Given a random string of 33 bits (0 or 1), how many (they can overlap) occurrences of two consecutive 0's would you expect? (i.e. "100101" has 1 occurrence, "0001" has 2 occurrences)
1. **Determine the probability of two consecutive bits being both 0:**
   - Each bit in the string can be either 0 or 1 with equal probability, i.e., \( \frac{1}{2} \).
   - The probability that two consecutive bits are both 0 is:
     \[
     \left( \frac{1}{2} \right)^2 = \frac{1}{4}
     \]

2. **Calculate the number of possible locations for a two-bit string in a 33-bit string:**
   - A 33-bit string has 32 possible locations for a two-bit substring (since each pair of consecutive bits can start at any of the first 32 positions).
   - This is because the first bit of the two-bit substring can be at any position from 1 to 32.

3. **Calculate the expected number of occurrences of two consecutive 0's:**
   - The expected number of occurrences is the product of the number of possible locations and the probability of each location having two consecutive 0's.
   - Therefore, the expected number of occurrences is:
     \[
     32 \times \frac{1}{4} = 8
     \]

The final answer is \(\boxed{8}\).