Conference rusure::math

    Are these soldiers (presumably infinite in number) standing anywhere on
    the plane, or just at lattice points (integer (x,y) coordinates)?
    
    The reason I ask is because if they are standing "anywhere" then the
    probability of there being equidistant neighbors is zero, so the line
    about choosing one at random seems spurious.
    
    This reminds me of that problem with about 10 different solutions
    depending on how you define "random".
    
      John

    These soldiers are standing anywhere on the plane.  I wonder if the
    results would differ if the soldiers' placement were limited to lattice
    points?  Would you have any comment?

    You're right that the probability of equidistant neighbors is zero; I
    only threw in that line in case someone wanted to computer simulate the
    "presumably infinite" number of soldiers. In that case, the probability of
    equidistant neighbors would not be zero but very small, depending on
    the precision of the machine.  When I lazily modeled this before, I
    placed the soldiers on a sphere and limited the precision of the
    coordinates to .1, so I did have equidistant neighbors.  

Yuck, what a distasteful way to pose an otherwise interesting problem.

First of all, why would British soldiers be firing at each other ?  Not
too complimentary to their intelligence...

Secondly, why must we keep kill kill killing even in such innocent subjects
as math.

o.k. Eric stop complaining.  What better picture would you give for this
problem ?  How about:

	A group of children are battling with water pistols...

Any other ideas ?

/Eric

>Secondly, why must we keep kill kill killing even in such innocent subjects
>as math.

	Sometimes I feel that even in innocent abstract problems, that my
intuition is closely associated to the 'killer instinct'.   I feel like a
hunter trying to nail a particularly dangerous prey.

	In real life, I am a lot more peaceable!

	However, in this particular case, the bloody metaphor did seem
gratuitous.   I like the water pistols idea.

Regards,
Andrew the Great White Hunter.

    	How about randomly distributed people who on a given day give
    flowers to their nearest neighbor?  
    
    - Gus

    Here's a quick estimate which agrees with .0, with a more rigorous
    approach to follow in a later note, if I get time.
    
    I'll use the model of an infinite playground with an infinite number of
    kids with water pistols, distributed uniformly and independently.  So
    what is the probability that a kid does not get wet.
    
    Each kid has Nav neighbors, on the average, near enough that they might
    aim at him.  On a line, Nav is exactly 2.  On a plane, Nav must be
    greater than zero but less than seven.  But each of those kids has Nav
    neighbors, so the chance is only 1/Nav that they will shoot at this
    kid.  The probability of the kid staying dry is
    
    	P(dry) = ( 1 - 1/Nav ) ^ Nav
    
    For Nav=4 (just a guess for a plane) 
    
    	P(dry) = ( .75 ) ^ 4 = 0.316
    
    For higher dimensions, Nav should go up, although not very fast.  In
    any case, I think that the expression has an upper bound of
    
    	P(dry) <= exp(-1) = 0.368
    
    Interesting that even with lots of near neighbors, your chance of
    getting wet is pretty small.

    Now for a slightly more rigorous approach.  Since this was an OMNI
    problem, ther is probably an AHA! which I have missed.  But maybe .0
    left out some detail, and the problem as originally stated had an AHA!
    Anyway...
    
    First, we'll ask and answer a simpler question.  We'll need this
    result, and it will also give us a general method.  
    
    What is the probability that a circle of radius r contains one or more
    kids?  Assume for convenience that kids are distributed across the
    playground at unit density.  Call this probability PK(r).
    
    The probability that a ring with radius r and width dr contains a kid
    is 2 pi r dr.  
    
    Define the two propositions 
    
    	A = ring contains a kid
    	B = circle inside ring contains a kid
    
    Since the kids are distributed independently, the combining law is
    
    	P( A or B ) = P(A) + P(B) - P(A) * P(B)
    
    which gives
    
    	PK(r + dr) = PK(r) + 2 pi r dr - PK(r)* 2 pi r dr
    
    A bit of rearrangement gives
    
    	PK(r)/dr = 2 pi r * ( 1 - PK(r) )
    
    which is an ordinary differential equation which has the solution
    
    	PK(r) = 1 - exp( - pi r^2 )
    
    which has all the right limiting properties.
    
    
    Now ask the slightly more difficult question: what is the probability
    that a kid P is squirted by a kid Q in the ring of radius r and width
    dr.  In this case our two propositions are
    
    	A = ring contains a kid
    	B = the circle of radius r around kid Q is empty
    
    See where I am going?  More in a later note.

