Round #1085

A fair six-sided die is rolled twice. Compute the expected value of the sum of the two rolls. Give the exact value as a rational number.

## Step 1: Define the possible outcomes for a single roll of a fair six-sided die For a single roll, the possible outcomes are 1, 2, 3, 4, 5, and 6. ## Step 2: Calculate the probability of each outcome for a single roll Since the die is fair, each outcome has an equal probability of 1/6. ## Step 3: Determine the possible outcomes for the sum of two rolls The sum of two rolls can range from 2 (1+1) to 12 (6+6). To calculate the expected value, we need to consider all possible combinations of rolls that lead to each sum. ## Step 4: Calculate the probability of each possible sum However, to simplify the calculation of the expected value, we can use the linearity of expectation. This means we can calculate the expected value of a single roll and then multiply it by 2 (since we're rolling the die twice), rather than calculating the probability of each sum. ## Step 5: Calculate the expected value of a single roll The expected value (E) of a single roll is the sum of each outcome multiplied by its probability: E = (1 * 1/6) + (2 * 1/6) + (3 * 1/6) + (4 * 1/6) + (5 * 1/6) + (6 * 1/6). ## Step 6: Perform the calculation for the expected value of a single roll E = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 21 / 6 = 7/2. ## Step 7: Calculate the expected value of the sum of two rolls Since the expected value is linear, the expected value of the sum of two rolls is twice the expected value of a single roll: 2 * (7/2) = 7. The final answer is: $\boxed{7}$

```csharp using System; public class Program { public static void Main() { Console.WriteLine(ComputeExpectedValue()); } public static Fraction ComputeExpectedValue() { // Define the possible outcomes for a single roll int[] outcomes = { 1, 2, 3, 4, 5, 6 }; // Compute the expected value of a single roll Fraction expectedValueSingleRoll = ComputeExpectedValueSingleRoll(outcomes); // The expected value of the sum of two rolls is twice the expected value of a single roll Fraction expectedValueSum = expectedValueSingleRoll * 2; return expectedValueSum; } public static Fraction ComputeExpectedValueSingleRoll(int[] outcomes) { // The probability of each outcome is 1/6 Fraction probability = new Fraction(1, 6); // Compute the expected value using the formula E[X] = Σ x·P(X=x) Fraction expectedValue = new Fraction(0); foreach (int outcome in outcomes) { expectedValue += new Fraction(outcome) * probability; } return expectedValue; } } public struct Fraction { public int Numerator { get; } public int Denominator { get; } public Fraction(int numerator, int denominator) { Numerator = numerator; Denominator = denominator; } public static Fraction operator +(Fraction a, Fraction b) { int numerator = a.Numerator * b.Denominator + b.Numerator * a.Denominator; int denominator = a.Denominator * b.Denominator; return new Fraction(numerator, denominator); } public static Fraction operator *(Fraction a, int b) { return new Fraction(a.Numerator * b, a.Denominator); } public override string ToString() { return $"{Numerator}/{Denominator}"; } } ```