Calculating the Probability of One Die Being the Average of the Other Two
An In-depth Exploration of Probabilistic Outcomes with Three Fair Six-Sided Dice
Key Takeaways
- Comprehensive Counting: Accurately identifying all favorable outcomes is crucial for determining probability.
- Avoiding Overcounting: Ensuring that overlapping cases are accounted for prevents inflated probability estimates.
- Mathematical Rigor: A step-by-step analytical approach leads to a precise calculation of the desired probability.
Understanding the Problem
Defining the Objective
We are tasked with finding the probability that, when rolling three fair six-sided dice simultaneously, one of the numbers rolled is the average of the other two. Mathematically, for outcomes represented as (a, b, c), we seek the probability that:
a = (b + c)/2
or
b = (a + c)/2
or
c = (a + b)/2
Total Possible Outcomes
Each die has six faces, so the total number of possible outcomes when rolling three dice is:
6 × 6 × 6 = 216
Methodology for Calculating Favorable Outcomes
Defining Favorable Outcomes
A favorable outcome is one where one of the three numbers is the average of the other two. This condition implies that the two numbers used to calculate the average must sum to an even number (since the average must be an integer between 1 and 6, inclusive).
Step-by-Step Calculation
1. Counting Favorable Triples
We need to determine all ordered triples (a, b, c) where one number is the average of the other two. This involves analyzing each possible scenario where a, b, or c satisfies the averaging condition.
2. Calculating for Each Averaging Condition
A. a is the average of b and c (a = (b + c)/2)
For a to be an integer, b + c must be even. We examine each possible value of a (from 1 to 6) and count the number of valid (b, c) pairs that satisfy the equation.
| a Value | Possible (b, c) Pairs | Number of Pairs |
|---|---|---|
| 1 | (1,1) | 1 |
| 2 | (1,3), (2,2), (3,1) | 3 |
| 3 | (1,5), (2,4), (3,3), (4,2), (5,1) | 5 |
| 4 | (2,6), (3,5), (4,4), (5,3), (6,2) | 5 |
| 5 | (4,6), (5,5), (6,4) | 3 |
| 6 | (6,6) | 1 |
| Total | 18 |
B. b is the average of a and c (b = (a + c)/2)
The logic mirrors that of part A. For each value of b from 1 to 6, we count the valid (a, c) pairs:
| b Value | Possible (a, c) Pairs | Number of Pairs |
|---|---|---|
| 1 | (1,1) | 1 |
| 2 | (1,3), (2,2), (3,1) | 3 |
| 3 | (1,5), (2,4), (3,3), (4,2), (5,1) | 5 |
| 4 | (2,6), (3,5), (4,4), (5,3), (6,2) | 5 |
| 5 | (4,6), (5,5), (6,4) | 3 |
| 6 | (6,6) | 1 |
| Total | 18 |
C. c is the average of a and b (c = (a + b)/2)
This scenario is analogous to parts A and B. For each value of c from 1 to 6, we determine the valid (a, b) pairs:
| c Value | Possible (a, b) Pairs | Number of Pairs |
|---|---|---|
| 1 | (1,1) | 1 |
| 2 | (1,3), (2,2), (3,1) | 3 |
| 3 | (1,5), (2,4), (3,3), (4,2), (5,1) | 5 |
| 4 | (2,6), (3,5), (4,4), (5,3), (6,2) | 5 |
| 5 | (4,6), (5,5), (6,4) | 3 |
| 6 | (6,6) | 1 |
| Total | 18 |
3. Summing Up Favorable Outcomes
At first glance, it might seem that the total number of favorable outcomes is simply the sum of the favorable outcomes from each scenario:
Total = 18 (from a) + 18 (from b) + 18 (from c) = 54
However, this count includes overlapping cases where multiple conditions are satisfied simultaneously. Specifically, triplets where all three numbers are equal (e.g., (2,2,2)) are counted three times—once in each condition. To avoid overcounting, we need to adjust for these overlaps.
4. Adjusting for Overcounting
Triplets where a = b = c satisfy all three conditions simultaneously. There are six such triplets: (1,1,1), (2,2,2), ..., (6,6,6).
Since each of these six triplets was counted three times in our initial total, we must subtract the extra counts to ensure each is only counted once:
Adjusted Total = 54 - (2 × 6) = 42
5. Final Calculation of Probability
The probability (P) that one of the numbers is the average of the other two is the ratio of favorable outcomes to total possible outcomes:
P = Favorable Outcomes / Total Outcomes = 42 / 216 = 7 / 36 ≈ 0.1944
Converting this to a percentage:
P ≈ 19.44%
Discrepancies in Source Calculations
Analyzing Source A and Source C
Both Source A and Source C arrived at a probability of 1/9 (≈11.1%). They approached the problem by identifying arithmetic sequences with distinct numbers and calculating permutations:
- Identified four distinct arithmetic sequences: (1,2,3), (2,3,4), (3,4,5), (4,5,6).
- Each sequence has 6 permutations, totaling 24 favorable outcomes.
- Ignored cases where all three numbers are equal, leading to a probability of 24/216 = 1/9.
However, this approach overlooks the triplets where all three numbers are the same. Since these triplets satisfy all three conditions simultaneously, they should be included in the favorable outcomes, bringing the total to 30. Even with this adjustment, it still falls short of the accurate 42 favorable outcomes calculated earlier.
Evaluating Source B
Source B concluded a probability of 1/2 (50%). Their methodology involved:
- Calculating valid (y, z) pairs for each possible x, resulting in 18 favorable (x, y, z) combinations.
- Multiplying by 6 to account for permutations, yielding 108 favorable outcomes.
This method incorrectly duplicates cases, as it fails to account for overlapping conditions where multiple averaging scenarios are satisfied by the same triplet. Consequently, it overestimates the number of favorable outcomes.
Comprehensive Conclusion
Final Probability Determination
Through meticulous analysis, considering all possible scenarios and adjusting for overlapping cases, the accurate probability that one of the numbers rolled on three fair six-sided dice is the average of the other two is:
Probability = 7/36 ≈ 19.44%
Key Insights
- Accurate probabilistic calculations require careful enumeration of all favorable cases, including those that may be counted multiple times.
- Overcounting can significantly inflate probability estimates, highlighting the importance of identifying and adjusting for overlapping cases.
- Understanding the underlying mathematical principles ensures precise and reliable probability assessments.
Mathematical Representation
To succinctly express the condition where one number is the average of the other two, consider the following equation forms:
-
a = (b + c)/2
-
b = (a + c)/2
-
c = (a + b)/2
These equations ensure that the sum of two numbers is twice the third, maintaining the requirement that the average is an integer between 1 and 6.
Reference Table of Favorable Outcomes
| Condition | Specific Equation | Number of Favorable Ordered Triples |
|---|---|---|
| a is the average | a = (b + c)/2 | 18 |
| b is the average | b = (a + c)/2 | 18 |
| c is the average | c = (a + b)/2 | 18 |
| Total Before Adjustment | 54 | |
| Adjustment for Overcounting (a = b = c) | -12 (since 6 triplets counted thrice) | |
| Total Favorable Outcomes | 42 | |
Final Thoughts
Probabilistic problems, especially those involving multiple conditions and permutations, necessitate a methodical approach to ensure accuracy. By thoroughly enumerating all possible favorable outcomes and meticulously adjusting for overlaps, we arrive at a precise probability calculation.
This exercise underscores the importance of not only following a structured methodology but also critically evaluating and reconciling differing approaches to achieve a reliable conclusion.
References
Last updated January 17, 2025