Ants On A Circle III

Question

Ten 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 if the number of ants is even (as in this problem), then, apart from the cases where all ants move clockwise or all move anticlockwise, the only way for each ant to return to its original position is if an equal number of ants initially move clockwise and anticlockwise.

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
$$
\sum_{i=1}^{10} 	heta_i = 0.
$$

Let $x$ be the number of ants that initially chose clockwise and $10-x$ 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,
$$
\sum_{i=1}^{10} 	heta_i
= x \cdot 360^\circ + (10-x)\cdot(-360^\circ)
= (2x - 10)\cdot 360^\circ.
$$

So we must have
$$
(2x - 10)\cdot 360^\circ = 0,
$$
which gives $x = 5$.

This shows that, for all ants to return to their original positions (assuming not all move clockwise or all move anticlockwise), it must be that $x = 5$, i.e., five ants move clockwise and five move anticlockwise.

However, we still need to show sufficiency. First, note that the initial and final sets of positions are identical (from Ants on a Circle I). Next, the circular order of the ants remains unchanged, since whenever two ants meet, they simply reverse directions rather than passing through each other, preserving their relative order around the circle. Finally, when five ants move clockwise and five anticlockwise, we have
$$
\sum_{i=1}^{10} 	heta_i = 0.
$$

Together, these facts imply that each ant must return to its original position when an equal number of ants move clockwise and anticlockwise.

Hence, the total number of configurations in which all ants return to their original positions is
$$
1 + 1 + \binom{10}{5},
$$
the first $1$ corresponding to all ants moving clockwise, the second to all moving anticlockwise, and $\binom{10}{5}$ representing the cases where five ants move clockwise and five move anticlockwise.

So, the required probability is
$$
\frac{1 + 1 + \binom{10}{5}}{2^{10}} = \frac{127}{512}.
$$