A Fair Die Is Rolled 14 Times. Let Be The Number Of Faces That Appear Exactly Three Times. Which Of The
When analyzing probability scenarios involving dice, understanding the likelihood of specific outcomes is essential for both enthusiasts and professionals in the fields of statistics, mathematics, and gaming. One interesting problem involves rolling a fair six-sided die 14 times and determining the probability distribution of faces that appear exactly three times. This problem combines combinatorial concepts, probability theory, and statistical reasoning to explore the possible configurations and their likelihoods.
In this comprehensive guide, we'll delve deeply into this problem, breaking it down into manageable sections. We'll review the fundamental principles of probability theory relevant to dice rolls, explore the combinatorial calculations involved, and analyze the expected outcomes. Whether you're a student preparing for examinations, a game designer considering probabilities, or a hobbyist interested in mathematical puzzles, this article provides a detailed, structured exploration of the problem.
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Understanding the Problem
The Scenario
- You roll a fair six-sided die 14 times.
- Each roll results in one face from 1 to 6.
- The goal is to analyze the distribution of faces that appear exactly three times.
The Key Question
Given this setup, which face(s) are likely to appear exactly three times? Additionally, how many faces can appear exactly three times, and what is the probability distribution of such outcomes?
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Fundamental Concepts in Probability and Combinatorics
Before addressing the specific problem, it's essential to review some core concepts that underpin the analysis:
- Probability of a Single Outcome
- For a fair six-sided die, the probability that any particular face appears on a single roll is \( \frac{1}{6} \).
- The probability of a sequence of outcomes depends on the individual probabilities, assuming independence.
- Multinomial Distribution
- When multiple independent trials are conducted with multiple possible outcomes, the multinomial distribution describes the probability of counts for each outcome.
- For dice rolls, the number of times each face appears over 14 rolls follows a multinomial distribution with parameters \( n = 14 \) and probabilities \( p_i = \frac{1}{6} \) for each face.
- Combinations and Permutations
- To count the number of sequences where certain faces appear a specific number of times, combinatorial calculations are necessary.
- The total number of possible sequences is \( 6^{14} \), since each of the 14 rolls can be any of 6 faces.
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Breaking Down the Problem: Faces Appearing Exactly Three Times
- What Does "Faces That Appear Exactly Three Times" Mean?
- For the 14 rolls, some subset of faces will occur exactly three times.
- The remaining rolls are distributed among other faces, which may appear more or fewer times, but we're focusing on faces with a count of exactly three appearances.
- Possible Numbers of Faces That Appear Exactly Three Times
- Since each face that appears exactly three times accounts for 3 rolls, the maximum number of such faces is limited by \( 14 \div 3 \), because:
\[
\text{Maximum number of faces appearing exactly three times} = \left\lfloor \frac{14}{3} \right\rfloor = 4
\]
- However, because 4 faces appearing exactly three times sum to 12 rolls, the remaining 2 rolls are distributed among other faces.
- Therefore, the possible counts of faces appearing exactly three times are:
\[
k = 0, 1, 2, 3, 4
\]
with the understanding that each configuration must fit into the total of 14 rolls.
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Calculating the Number of Faces That Appear Exactly Three Times
- General Approach
To determine the number of faces appearing exactly three times, we need to:
- Choose which faces will appear exactly three times.
- Assign counts to the remaining faces, ensuring the total adds up to 14.
- Count the number of sequences (outcomes) consistent with these counts.
- Step-by-Step Calculation
Let's formalize this process:
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Step 1: Choose the Faces That Appear Exactly Three Times
- For \( k \) faces to appear exactly three times, select \( k \) faces out of 6:
\[
\text{Number of ways} = \binom{6}{k}
\]
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Step 2: Assign Counts to the Selected Faces
- Each of these \( k \) faces appears exactly 3 times, contributing \( 3k \) rolls.
- Remaining \( 14 - 3k \) rolls are distributed among the remaining \( 6 - k \) faces.
- These remaining counts can vary, but for each configuration, the counts must be positive integers (or zero) summing to \( 14 - 3k \).
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Step 3: Distribute Remaining Rolls
- For the remaining \( 6 - k \) faces, the counts \( c1, c2, ..., c_{6 - k} \) satisfy:
\[
c1 + c2 + \cdots + c_{6 - k} = 14 - 3k
\]
\[
c_i \geq 0
\]
- The number of solutions is given by the stars-and-bars theorem:
\[
\binom{(14 - 3k) + (6 - k) - 1}{(6 - k) - 1} = \binom{(14 - 3k) + (6 - k) - 1}{5 - k}
\]
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Step 4: Count the Number of Sequences
- For a fixed set of counts, the number of sequences is:
\[
\frac{14!}{(3!)^{k} \prod{i=1}^{6 - k} ci!}
\]
- Summing over all possible distributions of counts \( c_i \), we obtain the total number of sequences for each \( k \).
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Probabilistic Calculations
- Probability of a Specific Configuration
- The probability of a particular configuration with counts \( c1, c2, ..., c_6 \), where exactly \( k \) faces appear exactly three times, is:
\[
P = \frac{14!}{\prod{i=1}^{6} ci!} \left(\frac{1}{6}\right)^{14}
\]
- Since the die is fair and rolls are independent, the multinomial probability applies.
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- Expected Number of Faces Appearing Exactly Three Times
- By symmetry, each face has an equal chance of appearing exactly three times.
- The expected number of faces with exactly three occurrences is:
\[
E = 6 \times P(\text{a particular face appears exactly 3 times})
\]
Calculating this precisely involves summing over all configurations, which can be complex, but the expectation can be approximated or calculated using the properties of the multinomial distribution.
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Practical Examples and Calculations
Let's illustrate the concepts with specific cases:
Example 1: Exactly 1 Face Appears Exactly Three Times
- Choose 1 face out of 6:
\[
\binom{6}{1} = 6
\]
- Remaining 11 rolls are distributed among 5 faces:
\[
c1 + c2 + c3 + c4 + c_5 = 11
\]
- Number of solutions:
\[
\binom{11 + 5 - 1}{5 - 1} = \binom{15}{4} = 1365
\]
- Number of sequences for each distribution:
\[
\frac{14!}{(3!) \times c1! c2! c3! c4! c_5!}
\]
- Summing over all solutions yields the total number of sequences where exactly one face appears three times.
Example 2: Exactly 2 Faces Appear Exactly Three Times
- Choose 2 faces:
\[
\binom{6}{2} = 15
\]
- Remaining 8 rolls among 4 faces:
\[
c1 + c2 + c3 + c4 = 8
\]
- Number of solutions:
\[
\binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = 165
\]
- For each, calculate the number of sequences and sum accordingly.
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Application: Calculating Probabilities for Specific Outcomes
- Probability that exactly one face appears exactly three times
- Sum over all configurations where one face appears three times and the remaining counts sum to 11 among the other faces.
- Multiply by the number of ways to choose the face and the counts, then divide by total possible sequences \( 6^{14} \).
- Probability that exactly two faces appear exactly three times each
- Similar calculation, considering the combinations of faces and distributions.
- Extending to other values of \( k \)
- The process is analogous for \( k=3 \) or \( k=4 \), with adjustments for the sum of remaining counts.
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Insights and Expected Outcomes
- Expected Number of Faces Appearing Exactly Three Times
- Due to symmetry and the multinomial distribution, the expected number of faces appearing exactly three times in 14 rolls is approximately:
\[
E \approx
