Conference rusure::math

I'd guess that the answer is larger than 1/3; 1/3 is the expected value of
the size of either range, so if you reduced the other range to a point you
would have a probability of intersection (= inclusion) of 1/3. Since the
other range is bigger than a point, the probability of intersection is
corresponingly bigger.

(Boy, it must be late - I know in my bones that the expected value of the
range size is 1/3, because that's why the "average" seek time of a disk is
the time to move the head through 1/3 of a full stroke, but the best way I
could find to derive it was to evaluate:

	1   x            1
	S ((S(x-y)dy) + (S(y-x)dy))dx	(how DO you make a good integral sign?)
	0   0            x

which yields 1/3 after too much bloody work)

Being too lazy right now to evaluate any more integrals, I'd be inclined to
guess 5/9 (= 1 - (2/3)^2), but I wouldn't put money on it....

    Aha! Here's an interesting argument: Let {a,b} and {c,d} be the
    end points of the two ranges. Assuming that each of a,b,c,d is randomly
    chosen from {0,1}, there are 4! = 24 orders in which the 4 numbers may
    appear in the interval. Of these, 2/3 occur with the ranges
    overlapping, so the desired probability is 2/3.
    
    Note that placing additional restrictions on a,b,c,d, e.g. a<b and
    c<d, has no effect on the probabilities, only on the number of cases
    to consider.

    How about two random rectangles on the unit square ?  What is the
    probability that they intersect ?
    
    How about two random cubes in three dimensions ?
    
    Obviously, we can generalize and ask for two random tesseracts in
    four dimensions etc.
    
    But wait !  Before getting too carried away, back to two dimensions
    for a spell:  What about two random CIRCLES !?  I don't know how
    to UNgeneralize this question back to one dimension.
    
    And now, of course, how about two random SPHERES . . .
    
    Gee, I almost have more fun ASKing these questions than bothering
    to figure out the answers.
    
    /Eric

Re .2:  I'm not yet convinced that the {a,b,c,d} 24-factorial method
    is valid.  Mightn't certain orderings of the four numbers produce
    unequally weighted probabilities, and hence merely counting which
    ones are overlapping might give an inaccurate result ?
    
    /Eric

    
    I believe .2 is correct.  Consider:
    
    	0-----w-----x-----y-----z-----1
    
    There is a 1/2 probability that w=R0 (since R0<R1 and S0<S1, the
    only other thing w could be is S0).  If w=R0, there's a 2/3 probability
    that they intersect (when y=R1 and when z=R1).  Therefore, the
    probability that w=R0 and they intersect is (1/2)*(2/3)=1/3.
    
    There's another 1/2 probability that w=S0.  If w=S0, there's the
    same 2/3 probability that they intersect (when y=S1 and when z=S1).
    The probability that w=S0 and they intersect is (1/2)*(2/3)=1/3.
    
    Since w=R0 or w=S0 by definition, the probability of intersection
    is 1/3 + 1/3 = 2/3.
    

    
    Given that there are n random ranges (as in 585.0), what's the
    probability that none will intersect? 
    

Let {a1,a2}, {a3,a4}, ... {a[2n-1], a[2n]} be non-overlapping ranges. These 
can be permuted in n! ways and each pair can be interchanged. So the 
number of non-intersecting cases is n!*2^n, out of a total of (2n)!
possible orderings.

... and that probability appears to be
		1
	---------------
	3*5*7...*(2n-1)

>   How about two random rectangles on the unit square ?  What is the
>   probability that they intersect ?

>   How about two random cubes in three dimensions ?

>   Gee, I almost have more fun ASKing these questions than bothering
>   to figure out the answers.

    Aha!  I see the solutions -- nice!

