Two (not necessarily different) numbers are chosen independently and at random from $\{1, 2, 3, \dots, 10\}$. On average, what is the product of the two integers? (Compute the expected product. That is, if you do this over and over again, what will the product of the integers be on average?)
1. Let \( X \) and \( Y \) be the two numbers chosen independently and at random from the set \( \{1, 2, 3, \dots, 10\} \).
2. We need to compute the expected value of the product \( XY \). The expected value of a product of two independent random variables is the product of their expected values. Therefore, we need to find \( E(X) \) and \( E(Y) \).

3. Since \( X \) and \( Y \) are chosen from the same set, \( E(X) = E(Y) \). The expected value of a uniformly distributed random variable over the set \( \{1, 2, 3, \dots, 10\} \) is given by:
   \[
   E(X) = \frac{1 + 2 + 3 + \cdots + 10}{10}
   \]

4. The sum of the first \( n \) positive integers is given by the formula:
   \[
   \sum_{k=1}^n k = \frac{n(n+1)}{2}
   \]
   For \( n = 10 \):
   \[
   \sum_{k=1}^{10} k = \frac{10 \cdot 11}{2} = 55
   \]

5. Therefore, the expected value \( E(X) \) is:
   \[
   E(X) = \frac{55}{10} = 5.5
   \]

6. Since \( X \) and \( Y \) are identically distributed, \( E(Y) = 5.5 \).

7. The expected value of the product \( XY \) is:
   \[
   E(XY) = E(X) \cdot E(Y) = 5.5 \cdot 5.5 = 30.25
   \]

Conclusion:
\[
\boxed{30.25}
\]