Question

Consider a game of dice in which each player gets one die and whoever rolls the higher number wins. If the number is equal, the players reroll until someone wins. There are three dice to choose from: - Die X: 1, 1, 3, 5, 5, 6 - Die Y:2,3,3,4,4,5 - Die Z: 1, 2, 2, 4, 6, 6 Using discrete math, answer the following:

a. What is the expected value of each die?
b. Consider each possible pair of dice. What is the probability of eacl die winning, tying, or losing a roll?
c. How can a player give themself an advantage?

Answer 1

a. The expected value of each die: Die X = 7/3, Die Y = 19/6 (approximately 3.17), Die Z = 2. b. Probability of winning, tying, or losing for each possible pair of dice. c. To give oneself an advantage, choose Die Y with the highest expected value.

Step-by-step explanation:

a. Expected value of each die:

- For Die X: The expected value can be calculated by summing up the products of each outcome and its corresponding probability. (1 * 2/6) + (3 * 1/6) + (5 * 2/6) + (6 * 1/6) = 7/3.

- For Die Y: The expected value can be calculated in the same way. (2 * 1/6) + (3 * 2/6) + (4 * 2/6) + (5 * 1/6) = 19/6 or approximately 3.17.

- For Die Z: (1 * 1/6) + (2 * 2/6) + (4 * 1/6) + (6 * 2/6) = 12/6 or 2.

b. Probability of winning, tying, or losing:

- For Die X and Die Y:

• Probability of winning for Die X: 11/36
• Probability of tying: 1/36
• Probability of losing: 24/36 or 2/3

- For Die X and Die Z:

• Probability of winning for Die X: 1/6
• Probability of tying: 1/36
• Probability of losing: 5/6

- For Die Y and Die Z:

• Probability of winning for Die Y: 1/3
• Probability of tying: 1/36
• Probability of losing: 2/3

c. Giving oneself an advantage:

A player can give themselves an advantage by choosing Die Y, which has the highest expected value of approximately 3.17. This increases the chances of winning compared to the other two dice.