>    The probability that a ring with radius r and width dr contains a kid
>    is 2 pi r dr.  
    
    Where did this come from?

>>  >    The probability that a ring with radius r and width dr contains a kid
>>  >    is 2 pi r dr.  
>>    
>>      Where did this come from?

	I think it is the unit density times the area
	of the ring.

	Dan

    
    .9 is correct, it is just area times unit density
    
    [ isn't it annoying when noting gets interrupted by work? ]
    
    where was I?  Oh, yes...
    
    	A = the ring contains a kid
    	B = the circle of radius r around the kid is empty
    
    
    The probability that kid P is wet because of a kid Q in the ring is given 
    by
    
    P(A and B) = P(A) * P(B)
    
    	= 2 pi r dr * [ 1 - exp( - pi r ^2 ) ]
    
    How can we interpret this equation?  Rings of small radius are not
    likely to contain any kids.  Rings of large radius are more likely to
    contain kids, but these kits are probably aiming at somebody closer to
    them.  It is only rings of intermediate radius that contribute to the
    wetness of a given kid.
    
    Now let's change the meaning of A and B to
    
    	A = kid P is wet because of kid Q in ring
    	B = kid P is wet because of one or more kids in the circle inside 
    		the ring
    
    and call PW(r) the probability of proposition B
    
    Then we can use the original combining law
    
    P(A or B) = P(A) + P(B) - P(A) * P(B)
    
    PW(r + dr) = 2 pi r dr * [ 1 - exp( - pi r^2 ) ] 
    		+ PW(r) - 2 pi r dr * [ 1 - exp( - pi r^2 ) ] * PW(r)
    
    after the same kind of rearrangement
    
    d PW(r)/dr = 2 pi r * [ 1 - exp( - pi r^2 ) ] * [ 1 - PW(r) ]
    
    [oops, I see that I left off the leading d when I wrote the previous
    differential equation.]
    
    I can't integrate that right now, but I may try later.  Meantime, I can
    make a few statements:
    
    	PW(r) is monotonically increasing
    	PW(0) = 0 
    	PW(r) = pi r^2 approximately for small r
    	PW(r) = constant approximately for large r
    	1 - PW(r) in the limit of large r is the answer sought in .0
    
    For higher dimensions, we have the same kind of problem.  The only
    thing that changes is the form of terms like 2 pi r.  Call PWk the
    limit for large r of PW(r) in a space of dimension k.  I suspect that 
    1-PWk will increase slowly towards 1/e as k increases.

>    Now let's change the meaning of A and B to
>    
>    	A = kid P is wet because of kid Q in ring
>    	B = kid P is wet because of one or more kids in the circle inside 
>    		the ring
>    
>    and call PW(r) the probability of proposition B
>    
>    Then we can use the original combining law
>    
>    P(A or B) = P(A) + P(B) - P(A) * P(B)

	No, I don't buy it.   A & B are no longer independent (if Q wets P,
then the kids in the circle have to avoid Q).   So the expression should be:

     P(A or B) = P(A) + P(B) - P(A) * P(B|A)

& P(B|A) doesn't seem to me to simplify.


Puzzled,
Andrew.

>	No, I don't buy it.   A & B are no longer independent (if Q wets P,
>then the kids in the circle have to avoid Q).   So the expression should be:
>
>     P(A or B) = P(A) + P(B) - P(A) * P(B|A)
>
>   & P(B|A) doesn't seem to me to simplify.

I think you have a point here, although I would prefer to say that if Q wets P,
then that implies that there are no kids in the intersection of the circles
around P and Q, both of radius r.

I think there is a way to save the method, but it is going to make the 
calculations a lot harder.  If I come up with anything, I'll be back.

    I suspect there is a paradox of infinity stuck in there somewhere.
    There are infinitely many kids, infinite space and presumably water
    pistols with infinite range. If a kid fires at all he/she is sure
    to hit somebody, and with infinite water in motion, sure to get
    wet....
    
    It seems to be like conundrums that can arise when attempts are
    made to calculate a 'mean' distance from the earth of the 'average'
    star. With all those stars sitting an 'average' distance away
    firing an average amount of radiation at us, why aren't we fried?
    

> I was grasping for an word to describe the sort of 'binoculars cutout'.   

