Here’s their solution. Same answer as me but arrived via the angular momentum concept. Frankly I thought my approach was simpler.
SOLUTION: We know from last week's puzzle, "Bugs on a Stick," that the final positions for the bugs collectively are the same as if they passed through one another instead of bouncing. If they pass one another, every bug simply travels once around the hoop and ends up where it began.
With bouncing, some bug will end up where Bugsy began, but will it be Bugsy? It certainly will be if it happens that all the bugs begin facing the same way, since then there will be no collisions. But that happens with probability only 2/(224) = 0.0000001192, approximately.
Wait, maybe there's another way. Note that the whole process preserves the angular momentum of the mass of bugs. Suppose, say, that 16 of the 24 bugs face clockwise; the remaining 8 are counterclockwise. Then the bugs will, as a whole, rotate clockwise at 1/3 cm/sec, and at the end, each bug will be, on average, 1/3 of the way around from where it began.
But since they never pass through one another, the bugs remain in the same order; thus every bug will end up (in this case) in the position of the bug that began 8 bugs in front of it, clockwise.
So all the bugs (including Bugsy) change position in this case, and indeed in every case unless the mass clockwise speed is an integer number of cm/sec. That integer can only be 1 (all bugs moving clockwise), -1 (all counterclockwise), or 0 (12 bugs in each direction).
The latter is quite a lively possibility, occurring with probability (24 choose 12)/(224) ≈ 0.1611802578. Adding the 0.0000001192 from before gives us a total probability of 0.1611803770 for the event that Bugsy ends up where he started.