Suppose you are given a fair coin and a sheet of paper with the polynomial $x^m$ written on it. Now for each toss of the coin, if heads show up, you must erase the polynomial $x^r$ (where $r$ is going to change with time - initially it is $m$) written on the paper and replace it with $x^{r-1}$. If tails show up, replace it with $x^{r+1}$. What is the expected value of the polynomial I get after $m$ such tosses? (Note: this is a different concept from the most probable value)
1. Let's denote the polynomial at the start as \( x^m \).
2. For each coin toss, if heads show up, we replace \( x^r \) with \( x^{r-1} \). If tails show up, we replace \( x^r \) with \( x^{r+1} \).
3. We need to find the expected value of the polynomial after \( m \) tosses.

To solve this, we can use the concept of a random walk on the exponents of \( x \). Each coin toss changes the exponent by \(\pm 1\) with equal probability.

4. Let \( X_i \) be a random variable representing the change in the exponent after the \( i \)-th toss. \( X_i \) can be either \( +1 \) (for tails) or \( -1 \) (for heads), each with probability \( \frac{1}{2} \).

5. The exponent after \( m \) tosses is given by:
   \[
   R = m + \sum_{i=1}^m X_i
   \]
   where \( m \) is the initial exponent and \( \sum_{i=1}^m X_i \) is the net change in the exponent after \( m \) tosses.

6. The expected value of \( R \) is:
   \[
   \mathbb{E}[R] = m + \mathbb{E}\left[\sum_{i=1}^m X_i\right]
   \]
   Since \( X_i \) are independent and identically distributed with \( \mathbb{E}[X_i] = 0 \):
   \[
   \mathbb{E}[R] = m + \sum_{i=1}^m \mathbb{E}[X_i] = m + m \cdot 0 = m
   \]

7. Therefore, the expected exponent after \( m \) tosses is \( m \).

8. The expected polynomial is then \( x^m \).

The final answer is \( \boxed{ x^m } \).