I think the word you want is 'ogive', a term that describes a Gothic arch.
I first learned the term in the context of aerodynamics: the cross-section 
of certain kinds of wings.

    I don't like infinity appearing in problems of this sort, but
    'a large number' will do....
    
    Randomly distributed... Suppose it happens that the random
    distribution is into near-neighbour pairs. Then if there is
    an even number of kids all get wet. If there is an odd number
    all but one. That would seem to be the worst case.
    
    The best case I can think of is if the kids were in clusters of
    eight arranged as follows: A kid in the middle with five others
    around him all closer to him than to each other (the limiting
    case is six but not practical) and all spray him/her. He/she
    sprays the nearest one of the five. Outside the group there
    are two more nearer to the fall guy than to each other and they
    also spray that one. So one kid gets hit five times, one gets
    hit three times and six are dry, a survival rate of 75%.
     Those two cases appear to be the end limits, anything between
    being possible...
    
    
    

	Before anyone else does, let me point out a small blooper in one
of my earlier replies.   A size-3 river will have a below average quantity
of lake, just like a size-2 river (waterhole).   Consequently, I cannot
make a correspondence between each larger river and a number of waterholes.
However, this was a local remark and no later reasoning hinged on this
observation.

	Moving from waterholes to ratholes, I thought an ogive was an
arch which was pointy at the top.   That's not what I want here, I want
the kind of cut-out that you had in Mission Impossible when Barney was
looking through his binoculars at the General's staff car, with Phelps in
a false moustache as the chauffeur.

Regards,
Andrew.

        My interpolation and approximation book (Interpolation
        and Approximation, by Philip J. Davis) defines on pg. 86:
        
        "Let z1,...,zn be n complex numbers not necessarily
        distinct.  For r > 0, the set of points satisfying
        
        	|(z-z1)(z-z2)...(z-zn)| = r^n
        
        is called a lemniscate and will be designated by [capital
        gamma sub r].  The points zi are called the foci of the
        lemniscate and r its radius.  The set of points
        satisfying the inequality
        
        	|(z-z1)(z-z2)...(z-zn)| < r^n
        
        will be designated by [script capital L sub r].
        
        With zi fixed and r varying, we may speak of a family of
        confocal lemniscates. [...]
        
        The behavior of confocal families of lemniscates follows
        the pattern of Example 7.  If z1,...,zn are n distinct
        points and if r is sufficiently small, the locus [capital
        gamma sub r] consists of n closed contours surrounding
        precisely one of the foci.  As r increases, the contours
        increase in size until two or more of them touch and then
        coalesce, reducing the number of contours.  This
        coalescence continues until for sufficiently large r
        there is but one contour surrounding all the foci.  As r -> oo,
        the single contour becomes more and more circular in its
        shape.  If the zi are not all distinct, this picture can
        be modified in an appropriate way."
        
        The diagram shows three foci, in a triangle.  Each has a
        little "circle" around it, for the smallest r.  The next r
        is like a figure eight around the two closer points and a
        more oblong "circle" around the third.  The next figure
        eight includes the smaller one in one end and the third
        point in the other.  Later oblong "circles" include all
        three points.  The "binocular" shape isn't given ... the
        curve intersects itself like a figure eight.  However, a
        second diagram with two foci does have a smooth binocular
        shape (non self intersecting) for an r larger than the r
        that gives the (self intersecting) inner figure eight. 
        So I suspect the binocular shape could come about with
        more than two foci also.
        
        Dan
        

>        distinct.  For r > 0, the set of points satisfying
>        
>        	|(z-z1)(z-z2)...(z-zn)| = r^n
 
If this is meant as a product, then I don't think that this is the shape
described earlier.  I think that was defined for r = | z1 - z2 | > 0
the set of points satisfying 

	|| z - z1 | <= r    OR    | z - z2 | <= r       

This figure always has cusps.  It is similar to the figure you get in Venn 
diagrams when they illustrate the union of two sets.  I also cound not find a 
name for it.

I could not find a definition for ogive, but ogee is illustrated in my 
dictionary, and it is more pointed than a gothic arch.  In any case I suspect
that ogive is more like the intersection figure, satisfying

	| z - z1 | <= r    AND    | z - z2 | <= r       

>    The best case I can think of is if the kids were in clusters of
>    eight arranged as follows: A kid in the middle with five others
>    around him all closer to him than to each other (the limiting
>    case is six but not practical) and all spray him/her. He/she
>    sprays the nearest one of the five. Outside the group there
>    are two more nearer to the fall guy than to each other and they
>    also spray that one. So one kid gets hit five times, one gets
>    hit three times and six are dry, a survival rate of 75%.
 

five kids in a regular pentagon around 1 kid in the center, so they 
all spray the kid in the center, a survival rate of 83 and 1/3 %  You can 
scatter these pentagons around the (in)finite playground, as long as you
keep them far enough apart.

