SPUG: Stick Riddle

[Don Shanks]

Am I missing something, or do three sticks joined end to end always make
a triangle?

-----Original Message-----
From: owner-spug-list at pm.org [mailto:owner-spug-list at pm.org] On Behalf
Of Brian Hatch
Sent: Thursday, 02 January, 2003 17:37
To: Fred Morris
Cc: spug-list at pm.org
Subject: Re: SPUG: Stick Riddle




> 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/

Every message PGP signed



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