From: mathwft@math.canterbury.ac.nz (Bill Taylor)
Subject: Re: polygonal pi puzzler [Spoiler]
Date: 5 Oct 1999 01:34:05 GMT
Newsgroups: sci.math,rec.puzzles
Keywords: geometric probability question can be counterintuitive

|> r.e.s. showed that if you pick a random point, the expected length of
|> the interval containing that point is 2/(n+1), while the length of an
|> interval not containing the point is 1/(n+1).

Ah, right.  This is for n intervals determined by n points chosen uniformly
and independently around a circle, right?    I didn't see the res proof,
but there is a very neat way of doing it highlighted once by Martin Gardner.

|> I continue to believe that there is a way to apply the "looking on both
|> sides argument", but the lack of independence makes it quite tricky.

Maybe it'll satisfy this query, though I'm not quite sure what you mean
by "lack of independence" here.


Anyway:-  we place n points at random on a circle through the origin; 
======    i.e. independently, with each one having uniform distribution.  
This partitions the circle into n arcs.     What is the expected length 
of the arc containing the origin; and of an arc not containing the origin.

Solution:  Place (n+1) points at random on a circle.  By symmetry, each 
--------   interval has expected length C/(n+1)   (for circumference C).
Now, (glib trick!), choose one of the points at random, and label it "O",
and slide the circle round till "O" is at the origin.  The remaining
points are generated as if they were the original n points in the problem.
Thus,  E[origin interval] = 2C/(n+1),   and E[other interval] = C/(n+1).

[Technical note for experts:-  we don't even need the independence of 
 the n points - just exchangeability of the n points with the origin.]


This is so glib it's bound to generate a lot of fury from the unbeliever,
but it's definitely OK.    Equivalently, and slightly less obviously;
if n points are chosen at random on a finite line, the expected length
of each of the (n+1) intervals is  L/(n+1).   It's the same problem
as the above, unrolled.  There is a tendency to feel that maybe the
END two intervals might have a slightly different expected length,
but it is not so.  Just unroll an (n+1)circle with a distinguished point. 


Neato kool, huh!?     Amazingly, it is still true in the discrete problem.

If you want to know the expected number of cards before the first Ace 
appears in a shuffled deck of cards, just imagine the deck rolled up
with an extra end-marker card shuffled in.  Now the expected number
of cards between any adjacent pair of the 5 special cards, is 48/5;
by symmetry.  Unroll at the end-marker, chuck it away, and still now
the expected # of cards till the first Ace is 48/5.     Sneaky plus!

It is also the expected # cards between the 1st and the 2nd Ace, and so on.


Probability - full of cute surprises!   I love it.

   _______
  /\ o   o\          
 /o \ o   o\_____      
<    >------> o /|    =====================================================
 \o /  o   /___/ |    God does not play dice with the universe, he plays Go
  \/______/   o| |    =====================================================
        |   o  |o|
        | o    | ;
        |______|/

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                Bill Taylor     W.Taylor@math.canterbury.ac.nz
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  William Tell, missing the apple but hitting his son in the eye 3 times:-

                     YES!   Look at that GROUPING!
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