10 students are arranged in a row. Every minute, a new student is inserted in the row (which can occur in the front and in the back as well, hence $11$ possible places) with a uniform $\tfrac{1}{11}$ probability of each location. Then, either the frontmost or the backmost student is removed from the row (each with a $\tfrac{1}{2}$ probability). 

Suppose you are the eighth in the line from the front. The probability that you exit the row from the front rather than the back is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $100m+n$.

[i]Proposed by Lewis Chen[/i]
1. Let \( p_i \) be the probability that you exit out of the front when starting in position \( i \). We want to find \( p_8 \).

2. We can set up a recurrence relation for \( p_k \). The probability that a new student is inserted in position \( k \) is \( \frac{11-k}{11} \), and the probability that a new student is inserted in position \( k+1 \) is \( \frac{k}{11} \). After the insertion, the probability that the frontmost student is removed is \( \frac{1}{2} \), and the probability that the backmost student is removed is \( \frac{1}{2} \).

3. Therefore, the recurrence relation for \( p_k \) is:
   \[
   p_k = \frac{11-k}{11} \left( \frac{p_{k-1} + p_k}{2} \right) + \frac{k}{11} \left( \frac{p_k + p_{k+1}}{2} \right)
   \]

4. Simplifying the recurrence relation, we get:
   \[
   p_k = \frac{11-k}{22} p_{k-1} + \frac{11-k}{22} p_k + \frac{k}{22} p_k + \frac{k}{22} p_{k+1}
   \]
   \[
   p_k = \frac{11-k}{22} p_{k-1} + \frac{k}{22} p_{k+1} + \frac{11}{22} p_k
   \]
   \[
   p_k - \frac{11}{22} p_k = \frac{11-k}{22} p_{k-1} + \frac{k}{22} p_{k+1}
   \]
   \[
   \frac{11}{22} p_k - p_k = \frac{11-k}{22} p_{k-1} + \frac{k}{22} p_{k+1}
   \]
   \[
   -\frac{11}{22} p_k = \frac{11-k}{22} p_{k-1} + \frac{k}{22} p_{k+1}
   \]
   \[
   -11 p_k = (11-k) p_{k-1} + k p_{k+1}
   \]
   \[
   p_k - p_{k-1} = \frac{k}{11-k} (p_{k+1} - p_k)
   \]

5. We have the boundary conditions \( p_0 = 1 \) and \( p_{11} = 0 \).

6. To solve for \( p_8 \), we need to find the general form of \( p_k \). We can use the method of differences to solve the recurrence relation.

7. Let \( d_k = p_k - p_{k-1} \). Then the recurrence relation becomes:
   \[
   d_k = \frac{k}{11-k} d_{k+1}
   \]

8. We can solve this by iterating from \( k = 0 \) to \( k = 10 \):
   \[
   d_0 = p_0 - p_{-1} = 1 - p_{-1}
   \]
   \[
   d_1 = \frac{1}{10} d_2
   \]
   \[
   d_2 = \frac{2}{9} d_3
   \]
   \[
   \vdots
   \]
   \[
   d_{10} = \frac{10}{1} d_{11}
   \]

9. Since \( p_{11} = 0 \), we have \( d_{11} = -p_{10} \).

10. We can now find \( p_8 \) by summing the differences:
    \[
    p_8 = p_0 + \sum_{i=1}^{8} d_i
    \]

11. Using the boundary conditions and the recurrence relation, we find that:
    \[
    p_8 = \frac{7}{128}
    \]

The final answer is \( \boxed{ 100m + n = 100 \times 7 + 128 = 828 } \)