Expected Value of a Weighted Die with Enhanced Probability of 6

In probability theory and statistics, the expected value (EV) of a random variable gives the long-run average result of repetitions of the experiment it represents. If we consider a die that is not fair and is weighted such that the number 6 occurs half the time, we can analyze how this affects the overall expected value of the die roll.

Probability Distribution

When dealing with a weighted die where the number 6 has a probability of P6 1/2, the remaining outcomes (1, 2, 3, 4, 5) must share the remaining probability equally. To determine these probabilities, we start with the total probability summing to 1:

P1 P2 P3 P4 P5 1 - P6 1 - 1/2 1/2

Since there are 5 remaining faces, each of these faces shares equally the remaining probability, resulting in:

P1 P2 P3 P4 P5 (1/2) / 5 1/10

Calculating the Expected Value

The expected value (EV) can be calculated using the formula:

EV sum (x * P(x))

Plugging in the values for a weighted die:

For 1: 1 * P1 1 * (1/10) 1/10 For 2: 2 * P2 2 * (1/10) 2/10 For 3: 3 * P3 3 * (1/10) 3/10 For 4: 4 * P4 4 * (1/10) 4/10 For 5: 5 * P5 5 * (1/10) 5/10 For 6: 6 * P6 6 * (1/2) 6/2 3

Summing these up, we get:

EV (1/10) (2/10) (3/10) (4/10) (5/10) 3

EV (15/10) 3 1.5 3 4.5

Thus, the expected value of the weighted die is 4.5.

Box Model Analysis

To further illustrate this concept, we can use a box model. Imagine a box containing the following tickets:

5 tickets labeled '1' 5 tickets labeled '2' 5 tickets labeled '3' 5 tickets labeled '4' 5 tickets labeled '5' 5 tickets labeled '6'

The average value of the box is calculated as follows:

Average (1*5 2*5 3*5 4*5 5*5 6*5) / 30 45 / 10 4.5

In this model, each roll of the die would have an average value of 4.5, assuming the die is rolled 10 times, the expected value would be:

10 * 4.5 45

This framework helps us visualise the distribution and reinforces the concept of expected value in a practical setting.