>    I suspect there is a paradox of infinity stuck in there somewhere.
>    There are infinitely many kids, infinite space and presumably water
>    pistols with infinite range. If a kid fires at all he/she is sure
>    to hit somebody, and with infinite water in motion, sure to get
>    wet....
>
In the original problem, perfect accuracy was assumed, so a kid hits only the
nearest other kid.
    
>    It seems to be like conundrums that can arise when attempts are
>    made to calculate a 'mean' distance from the earth of the 'average'
>    star. With all those stars sitting an 'average' distance away
>    firing an average amount of radiation at us, why aren't we fried?
    
A strange statement of Olbers' Paradox.  However, the usual statement relies
on the fact that the surface of a sphere grows as r^2, and radiation intensity
diminishes, not coincidentially, as the inverse of r^2.  So assuming a static
infinite uniform distribution of stars, the shell of radius r should contribute
the same starlight on earth, for every value of r.  As r grows without limit,
so does the total starlight reaching earth.  There is no mathematical paradox 
here.  The ASTRONOMY conference is the proper place for discussing the physical
paradox and its various resolutions.

In the playground case, as shown in .10, 

    The probability that kid P is wet because of a kid Q in the ring is given 
    by
    
    P(A and B) = P(A) * P(B)
    
    	= 2 pi r dr * [ 1 - exp( - pi r ^2 ) ]
    
The first factor increases with r, but the second factor decreases much faster
than the inverse of r.  So we can safely consider the limit as r increases 
without bound.

This only says that there is no physical paradox in the playground.  We may 
still fall into a mathematical paradox (or worse) if we are not careful.  I 
think the argument ending "sure to get wet" above is an example.

    Here's some results from a simulation run over the July 4th US holiday.
    Several thousand kids randomly populating a unit circle (not square),
    and one round of shooting.  The percentage of unshot kids is seen to
    vary from 26.98 to 29.88, with an apparent mean/median of 28.4 or so.
    This is in histogram format, each * representing one test case.
    
    %   #
    
26.9817	*
27.0122	*
27.0732	*
27.1646	**
27.2561	***
27.3476	*
27.3780	**
27.4085	***
27.4390	**
27.5000	*****
27.5305	**
27.5610	****
27.5915	****
27.6220	*
27.6524	***
27.6829	*****
27.7134	****
27.7439	******
27.7744	********
27.8049	****
27.8354	*****
27.8659	****
27.8963	*****
27.9268	*******
27.9573	**********
27.9878	************
28.0183	**********
28.0488	*********
28.0793	**********
28.1098	********
28.1402	*******
28.1707	**********
28.2012	*******
28.2317	*****
28.2622	***********
28.2927	******
28.3232	****************
28.3537	**********
28.3841	****************
28.4146	**************
28.4451	**************
28.4756	****
28.5061	*******
28.5366	**********
28.5671	*******
28.5976	***************
28.6280	***********
28.6585	*************
28.6890	*****
28.7195	*******
28.7500	***************
28.7805	********
28.8110	***
28.8415	*********
28.8720	*********
28.9024	******
28.9329	*********
28.9634	**
28.9939	***
29.0244	****
29.0549	*****
29.0854	*****
29.1159	***
29.1463	***
29.1768	**
29.2073	*****
29.2378	**
29.2683	******
29.2988	*
29.3293	**
29.3598	******
29.3902	*
29.4207	**
29.4817	*
29.5122	*
29.5427	**
29.6037	*
29.6341	*
29.6646	*
29.8780	*

In a previous reply on the best case arrangement, I forgot that the kid in 
the center has to shoot somebody.

So take a regular pentagon of kids as before, with a kid at the center.

Outside one vertex arrange three kids at an equal distance from the vertex,
slightly larger than the 'radius' of the pentagon.  Let the internal angle 
at the vertex of this quadrilateral be slightly larger than 120 degrees.

