Conference rusure::math

	Before plunging into the geometry of the problem, I've gotta confess
that I've not fully resolved what probability *means* in the context of
this kind of question.

	Let me give a simpler example to show where I'm hung up.

	Suppose that we tile the upper right quadrant of the Cartesian
plane with unit squares, which are black/white with probability 50%/50%.
Label each tile by the co-ordinates of its bottom left hand corner.

	What's the probability that there is an entire column which is
white?   ie: the tiles (x,0),(x,1),(x,2)... are all white.   Is this even
a meaningful question?

	How would I tackle this problem?   The probability that the bottom
n tiles in a column are all white is 2^(-n).   So the probability that one
of the first m columns contains a tower of at least n tiles is:
	1 - (1-2^(-n))^m.

And we want the limit of this as m & n both go to infinity.   But how
exactly do we take this limit.   Let's say that we're going to walk out
on a path (n,m(n)), where m(n)=k(n)2^n, and k is a function we are free to
define.

	lim (x->0)   (1-x)^(1/x) = e^(-1).
So
	lim (m->infy,n->infy)    1 - (1-2^(-n))^m
=	lim (n->infy)	   1 - e^(-k(n))

So, and we can now converge this to whatever we like by defining k
appropriately.   What this tells us is that no well-defined value of
	lim (m->infy,n->infy)    1 - (1-2^(-n))^m
exists.   I can believe that.

	But for the hexagonal problem we are asked to discuss, how is
probability defined so that we don't run into exactly this kind of
problem?
	
Cheers,
Andrew.

    The assignment of probabilities is not cut-and-dried, as you have
    noticed.  What we usually do is to read the original question over and
    over until we have an 'iron-clad' interpretation and then use that to
    define the model.  :-)
    
    In practical, not puzzle, problems, nature gives a hand.  We develop a
    model based on the theoretical concepts we are testing.  For cases with
    denumerable infinities like these, we could proceed as follows:
    
    1) Assume a non-terminating process which enumerates the cells in some
    fashion.
    
    2) Assume that as each cell is enumerated, it is independently colored.  
    
    3) The operation of 1) and 2) describe the way in which the Black and
    White subsets are developed, also via non-terminating processes.
    
    4) Develop a sequential version of the geometric question (this may be
    tough) and calculate the probabilities in conformance with the question
    and 3).
    
    /rab

        I would interpret the probability in the sense of measure
        over the space of subsets of a countable set.
        
        Dan

>        I would interpret the probability in the sense of measure
>        over the space of subsets of a countable set.

	Can you explain that statement simply?

-------------------------------------------------------------------------------

	I intuit that the way forward with this is to exploit the fractal
map which maps each clump of 7 hexagons to a single hexagon.   There is a well-
known tiling of clumps of 7 hexagons which is a hexagonal tiling itself.

	I would seek to repeatedly apply this mapping, getting bigger and
bigger hexagons at each stage, and I would be asking at each stage:

	"What are the probabilities that there is a black path between edge
i and edge j, for all i,j in {1,2,3,4,5,6}?"

	Then I look at the limit of these probabilities as I repeatedly
apply the fractal map.

Andrew.

>>	>        I would interpret the probability in the sense of measure
>>	>        over the space of subsets of a countable set.
>>
>>	Can you explain that statement simply?
        
        Pick a bijective correspondence between the tiles and the
        nonnegative integers \omega = {0,1,2,3,...}.  Each
        coloring can be characterized by the subset of black
        tiles, which corresponds to a subset of \omega, which
        corresponds to an element of 2 ^ \omega = {0,1} ^ \omega. 
        So to the set of colorings with an infinite contiguous
        set of tiles that is all one color there corresponds a
        set of elements of 2 ^ \omega.  If that set is
        measurable, then take its measure to be the probability.
        
        Define the measure as follows.  For each finite subset E
        of \omega and each g:E -> {0,1}, define the measure of {f
        such that f:\omega -> {0,1} and f restricted to E = g} to
        be 1/2^|E|.  For example, the measure of the set of f
        such that f(7) = 0, f(17) = 1, and f(32) = 1 is 0.125. 
        Extend this measure to the \sigma-algebra of subsets of
        2^\omega which is generated by the basic subsets defined
        above.
        
        Alternatively, ignore the colorings with only finitely
        many black tiles (there are only countably many such, a
        set of measure zero) and map the remaining elements of
        2^\omega into the closed interval [0.0, 1.0] of the real
        line by sending f into sum(f(i)/2^(i+1), i = 0,1,2,...}.
        Then use Lebesgue measure on the subset of [0,1]
        corresponding to colorings with infinite monochromatic
        contiguous sets.
        
        Dan

>        Then use Lebesgue measure on the subset of [0,1]
>        corresponding to colorings with infinite monochromatic
>        contiguous sets.
    
    How do you know that's a measurable set?
    
    Can we estimate the answer via software?

        re .6 "How do you know that's a measurable set?"
        
        This previous statement from .5 was still operative at
        that point.
        
> .5					 If that set is
>	measurable, then take its measure to be the probability.
        
        re .6 "Can we estimate the answer via software?"
        
        I'm not sure how to do that.  Use a theorem prover? :-)
        
        Dan