Coin Flips And Probability Calculating 'THTH' Pattern Occurrence

In the realm of probability and combinatorics, coin flip problems often serve as excellent examples to illustrate fundamental principles. This article delves into a specific coin flip scenario, exploring the probability of observing a particular pattern within a sequence of flips. We will discuss the question, provide a comprehensive solution using the inclusion-exclusion principle, and highlight the underlying concepts involved. This will show the intricacies of probability calculations and combinatorial reasoning.

The Coin Flip Challenge: Probability of "THTH" Pattern

Let's consider the following problem, which blends probability and combinatorics:

Problem Statement: Imagine you flip a fair coin 11 times, resulting in a sequence of 4 heads (H) and 7 tails (T). What is the probability that the pattern "THTH" appears at least once within this sequence?

This problem challenges us to not only enumerate the possible sequences of coin flips but also to account for the overlapping occurrences of the target pattern. We will explore how to systematically approach this calculation, making our first step towards understanding the solution by focusing on counting favorable outcomes rather than directly calculating probabilities.

Deconstructing the Problem: Applying Combinatorial Principles

To tackle this problem, we'll employ a strategic approach rooted in combinatorial principles. The core idea revolves around counting the number of sequences where the pattern "THTH" appears at least once. However, a direct count can be tricky due to potential overlaps. For instance, a sequence might contain "THTHTH", where the pattern "THTH" appears twice. To avoid overcounting, we'll leverage the inclusion-exclusion principle, a powerful tool in combinatorics.

Step 1: Counting Total Possible Sequences

Before diving into the target pattern, we need to establish the total number of possible sequences with 4 heads and 7 tails. This is a classic combinatorial problem that can be solved using combinations. We have 11 positions, and we need to choose 4 of them to be heads (the remaining 7 will be tails). The number of ways to do this is given by the binomial coefficient:

$$\binom{11}{4} = \frac{11!}{4!7!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$$

This means there are 330 distinct sequences of 11 coin flips with exactly 4 heads and 7 tails. This number forms the denominator in our probability calculation, grounding our understanding in the fundamental principles of combinations. It’s essential to calculate this accurately as it forms the basis for further computations.

Step 2: Identifying Potential "THTH" Occurrences

Now, let's identify the possible positions where the pattern "THTH" can occur within the 11-flip sequence. The pattern "THTH" has a length of 4, so it can start at positions 1, 2, 3, 4, 5, 6, 7, or 8. We define events ${A_i}$ as the event that the pattern "THTH" starts at position ${i}$. For example:

• ${A_1}$: "THTH" starts at position 1 (THTH........)
• ${A_2}$: "THTH" starts at position 2 (.THTH.......)
• ... and so on.

Step 3: Applying the Inclusion-Exclusion Principle

The inclusion-exclusion principle is crucial for accurately counting the sequences with at least one occurrence of “THTH.” The principle states that to find the number of elements in the union of several sets, we sum the sizes of the individual sets, subtract the sizes of the pairwise intersections, add back the sizes of the three-way intersections, and so on. This systematic approach ensures we avoid overcounting elements that belong to multiple sets.

In our case, we want to find the number of sequences in the union of the events ${A_1, A_2, ..., A_8}$. The inclusion-exclusion principle gives us:

$$|A_1 \cup A_2 \cup ... \cup A_8| = \sum_{i=1}^{8} |A_i| - \sum_{1 \le i < j \le 8} |A_i \cap A_j| + \sum_{1 \le i < j < k \le 8} |A_i \cap A_j \cap A_k| - ...$$

This formula might look daunting, but it breaks down into manageable parts. Let's tackle each term systematically.

Term 1: Single Occurrences (${\sum_{i=1}^{8} |A_i|}$)

We need to count the number of sequences where "THTH" appears at least once starting at position ${i}$. If “THTH” starts at position ${i}$, we have fixed four positions. The remaining 7 positions need to accommodate 3 heads and 4 tails. The number of ways to arrange these is:

$$\binom{7}{3} = \frac{7!}{3!4!} = 35$$

Since there are 8 possible starting positions for “THTH”, the sum of single occurrences is:

$$\sum_{i=1}^{8} |A_i| = 8 \times 35 = 280$$

This part of the calculation provides us with an initial count of sequences containing “THTH,” focusing on instances where we consider each occurrence independently. It serves as the foundation upon which we'll refine our count using subsequent steps of the inclusion-exclusion principle.

Term 2: Pairwise Intersections (${\sum_{1 \le i < j \le 8} |A_i \cap A_j|}$)

Now, we need to consider cases where "THTH" appears at two different positions. This is where things get a bit more intricate, as the overlaps between the pattern occurrences influence our calculations. We need to analyze different pairs of positions and their implications on the arrangement of remaining flips.

Let's analyze different scenarios for the positions ${i}$ and ${j}$:

• Case 1: ${j = i + 1}$

  If "THTH" appears consecutively (e.g., starting at positions 1 and 2), the sequence would look like "THTHTH...". This means we have used 6 positions. The remaining 5 positions must accommodate 2 heads and 3 tails. The number of ways to do this is:

  $$\binom{5}{2} = \frac{5!}{2!3!} = 10$$

  There are 7 such pairs (1 & 2, 2 & 3, ..., 7 & 8).

• Case 2: ${j = i + 2}$

  If “THTH” patterns are two positions apart (e.g., starting at positions 1 and 3), the sequence would look like “THTH.THTH…”. This uses 8 positions, leaving 3 positions for 1 head and 2 tails. The number of ways to arrange this is:

  $$\binom{3}{1} = \frac{3!}{1!2!} = 3$$

  There are 6 such pairs (1 & 3, 2 & 4, ..., 6 & 8).

• Case 3: ${j = i + 3}$

  If "THTH" patterns are three positions apart (e.g., starting at positions 1 and 4), the sequence would look like "THTH..THTH...". This uses 10 positions, leaving only 1 position for 0 heads and 1 tail. The number of ways to arrange this is:

  $$\binom{1}{0} = 1$$

  There are 5 such pairs (1 & 4, 2 & 5, ..., 5 & 8).

• Case 4: ${j \ge i + 4}$

  It's not possible for this case since we can't fit 4 Heads in remaining flips

Summing up the pairwise intersections:

$$\sum_{1 \le i < j \le 8} |A_i \cap A_j| = (7 \times 10) + (6 \times 3) + (5 \times 1) = 70 + 18 + 5 = 93$$

Term 3: Triple Intersections (${\sum_{1 \le i < j < k \le 8} |A_i \cap A_j \cap A_k|}$)

Now, let’s consider cases where “THTH” appears at three different positions. We need to analyze the possible scenarios and their implications.

• Case 1: i, j, k are consecutive (j = i + 1, k = j + 1)

  Example: Positions 1, 2, and 3. The sequence would start like “THTHTHTH…”. 10 positions are used, leaving 1 position for 1 tail and 0 heads. Number of arrangements: 1. There are 6 such triplets (1, 2, 3), (2, 3, 4), ..., (6, 7, 8).

• Case 2: Other triplets

  There are no other triplets possible since we only have 4 Heads.

Summing up the triple intersections:

$$\sum_{1 \le i < j < k \le 8} |A_i \cap A_j \cap A_k| = 6 \times 1 = 6$$

Term 4: Quadruple Intersections

It is impossible to have four