The Setup

  • A 1-foot string has 500 ants, each placed uniformly at random in \( [0, 1] \).
  • Each ant independently chooses to walk left or right with equal probability (\( \frac{1}{2} \)).
  • Ants walk at a constant speed of 1 ft/min.
  • When two ants meet, they reverse directions—or equivalently, pass through one another unchanged.
  • Ants fall off the string upon reaching either end.

Key Insight

The reversal behavior upon meeting is indistinguishable (in terms of final outcomes) from the ants passing through each other. So we may treat each ant as continuing in its original direction without interaction.

Thus, the last ant to fall off is simply the ant that takes the maximum time to reach an endpoint.

Time to Fall Off

Each ant:

  • Starts at a position \( x \in [0,1] \)
  • Goes left with probability \( \frac{1}{2} \): time to fall off = \( x \)
  • Goes right with probability \( \frac{1}{2} \): time to fall off = \( 1 - x \)

So the time for a given ant to fall off is:

\[ T = \begin{cases} x & \text{with probability } \frac{1}{2}
1 - x & \text{with probability } \frac{1}{2} \end{cases} \]

Thus, the time for all ants to fall off is:

\[ \max_{i=1}^{500} T_i \]

Each \( T_i \) is uniformly distributed over \( [0,1] \) (as a symmetric combination of \( x \) and \( 1 - x \)).

Expected Maximum of 500 Uniform(0,1) Variables

Let \( T_i \sim \mathrm{Uniform}(0,1) \). Then:

\[ \mathbb{E}\left[\max_{i=1}^{500} T_i\right] = \frac{500}{501} \]

Final Answer

\[ \boxed{\frac{500}{501}} \text{ minutes} \]

This is the expected time until all ants have fallen off the string.

Reference