There are $24$ ants scattered around a hula hoop of circumference $3$ meters . They each randomly, and independently, choose to face clockwise or counterclockwise, and then simultaneously start marching at $1$ cm/sec. When two ants meet, they instantaneously reverse direction.

Let's say that one of the ants is named Alice. After $5$ minutes of marching, what is the probability that Alice is back where she started?

Note: I have changed the number of ants from $25$ to $24$. This doesn't affect how you approach/solve the problem, but makes the answer more interesting.

  • $\begingroup$ How are the ants initially positioned? Are they evenly spaced? If there's a random component to their initial positions, what's the distribution? $\endgroup$ – user2357112 supports Monica Apr 28 '15 at 3:23
  • $\begingroup$ @user2357112 All you know about the initial positioning is that no ants are on top of one another. $\endgroup$ – Mike Earnest Apr 28 '15 at 3:25

I'll say

about $16\%$.

First note that

as is typical of these ant puzzles, you can think of the ants as passing through each other instead of changing directions, and the ant locations remain the same (though not the individual ant identities). Thus, after 5 minutes or 300 seconds, the ants have gone around the hula-hoop and are in the same locations they started.

Next note that

order of the ants never changes. If Alice ends in her same spot, then her neighbors end in their same spots, as do their neighbors, etc., all the way around the hoop. So if Alice ends in her spot, nobody has moved overall. If Alice has moved over one ant position, the all move over one ant position.

Finally, we have

if there are $k$ ants going clockwise, then the average net clockwise speed of the ants is $\frac{k - (24 - k)}{24} = \frac{k}{12} - 1$. So on average, after 5 min, one of the ants has moved $25k -300$ centimeters in the clockwise direction. Each ant ends on its starting value exactly when the average distance moved is $-300$, $0$, or $300$. This corresponds to $k = 0, 12, 24$. Thus $(2 + {24 \choose 12})/2^{24}$ fraction of the starting directions lead to Alice coming home.

| improve this answer | |
  • $\begingroup$ Well done, very quick! $\endgroup$ – Mike Earnest Apr 28 '15 at 4:36
  • 1
    $\begingroup$ @MikeEarnest: You post fantastic math puzzles -- thanks for your contributions! $\endgroup$ – Tyler Seacrest Apr 28 '15 at 5:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.