Puzzle time - more bugs on a meter stick
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As a hint here's a suggested approach:
Imagine Bugsy has a baton and passes it to whatever bug he comes into contact with, who then passes it along similarly, etc.
It's clear from our last problem that the baton will make a revolution of the circle in 100s, ending up at Bugsy's starting point.
How many 'contacts' or 'handoffs' must occur for the baton to end up back in Bugsy's hands in the end?
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It seems clear to me that, for the baton to return to Bugsy, the number of 'handoffs' needs to be a multiple of 24, since the relative position of the bugs doesn't change.
What does that imply about the starting position of the bugs?
"You never know what worse luck your bad luck has saved you from."
-Cormac McCarthy -
It seems clear to me that, for the baton to return to Bugsy, the number of 'handoffs' needs to be a multiple of 24, since the relative position of the bugs doesn't change.
What does that imply about the starting position of the bugs?
@jon-nyc said in Puzzle time - more bugs on a meter stick:
It seems clear to me that, for the baton to return to Bugsy, the number of 'handoffs' needs to be a multiple of 24, since the relative position of the bugs doesn't change.
What does that imply about the starting position of the bugs?
So what's your answer?
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There are two handoff events for every bug traveling in the opposite direction of the baton. So we need 12 bugs heading one way and 12 the other. So you get C(24,12) * (1/2) ^ 24 ~= 0.16118
Well, that's the engineer's answer. Bugsy also gets to the same spot if the bugs are all heading in the same direction, so for the mathematician's answer you need to add (1/2)^23 to the above.
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Interestingly the originator of the puzzle sent out the following hint today:
"Consider the angular momentum of the mass of bugs"
That still seems to be pointing you to 12 bugs going one way and 12 the other leading to an angular momentum of zero, but I'm not sure exactly how.
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Thinking about their hint, the bugs must shift positions as a function of #clockwise - #counterclockwise, since they can't shift relative position and must end in a state with a bug in each of the original starting positions.
I'll get their exact reasoning on Saturday
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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.