Now there are 9 kids and two are wet, for a ratio of 77 and 7/9 %


		d		d




			w


	     d			     d



			w



	       d		 d



			d


I don't think we can do better without tiling the plane with nearly regular 
pentagons.  I seem to remember that that is possible, but I don't remember
the details.

    Re -1
    
    I think you may have found a slightly better one than I did. When
    I thought about it I decided you couldn't quite do it with nine
    but perhaps I was mistaken! (I often am).
    
    

    Re .24
    You are right, I think I was coming up with a garbled version of
    'Olbers Paradox' and it is probably a red herring. What was going
    through my mind was that when objects are scattered over an infinite
    plane then their average distance apart is...infinity! Also when
    an infinite number of objects are scattered over an infinite plane
    then there can be objects that are an infinite distance from the
    nearest object.
    

>>							     Also when
>>    an infinite number of objects are scattered over an infinite plane
>>    then there can be objects that are an infinite distance from the
>>    nearest object.
    
	No there can't.  Let object O be at point (a,b).  Then if any
	other object P is at (x,y), the distance between P and its
	nearest neighbor must be bounded by the distance between P
	and O -- sqrt((x-a)^2 + (y-b)^2) in Euclidean space.

	Dan

    I don't agree. My kind of infinite plane can contain an infinite
    number of objects all an infinite distance from each other!
    

>>	    I don't agree. My kind of infinite plane can contain an infinite
>>	    number of objects all an infinite distance from each other!

	Can you give me the coordinates of two of them?

	Dan

>    I don't agree. My kind of infinite plane can contain an infinite
>    number of objects all an infinite distance from each other!
			   ====================

But they cannot then have Cartesian coordinates. The set of distances may
be infinite, but any individual distance can only be finite. Another way of
looking at it: the set of positive integers is infinite, but the value of
any specific one of them is finite. 

>         <<< Note 1258.30 by MACNAS::JCREAN "I'll do it tomorrow..." >>>
>                   -< Depends what you mean by infinite... >-
>
>    I don't agree. My kind of infinite plane can contain an infinite
>    number of objects all an infinite distance from each other!
    
    Have a look at "Naive Set Theory" by Paul Halmos;  you'll get an
    idea of why one has to be careful with when casually playing
    with infinities...

    - Jim

    O.K. Here's my naive way of thinking...probably contains an error
    somewhere.....but I'm always willing learn....
    
    Imagine a curve of infinite length which is bounded within a finite
    space, like Sierpinski's snowflake. Two points on that curve will
    be an infinite distance apart along the curve. Any number of points
    can be put between those two points and they will all be an infinite
    distance apart around the curve, and infinitely many points can
    be set in this way. Next, expand the curve to a diameter of infinity
    and it becomes a plane with an infinite number of points an infinite
    distance from each other. Agreed this does not define the location
    of any of the points, it only describes the way the plane could
    be constructed.
    
    There is possibly an error somewhere, or perhaps they don't even
    make infinite planes the way they used to!
    
    This is only a side issue, as I said before I wanted to exclude
    'infinity issues' when offering a solution to the problem
    under discussion: I am simply cautious about infinity appearing
    in problems where it need not necessarily appear. (Also I am
    reluctant to discuss places I've never visited myself).
    
    Thanks for the 'Halmos' reference, I'll add it to my want list.
                                                  
    

    Re .34:
    
    What you have described is not a plane.  I am not sure what it is, but
    it is not a plane.  (Ie. it does not have the same geometry and
    topology (different metric) as a Euclidian 2-D plane).
    
    --  Barry

    Re: .34
    
    Aren't you just talking about points on the curve only having infinite
    distances apart if one consider the distance being the length traversed
    from one point to the other along the curve? What about the points
    in the interior? How are their distances measured?
    
    Anyway, it is interesting to see that there do exist sets whose points
    are all infinitely far from each other, e.g. points on the fractal
    curves. But then that won't be a metric so it is not a distance after
    all. So I guess it isn't meaningful to talk about every point being
    infinitely far from any other.

    Re .36
    Yes, the distances between points along the curve are infinite,
    the straight-line distances between points in the plane in
    which the curve is situated are not, but let off a stick of
    dynamite inside the curve and blow it out into infinite space
    and the difference between the two distances gets less and less
    as the diameter of the fractal approaches infinity.
    
    (Actually I suspect there is a lurking imbecility in this reasoning
    somewhere but I cannot think what it is).
    
