Expected Value and Probability Decisions

Expected value is a fundamental concept in probability that helps us predict the long-term average outcome of a random event. By calculating the expected value, we can make mathematically informed decisions about investments, insurance, lotteries, and business strategies.

What is Expected Value?

The expected value, denoted as $E(X)$, is the sum of all possible outcomes of a random variable, each multiplied by its probability of occurring.

The formula is: $E(X) = x_1 P(x_1) + x_2 P(x_2) + \dots + x_n P(x_n) = \sum_{i=1}^{n} x_i P(x_i)$

Where:

• $x_i$ is a specific numerical outcome.
• $P(x_i)$ is the probability of that outcome occurring.

Making Decisions and Fair Games

When evaluating a financial decision or a game of chance, we often look at the expected net gain. This is the expected value of your profit (winnings minus the cost to play).

• If $E(X) > 0$, the decision is profitable in the long run.
• If $E(X) < 0$, you will lose money in the long run.
• If $E(X) = 0$, the situation is considered a fair game. In a fair game, neither side has a mathematical advantage.

Example 1: Evaluating a Lottery

Problem: In a lottery, 1000 tickets are sold at \2$each. There is one prize of$$500$and five prizes of$$50$. Find the expected net gain of buying one ticket.

Solution: First, determine the net gain ($x$) and probability ($P(x)$) for every possible outcome.

1. Win the Grand Prize:
   • Net gain: \500 - $2 = $498$
   • Probability: $\frac{1}{1000} = 0.001$
2. Win a Small Prize:
   • Net gain: \50 - $2 = $48$
   • Probability: $\frac{5}{1000} = 0.005$
3. Lose (Win Nothing):
   • Net gain: \0 - $2 = -$2$
   • Probability: $\frac{1000 - 1 - 5}{1000} = \frac{994}{1000} = 0.994$

Now, calculate the expected net gain: $E(X) = (498)(0.001) + (48)(0.005) + (-2)(0.994)$ $E(X) = 0.498 + 0.240 - 1.988$ $E(X) = -1.25$

Conclusion: The expected net gain is -\1.25$. On average, you lose$$1.25$for every ticket you buy. Because$E(X) \neq 0$, this is not a fair game.

Example 2: The Insurance Decision

Problem: Should you insure a \200$item if insurance costs$$15$and the probability of losing the item is$P(\text{loss}) = 0.05$?

Solution: To make this decision, compare the guaranteed cost of insurance against the expected financial loss if you remain uninsured.

Option A: Buy Insurance

• Your cost is fixed at -\15$.

Option B: Do Not Buy Insurance Let $X$ represent your financial change.

• Item is lost: Outcome is -\200$, with$P = 0.05$.
• Item is safe: Outcome is \0$, with$P = 0.95$.

Calculate the expected value of remaining uninsured: $E(X) = (-200)(0.05) + (0)(0.95)$ $E(X) = -10 + 0 = -10$

Conclusion: The expected cost of not buying insurance is \10$, which is less than the$$15$ cost of buying the insurance. From a strictly mathematical standpoint, you should not insure the item, as your expected loss is smaller without it. (Note: Insurance companies rely on this principle; they charge a premium higher than the expected loss to ensure their own expected net gain is positive)..