1. A Simple Random Variable, A Fair D6 and the Inheritance of Probabilities through Random Variables

2. The Fair d6 Model Face Values: 1,2,3,4,5,6 Fair Model: Equally likely face values – 1/6 per face value Pr{d6 Shows 1} = (1/6) @.1667 or 16.67% In long runs of tosses, approximately 1 toss in 6 shows “1”. Pr{d6 Shows 2} = (1/6) @.1667 or 16.67% In long runs of tosses, approximately 1 toss in 6 shows “2”. Pr{d6 Shows 3} = (1/6) @.1667 or 16.67% In long runs of tosses, approximately 1 toss in 6 shows “3”. Pr{d6 Shows 4} = (1/6) @.1667 or 16.67% In long runs of tosses, approximately 1 toss in 6 shows “4”. Pr{d6 Shows 5} = (1/6) @.1667 or 16.67% In long runs of tosses, approximately 1 toss in 6 shows “5”. Pr{d6 Shows 6} = (1/6) @.1667 or 16.67% In long runs of tosses, approximately 1 toss in 6 shows “6”.

3. Map D6 to D3 and Note D3 Face Value D6 Face Value  D3 Face Value 1, 2  1 3, 4  2 5, 6  3 This is a simple random variable.

4. The Fair d3 Model Nested within a Fair d6 Model FV: Face Values: 1(1,2), 2(3,4), 3(5,6) Fair Model: Equally likely face values – (2/6 =)1/3 per face value. Pr{d3 shows “1”} = Pr{d6 Shows 1} + Pr{d6 Shows 2} = (1/6) + (1/6) = 2/6 = 1/3 In long runs of tosses, approximately 1 toss in 3 shows “1”. Pr{d3 shows “2”} = Pr{d6 Shows 3} + Pr{d6 Shows 4} = (1/6) + (1/6) = 2/6 = 1/3 In long runs of tosses, approximately 1 toss in 3 shows “2”. Pr{d3 shows “3”} = Pr{d6 Shows 5} + Pr{d6 Shows 6} = (1/6) + (1/6) = 2/6 = 1/3 In long runs of tosses, approximately 1 toss in 3 shows “3”.