0.60 + (-\$2,000) \cdot 0.40 \)
\( E(\text{Investment}) = \$3,000 - \$800 = \$2,200 \)
The expected value of this investment is \$2,200. Since the expected value is positive, it suggests that, on average, this investment is likely to be profitable in the long run. Based solely on expected value, it would be a rational decision to invest. However, in real-world decision making, factors beyond just expected value (like risk tolerance, need for immediate capital, etc.) also play a crucial role. Expected value provides a quantitative basis for decision making under uncertainty, but it is not the only factor to consider.
4) Examples (Detailed) ๐
Example 5: Expected Value with Multiple Outcomes - Raffle
Let's consider a more complex example involving multiple outcomes and expected value.
Challenge 1: In a raffle, 1000 tickets are sold at \$1 each. There is one grand prize of \$500, two 2nd prizes of \$100 each, and five 3rd prizes of \$20 each. What is the expected value of buying one ticket?
**Let's solve the challenge:**
Solution to Challenge 1: Expected Value of Raffle Ticket
First, we need to identify all possible outcomes and their net payoffs (prize minus ticket cost) and probabilities.
- Outcome 1: Win Grand Prize. Prize = \$500. Net payoff = \$500 - \$1 (ticket cost) = \$499. Probability = \( \frac{1}{1000} \) (1 grand prize out of 1000 tickets).
- Outcome 2: Win 2nd Prize. Prize = \$100. Net payoff = \$100 - \$1 = \$99. Probability = \( \frac{2}{1000} \) (2 second prizes out of 1000 tickets).
- Outcome 3: Win 3rd Prize. Prize = \$20. Net payoff = \$20 - \$1 = \$19. Probability = \( \frac{5}{1000} \) (5 third prizes out of 1000 tickets).
- Outcome 4: Win No Prize (Lose). Prize = \$0. Net payoff = \$0 - \$1 = -\$1. Probability = \( 1 - (\frac{1}{1000} + \frac{2}{1000} + \frac{5}{1000}) = 1 - \frac{8}{1000} = \frac{992}{1000} \) (remaining tickets are losers).
Now we calculate the expected value using the formula: \( E(X) = \sum_{i=1}^{n} x_i \cdot P(X=x_i) \).
\( E(\text{Raffle Ticket}) = (\$499) \cdot \frac{1}{1000} + (\$99) \cdot \frac{2}{1000} + (\$19) \cdot \frac{5}{1000} + (-\$1) \cdot \frac{992}{1000} \)
\( E(\text{Raffle Ticket}) = \frac{499 + 198 + 95 - 992}{1000} = \frac{792 - 992}{1000} = \frac{-200}{1000} = -\$0.20 \)The expected value of buying one raffle ticket is -\$0.20 or -20 cents. This means that on average, for each ticket you buy, you are expected to lose 20 cents. This is a negative expected value for the player, making the raffle favorable to the organizers (who collect \$1 per ticket and on average pay out less in prizes per ticket sold).
5) Practice Questions ๐ฏ
5.1 Fundamental โ Build Skills
1. Define "Expected Value" in your own words. What does it represent?
2. Write down the formula for calculating the expected value of a discrete random variable. Explain each component of the formula.
3. Calculate the expected value when you flip a fair coin. Let Heads = 1, Tails = 0.
4. In a game, you win \$5 if you roll a 5 or 6 on a fair die, and lose \$2 if you roll 1, 2, 3, or 4. What is the expected value of playing this game once?
5. A random variable \(X\) has the following probability distribution: \(P(X=1) = 0.2, P(X=2) = 0.3, P(X=3) = 0.4, P(X=4) = 0.1\). Calculate \(E(X)\).
6. Is a game with an expected value of \$0 considered fair? Explain why.
7. You sell 200 raffle tickets at \$5 each for a prize worth \$600. What is the expected value if you buy one ticket?
8. What does a negative expected value in a game typically imply for the player?
9. Consider a spinner divided into 4 equal sections colored red, blue, green, yellow. You win \$10 if it lands on red, \$8 on blue, and \$5 on green, and lose \$3 if it lands on yellow. What is the expected winnings in one spin?
10. Explain how expected value can be used in making investment decisions.
5.2 Challenging โ Push Limits ๐ช๐
1. A company is considering launching a new product. There is a 60% chance of high demand which would lead to a profit of \$500,000, a 30% chance of medium demand leading to a profit of \$200,000, and a 10% chance of low demand resulting in a loss of \$100,000. What is the expected profit from launching this product?
2. In a lottery game, you choose 5 numbers from 1 to 50. The jackpot prize is \$1,000,000, and the probability of winning the jackpot is \( \frac{1}{2,000,000} \). A ticket costs \$2. What is the expected value of buying one lottery ticket? Is it a fair game?
3. Word Problem: An insurance company sells a one-year term life insurance policy to a 50-year-old woman. The policy is for \$100,000, and the annual premium is \$500. The probability of a 50-year-old woman dying within the next year is approximately 0.008. What is the insurance company's expected profit from selling this policy?
4. (Conceptual) Discuss the limitations of using expected value as the sole criterion for decision making. In what situations might relying only on expected value be misleading or insufficient? Consider factors like risk aversion.
5. (Advanced Problem) Consider a game where you roll a fair die repeatedly until you roll a 6. Let \(N\) be the number of rolls needed to get the first 6. (a) What is the probability that \(N=1, N=2, N=3\)? (b) (Challenge) Can you find a pattern and determine the expected value of \(N\)? (This introduces the concept of expected value for a geometric distribution).
6) Summary ๐
- Expected Value (Expectation) \(E(X)\) is the average outcome of a random experiment over many trials.
- Formula for Expected Value: \( E(X) = \sum_{i=1}^{n} x_i \cdot P(X=x_i) \), sum of (outcome ร probability of outcome) for all outcomes.
- A Fair Game has an expected value of zero.
- Expected value is a crucial tool for decision making under uncertainty, helping to compare options based on average outcomes.
- In real-world decisions, while expected value provides a quantitative measure, other factors like risk tolerance are also important.
Fantastic job! You've now learned about Expected Value, a cornerstone concept in probability and decision theory. You can now calculate expected values, interpret them in the context of fair games, and understand how they can be used in decision-making. As you continue to explore probability, you'll find expected value to be an invaluable tool in analyzing uncertain situations and making informed choices. Keep up the excellent work! ๐ Level 3 Probability and Combinatorics continues in the upcoming topics!
โ Back to Previous Topic (Part 3 - Conditional Probability) View Level 3 Topics Overview โ