91Ó°ÊÓ

When working with dice or any random phenomena, it is essential to understand probability distribution. This is a mathematical function that provides the probabilities of occurrence of different possible outcomes. For our specially loaded die, this means each face does not have an equal chance of appearing when rolled. Instead, the probability of landing a particular face is directly proportional to the number on that face.

To create a valid probability distribution, the sum of all probabilities must equal 1. In this problem:

• For face 1: Probability is proportional to 1, denoted as \(k \cdot 1\).
• For face 2: Probability is proportional to 2, denoted as \(k \cdot 2\).
• Continuing similarly up to face 6: Probability is \(k \cdot 6\).

This gives us an equation: \(k(1+2+3+4+5+6) = 1\). Solving, we find \(k = \frac{1}{21}\). This means:

• Probability for face 1: \(\frac{1}{21}\)
• Probability for face 2: \(\frac{2}{21}\)
• ... and so on, up to face 6.

By setting up the probability in this way, we have a custom distribution where larger numbers are more likely to appear when rolling the die.

The expected value, often known as the mean or average, is a key concept in probability that tells us the average outcome if we were to repeat an experiment many times. This is particularly important in scenarios like this loaded die, where outcomes are weighted differently.

To find the expected value, we multiply each possible outcome by its probability and sum all those results. Using our custom probability distribution:

• The face rolled can be 1 with probability \(\frac{1}{21}\), so it contributes \(1 \times \frac{1}{21}\).
• Face 2 contributes \(2 \times \frac{2}{21}\), 3 contributes \(3 \times \frac{3}{21}\), and so on.

The calculation becomes:\[ E(X) = \frac{1}{21}(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) \]Evaluating the sums, it works out to be \(\frac{91}{21} = \frac{13}{3}\).

Therefore, when rolling this loaded die, the average outcome is \(\frac{13}{3}\) or approximately 4.33. This is higher than the average face value of an ordinary fair die because of the increased probabilities attached to larger numbers.

Variance is a statistical measurement that describes the spread of numbers in a data set—in this case, how much the die roll outcomes are spread out from the expected value. It is especially relevant here because our die is not fair, making the variance a useful tool for measuring the consistency of outcomes.

The formula for variance is \(V(X) = E(X^2) - (E(X))^2\). We have already found \(E(X)\), so next, we calculate \(E(X^2)\), the expected value of the squares of the outcomes:

• Square the face value and multiply by its probability.
• Add all these values for faces 1 through 6.

\[ E(X^2) = \frac{1}{21}(1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3) \]This simplifies to 21. Now, substituting into the variance formula gives:\[ V(X) = 21 - \left(\frac{13}{3}\right)^2 = \frac{20}{9} \]

This value, \(V(X) = \frac{20}{9}\), demonstrates the extent to which the outcomes deviate from the average. A higher variance in this context means a more "unpredictable" result with spread out outcomes, but our specific scenario has a moderate level of spread.