SPUG: Stick Riddle

[Brian Hatch]

> This is the crux of the djinn's counsel: to break the stick once and then
> again (two whacks with one blade), instead of taking a double-headed
> cleaver to it (one whack with two blades).

But what is the effective difference?

Never was it said which piece should be cut again.  should it
be the big or the small one?  The left or the right one?  Or
should it be random?  This is assumption one if you break in two
phases.

(If you choose utterly random, there's no difference from using a
double-headed cleaver.  Although I suppose you could argue that
very very very very infrequently you get the same random number
twice in a row and end up cutting the stick in the same place,
but I think we can rule that out for a large enough sample size.)


The second assumption is "is a stick likely to break at every
point evenly".  If I bend the stick to break it, it's much more
likely to break near the middle than under my thumb.  I think
building in stick-breakage-vs-distance logic requires research
beyond the scope of the question.

I'm sticking with 25% as my answer.  I'd love to hear the mathematics
that yielded 50%.  (Not counting the averaging of random data, which
yields very bad random data.)


> (Do you know the apocryphal story of how Gauss added up the numbers from 1
> to 100?)

Would that be
	
	1+100 + 2+99 + 3+98 ... + 49+52 + 50+51
	
	==

	101   + 101  + 101  ... + 101   + 101


	== 101 * 50 == 505

--
Brian Hatch                  I used to be a lumberjack,
   Systems and                but I just couldn't hack
   Security Engineer          it, so they gave me the ax.
http://www.ifokr.org/bri/

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