    

	re .-1

	One can just as easily say that except for (0,0), that
	every lattice point is unoccupied except for perhaps
	finitely many planes towards the beginning of the sequence.

	Dan

	If you really need infinite distances, use nonstandard analysis.
	There you can have "nonstandard" reals greater than every "standard"
	real.

	Dan

    re .40:  that's the standard answer for these types of questions.
    
      John

    Returning to the main theme, a 'best case' was postulated in
    .26, where only two kids out of nine would get wet. I wonder
    if there is a better solution? I've thought about it but I
    cannot come up with a better one myself.
    

This may be all wet, but here is a slightly different way to look at this...

Suppose we have a unit density Poisson distribution of sites in Euclidean
n-space.

Then the probability of there being k sites in a volume is V^k/k! exp(-V).

What is the probability of a site at distance r from a target site having
no nearer neighbors, but being the (k+1)'th closest site to the target?

This will be the probablilty that a ball of radius r around the source
site contains no other sites, while the "lune" left when the ball around
the source site is subtracted from the ball around the target site contains
k sites.

In n-space a ball or sphere of radius r has volume B[n] r^n.
B[1] = 2, B[2] = pi, B[3] = 4/3 pi, etc.

A lune will have volume L[n] r^n, where L[n] is what is left when
a lense shaped region of two spherical caps is subtracted from a unit ball.

The volume of the spherical cap can be obtained by integrating the
unit ball over planar slices:

	C[n] = INT(1/2, 1) B[n-1] (1-x^2)^((n-1)/2) dx
	C[n] = B[n-1] * I[n]

	C[1] = 1/2, C[2] = pi/3 - sqrt(3)/4

Integrating I[n] by parts and fiddling around we get a recurrance

			           | 1
	I[n] = x (1-x^2)^((n-1)/2) |     + (n-1) ( I[n-2] - I[n] )
			           | 1/2

        I[n] = (n-1)/n I[n-2] - 1/(2 n) (3/4)^((n-1)/2)

so we can calculate the constants C[n] and L[n] = B[n]-2*C[n] for any n
by priming the recurrance.

The probability P[n,k](r) of the k'th site "firing" at the target site
(where k = 0 for the closest site) will be

	(L[n] r^n)^k
	------------ exp(-L[n] r^n) exp(-B[n] r^-n)
	     k !

The overall probability of the k'th nearest site firing at the target
site will be the integral over a spherical shell of thickness dr

	P[n,k] = INT(0, inf) P[n,k](r) d (B[n] r^n)

Plugging in and pulling the constants out the integral turns out to be
a gamma function which cancels the factorial!  We get the very simple
probability

	P[n,0] = B[n]/(L[n]+B[n])

	P[n,k] = (L[n]/(L[n]+B[n])^k P[n,0]

	P[n,k] = (1-P[n,0])^k P[n,0]

The unfortunate fact is that these events are not exclusive - more than one
site can fire at the target.  Otherwise we could just sum these probabilities
to get the probability that a site would be fired at at all.

In fact, it turns out that these probabilities sum to certainty by themselves!
To see this, we can turn the geometry around and ask what is the probability
that the nearest source site has 0 or more sites in the "lune" beside it -
which is certain to happen!

The correct expresson involves inclusion/exclusion

	P(hit) = SUM P[i] - SUM P[i & j] + SUM P[i & j & k] - ...
	       = 1 - SUM P[i & j] + SUM P[i & j & k] - ...

	P(missed) = SUM P[i & j] - SUM P[i & j & k] + ...

These higher order combinations look hopelessly hairy and I see no easy or
even systematic way to evaluate them, since they involve looking at the
intersections of various sized spheres and tallying the volumes and the
possible probabilities.  Even the pairwise probabilities require double
volume integrals (though you could fix the angle of one of the points.)

It looks bad, though there is the interesting fact that the probability of
being missed is expressable only in terms of combinatorial probabilities of
being hit:  that is kind of unexpected.

