Originally published at: http://boingboing.net/2016/11/14/the-best-place-to-sit-in-a-s.html
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I guess “not sitting in a suicide circle” is not an option?
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“Be right back, I have to use the bathroom.”
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“today we’ll be going counter-clockwise”
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I like numberphile videos, but that one is a home run.
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That would be my first pick.
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“No, carry on without me.”
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On the plus side, as long as you’re one of the last two standing, you only have to win one sword fight.
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Trick question. The circle is a single point.
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“OK, first we’re going to go through it once, and I’ll gather data before I decide where to stand.”
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Don’t you mean dress rehearsal to make sure we all know our parts?
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I thought Josephus’ actual trick was to go over to the Romans in the first place.
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A related (OK, kinda completely different) puzzle that I like:
You and your two arch-nemeses agree to have a three-way pistol duel. You stand in a triangle and agree to take turns shooting at each other. You’ll go round and round in order, A, B, C, A… shooting until only one of you is left standing. Everyone wants to win and acts rationally.
Everyone knows how well the others shoot. Specifically, A (you) have a 1/3 chance of hitting your target, B has a 2/3 chance, and C “the sharpshooter” always hits.
You, you lucky dog, are person “A” and get to shoot first.
How should you shoot to maximize your chance of winning? (And what are your odds?)
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I’ll make sure your parts are kept close together, but I want to get the maths right.
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Oh boy, leg cramp. Go on without me. I’ll be along shortly.
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I was thinking the same thing. Counter-clockwise negates the need to skip a man.
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[note to self: don’t join a doomed army with @knoxblox]
“Please select your first choice for unit assignment.”
41st Number Theorists Brigade
Spatially Intuitive Artists Company
Boing Boing Regulars
Army Fencing Team
Disorienteers 
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The justification for the binary trick is that you’re just doing the W(n) = 2l+1 in binary.
Starting from the number total n to get to W(n):
- Drop the digit signifying the part of the total number that is the largest power of 2. Obviously, this digit will be is a 1. This subtracts the largest power of 2. You are now left with l.
- Left shift. This is the times 2l.
- Add a one in the 2^0 position (supposedly the 1 you removed from the front in step 1, haha). This increments by 1, getting us to 2l+1.
Voila.
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i would shoot at my foot first. I only have a 1/3 chance of hitting and the other two shooters will hopefully not think I’m a threat and focus on each other.
My second shot will be at whoever is left over.
I don’t know if that’s the correct answer, it’s just what I would do.
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First shot into the air. Second shot at whichever one is standing?
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