Madhu and Ankit each flip a fair coin 5 times. Suppose that Madhu has flipped more heads than Ankit after each has flipped the coin 2 times. The probability that Madhu has flipped more heads than Ankit after each has flipped the coin 5 times is \frac{m}{n} , where m and n are relatively prime positive integers. Find |m – n|.

Given the problem, we need to find the probability that Madhu has flipped more heads than Ankit after each of them has flipped the coin 5 times. The probability can be expressed as \(\frac{m}{n}\), where \(m\) and \(n\) are relatively prime positive integers, and then determine \(|m - n|\).

### Step 1: Analyze the Given Condition

Since each flips the coin 5 times, we know that Madhu has more heads than Ankit after the first 2 flips. We want to find the probability that this condition holds after all 5 flips.

### Step 2: Consider Paths in a Lattice Diagram

Each possible sequence of flips can be represented as a walk on a grid, where each step represents a coin flip resulting in either a head or a tail. Since Madhu must always have more heads than Ankit, we restrict our paths to those that do not cross the line where Madhu and Ankit have the same number of heads.

### Step 3: Apply the Reflection Principle and Combinatorics

Using the reflection principle and the concept of restricted lattice paths, the number of such paths can be calculated. These are Catalan-type problems, but with an additional constraint that Madhu has more heads after 2 flips.

Let's first calculate the total number of sequences where Madhu has more heads than Ankit after all flips (ignoring the condition after 2 flips). The problem can be solved using the ballot theorem.

#### Ballot Theorem:

The theorem states that if candidate \(P\) gets \(p\) votes and candidate \(Q\) gets \(q\) votes (\(p > q\)), the number of ways \(P\) will stay strictly ahead of \(Q\) throughout the counting process is given by:

\[
\text{Number of valid sequences} = \frac{p-q}{p+q} \binom{p+q}{q}
\]

For our problem, \(p = 5\) and \(q = 0, 1, 2, 3\) for different possible scenarios.

The probability that Madhu has more heads than Ankit after 5 flips is:

\[
\text{Probability} = \frac{\text{Number of valid sequences}}{\text{Total sequences}}
\]

However, considering that Madhu has more heads after 2 flips, we must exclude sequences where they are tied or where Ankit has more heads after 2 flips.

### Step 4: Calculation

The total number of sequences with 5 heads and 0 tails is \(\binom{10}{5}\). From these, calculate the ones that meet our criteria.

- Total sequences after 5 flips: \(2^{10} = 1024\)
- Favorable sequences (Catalan number for 5 steps): Consider ballot theorem with restriction after 2 flips.

Given the ballot theorem combined with conditions:

\[
\text{Favorable paths} = \frac{6}{14} \times \binom{10}{5} = \frac{6}{14} \times 252 = 108
\]

Thus, probability is \( \frac{108}{1024} = \frac{27}{256}\).

Thus, \(m = 27\) and \(n = 256\).

### Step 5: Find \(|m - n|\)

\[
|m - n| = |27 - 256| = 229
\]

So, \(\boxed{229}\) is the final answer.

Calculate the probability that Madhu flips more heads than Ankit after 5 coin flips using combinatorics and probability theory. Includes step-by-step solution and detailed explanation of the Ballot Theorem and Catalan numbers.

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