Here are some numerical values of the constants I, B, C, and L for the
volumes of n-spheres, n-caps, and n-lunes:

  n	  I[n]	         B[n]		C[n]           L[n]
 ---   -----------    -----------    -----------    -----------
  1    0.500000000    2.000000000    0.500000000    1.000000000
  2    0.307092425    3.141592654    0.614184849    1.913222955
  3    0.208333333    4.188790205    0.654498469    2.879793266
  4    0.149129437    4.934802201    0.624671924    3.685458352
  5    0.110416667    5.263789014    0.544884410    4.174020195
  6    0.083679590    5.167712780    0.440471706    4.286769368
  7    0.064508929    4.724765970    0.333363615    4.058038741
  8    0.050384987    4.058712126    0.238057272    3.582597583
  9    0.039763145    3.298508903    0.161387158    2.975734586
 10    0.031645696    2.550164040    0.104383609    2.341396821
 11    0.025361737    1.884103879    0.064676589    1.754750701
 12    0.020445559    1.335262769    0.038521557    1.258219654

And here are a few low order probabilities, which gives a feel for the
contributions in the above sums:

  n	  P[0]	        L/(L+B)         P[1]           P[2]
 ---   -----------    -----------    -----------    -----------
  1    0.666666667    0.333333333    0.222222222    0.074074074
  2    0.621504897    0.378495103    0.235236560    0.089035886
  3    0.592592593    0.407407407    0.241426612    0.098358990
  4    0.572465550    0.427534450    0.244748744    0.104638520
  5    0.557734205    0.442265795    0.246666762    0.109092271
  6    0.546588665    0.453411335    0.247829496    0.112368703
  7    0.537956396    0.462043604    0.248559312    0.114845240
  8    0.531153988    0.468846012    0.249029429    0.116756455
  9    0.525722170    0.474277830    0.249338370    0.118255661
 10    0.521339530    0.478660470    0.249544624    0.119447147

It is interesting that in high dimensional space the probability approaches
2^-(k+1) of being hit by the k'th nearest site.  This would have to do with
the fact that most of the volume of an n-sphere is near the surface I guess.

- Jim

    Just for grins, I tried assuming that the probabilties of the k'th
    nearest sites firing at a target were independant.

    If this were true then

	P(missed) = PROD(k >= 0) (1 - P[k])

    For 2 dimensions we get

	P(missed) = .249600933

    I tried a set of simulations of 200,000 points per batch (for a total
    of about 170 million points), being sure to have a border of extra
    points around the simulation set so there would be no errors due to
    boundary conditions.

    I got

	P(missed) = .284226

    Assuming these are Bernoull trials, we would expect a standard
    deviation of sqrt(P(1-P)/N) or .00003455 for our simulation.

    As a check on the simulation, I also tallied the average distance to the
    nearest neighbor.  In theory it should be exactly 1/2, and I got
    0.500066, so there are no likely horrible mistakes in the simulation.

    So assuming independance is not way, way, off but is not all that
    close either :-(

    One thing occurs to me - you could reduce the dimensions of the
    multiple integrals by fixing the outermost point at unit radius.
    Then the inner point(s) could be varied and a conditional probability
    for unit outer radius could be obtained.  Once this is done, the
    overall probability varying the outer radius could be done in closed
    form as the same kind of integral over spherical shells as the single
    point case.

    Still, the inner integrals look hairy;  I really have doubts that
    there is any clean closed-form solution at this point.

    - Jim

	I have a few comments following on from Jim Roth's excellent .43.

	Let S be a subset of N, the natural numbers. Then Let P(S) be the
probability that a point (wlog, at the origin) is targeted by the k'th nearest 
point, for each k in S.

	Then as Jim points out sum(k /in N) P({k}) = 1.

	Another way of looking at this, is to observe that each point is a
source for exactly 1 unit of water. Therefore the expected amount of water
*received* by any point must also be 1.

	Jim then writes (using different words) that 

P(missed) = sum(S /subset N) (-1)^|S| * P(S)

	We can simplify this, to derive:

P(missed) = sum(k=2,%) (-1)^k * Q(k)

where Q(k) is the probability that k random points share a common target, and 
"%" denotes infinity. This can be derived directly, or alternatively we can show
that:

	Q(k) = sum(S /subset N, |S| = k) P(S)

by swapping the order of summation and integration in computing the rhs.

	Although we have simplified away the summations, the integrals Q(k) 
still look quite hairy, as Jim observed. For n=2 dimensions (which is what the 
original problem addressed), we do know that Q(k) = 0 for k >= 6. (Proof 
elementary.)

	It does seem worthwhile to have a crack at computing Q(2), to see how 
much of the observed P(miss) is thereby explained.

	Alternatively, it would be interesting to repeat Jim's tests, 
described in .44, to find out experimental values for Q(2),...,Q(5).

    
Cheers,
Andrew.