Ants On A Circle II

Question

Nine ants are placed at equal spacing around a circle. Each ant independently chooses clockwise or counterclockwise and then moves at constant speed so that, if uninterrupted, each would make exactly one full revolution in one minute. When two ants meet they instantly reverse direction and continue at the same speed. The ants are distinguishable. What is the probability that after one minute every ant is exactly at its own starting point? Here, since the ants are distinguishable, we require every ant to return to its own original position.

QuantSoc · Accepted Answer

If all ants move clockwise or all move anticlockwise, they will each return to their original positions. We now show that they cannot return to their original positions in any other case when there are 9 ants.

Note that an ant can return to its original position only if it moves a full $360^\circ$ clockwise, a full $360^\circ$ anticlockwise, or has a net movement of $0^\circ$. Moreover, if the ants are not all moving in the same direction, each ant will collide with at least one other and reverse its direction. Consequently, every ant must have a net movement of $0^\circ$ for all of them to return to their original positions.

Let the net angular displacement of ant $i$ after one minute be $	heta_i$, measured positive for clockwise and negative for counterclockwise. For every ant to be at its original location, we must have
$$
	heta_i = 0 \quad 	ext{for every } i,
$$
hence
$$
\sum_{i=1}^{9} 	heta_i = 0.
$$

Let $x$ be the number of ants that initially chose clockwise and $9-x$ the number that initially chose counterclockwise. The quantity $x$ is invariant in time because at each collision two ants reverse direction, so one clockwise ant becomes counterclockwise and one counterclockwise becomes clockwise.

Thus,
$$
\sum_{i=1}^{9} 	heta_i
= x \cdot 360^\circ + (9-x)\cdot(-360^\circ)
= (2x - 9)\cdot 360^\circ.
$$

For all ants to return to their original positions, we require
$$
(2x - 9)\cdot 360^\circ = 0,
$$
which is impossible for $x \in \{1,2,\dots,8\}$.

Since there are $2^9$ equally likely initial direction assignments and exactly two favorable assignments (all clockwise or all anticlockwise), the required probability is
$$
\frac{2}{2^9} = \frac{1}{256}.
$$

Hence, the probability that every distinguishable ant is exactly at its own starting point after one minute is
$$
\boxed{\frac{1}{256}}.
$$