Discrete Random Variables

Question

Suppose I roll a standard 6-sided die twice. Let X = the difference in values between my 2 rolls (highest minus lowest). i.e. if I roll (4,2), then X=2, if I roll (3,3) then X=0.

1. Create a tabular pdf listing all possible outcomes (x) of the experiment along with all outcome probabilities?
2. Find the mean and standard deviation of X?
3. What is the probability that X is at least 3?

Answer 1

When rolling two 6-sided dice, there are 6×6=36. The difference X=highest value−lowest value can range from 0 (if both dice are equal) to 5 (if one die rolls a 6 and the other rolls a 1). Let’s create the table listing all possible outcomes of X and their probabilities.

First, we count the number of ways each value of X can occur.

1. X = 0: This happens when both dice roll the same number: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)(. There are 6 outcomes.
2. X = 1: This occurs for pairs where the difference is 1, e.g., (2,1), (3,2), etc., and their reverse. The pairs are (2,1),(3,2),(4,3),(5,4),(6,5) and their reverse (1,2),(2,3),(3,4),(4,5),(5,6). There are 5×2=10 outcomes.
3. X = 2: This occurs when the difference is 2, e.g., (3,1),(4,2),(3,1), (4,2), etc. The pairs are (3,1),(4,2),(5,3),(6,4)(3,1), (4,2), (5,3), (6,4) and their reverse (1,3),(2,4),(3,5),(4,6), so there are 4×2=84 \times 2 = 8 outcomes.
4. X = 3: The pairs are (4,1),(5,2),(6,3) and their reverse, making 3×2=6 outcomes.
5. X = 4: The pairs are (5,1),(6,2) and their reverse, making 2×2=4 outcomes.
6. X = 5: The pairs are (6,1) and its reverse, making 1×2=2 outcomes.

Now we can tabulate the probability distribution:

X (Difference)	Count of Outcomes	Probability P(X)
0	6	6/36=1/6
1	10	10/36=5/18
2	8	8/36=2/9
3	6	6/36=1/6
4	4	4/36=1/9
5	2	2/36=1/18

Now you can compute E[X], Var[X] and P[X.>=3]