    Well, when I first thought about these questions, my instinct was
    to multiply
    
    	the area of one rectangle by the area of the other rectangle
    	in the case of the unit square,
    
    and
    
    	(modifying the 2nd problem slightly) the volume of one 
    	rectangular solid by the volume of the other rectangular
    	solid in a unit cube case;
    
    however, I backed off on this.  I want to answer a different question,
    first: 
    
    What is the probability that a point (a,b) is within a rectangle
    of random dimension, orientation, and location if the point and the
    rectangle are both entirely inside a unit square?
    
    I started to work this problem, but quickly got nowhere.  So, I
    simplified it:
    
    What is the probability that a point (a,b) is within a circle of
    random dimension, (orientation), and location if the point and the
    circle are both entirely inside a circle of radius 1?
    
    I haven't had much opportunity to work on this, but I think the
    solution should be simple.
    
    

    re .11:
                          
    > What is the probability that a point (a,b) is within a rectangle
    > of random dimension, orientation, and location if the point and the
    > rectangle are both entirely inside a unit square?

    seems to me the probability of the point being in the rectangle
    is exactly equal to the area of the rectangle (since the sample
    space is a unit square). 
    
    Am I overlooking something?
                                                   
                  /
                 (  ___
                  ) ///
                 /
      
    

    BTW, A random rectangle in a unit square may be chosen by randomly
    choosing two points in the square.

    re .13:
    
    only an orthogonal rectangle can be chosen this way, not a diagonal
    one.
    
                                                   
                  /
                 (  ___
                  ) ///
                 /
    

To deal with this problem you first have to resolve a philosophical issue:
Do you care what the a priori probability of the objects being in the 
surrounding figure is? Another way of saying this is to rephrase the 
questions to "given a surrounding figure and a random point in its 
interior, what is the probability of throwing another figure into the area 
so that it encloses the point?", and to do that you need to proscribe the 
universe of possible places the second figure can land.

If, conversely, you assume that THE SECOND FIGURE IS ALREADY THERE, the
solution is quite simple: the probability that a point falls inside both 
figures is simply the ratio of their areas. The shapes in this case are not 
relevant.

The philosopical issue is not at all easy to resolve, and depending on what 
model you choose for a random placement of the smaller figure, you can get 
an unbelieveable range of probabilities out. There is a nice example of 
this in E. P. Northrup's book, "Riddles in Mathematics" (D. van Nostrand, 
1944). He discusses the problem, 

	"A chord is drawn at random in a given circle. What is the 
	probability that the chord is longer than one side of an
	equilateral triangle described in the circle?"

and summarizes:

	... "If we assume that the chord, passing through a [random] point
	on the circumference... is as likely to make one angle with the 
	tangent as another, then the probability is 1/3. If we assume that
	the chord, drawn perpindicular to a diameter of the circle, is as
	likely to pass through one point on the diameter as another, then
	... 1/2. If ... the midpoint of the chord is as likely to be one 
	point in the circle as any other, then ... 1/4."

A related problem is: given a probability 0<=p<=1, does there exist a
rational definition of randomness for the problem in question that produces
a probability p as the outcome? In three dimensions, the answer is probably
'yes'! 

    Oops.  Right.  Let's assume these are aligned rectangles.

    Is the probability that the point is within the random rectangle
    the same whether the point is in one corner or in the center?
    
    It would seem that if, for example, the random rectangle has an
    area of .99, the point in the center is certainly contained.  But
    a corner point has less than certainty.
    

    Hypothesis:  The probability that two random rectangles, r1 and r2,
    both contained within a unit square, have a probability of intersecting
    equal to area(r1)*area(r2).
    
    Counter example:  area(r1)=0.5, area(r2)=0.6.  These two rectangles
    certainly intersect.  Yet, the product of their areas is less than one.
    In general, if area(r1)+area(r2)>=1 then it's certain that they
    intersect.
    
    Peter, if there's really an "Aha!" for this problem, it's got to
    be the best I've ever seen.
    

    Spoiler follows.
    
    What is the probability the projections of the rectangles onto one side
    of the unit square intersect?  What is the probability the projections
    of the rectangles onto another side of the square, normal to the first,
    intersect?
    
    
    				-- edp