Conference rusure::math

    Re .0:
    
    It is not possible with two cubic dice, because there are 36
    equiprobable combinations of faces that could come up, and 36 cannot be
    divided into sets of 11.
    
    Is there a regular polyhedron with a number of sides that is a multiple
    of 11?  I'm not familiar with them all, but I don't recall any with
    such a number of sides.  That rules them out as well.
    
    If you want to make dice that aren't regular polyhedra, you might get
    somewhere . . .  How about a pyramid with a regular 11-gon as its base. 
    You could spin it on its peak, so that it would come to rest on one of
    the 11 sides.
    
    
    				-- edp

>    It is not possible with two cubic dice, because there are 36
>    equiprobable combinations of faces that could come up, and 36 cannot be
    
    These are unfair dice, so your response is not germane.
    
    I'd answer the question but am busy researching my mother's drinking
    habits before I was born.
    
      John

    My first impression is that there are 10 constraints and 10 variables,
    so the answer should be yes.  Will think more about it later.  On the
    other hand, the variables have constraints of their own.  An interesting
    if somewhat perverted linear programming problem. :-)  Have been doing
    a lot of those for work these days.
    
    Eugene

        Let the first die have probabilities p1,...,pm of results
        x1,...,xm, represented by the z-transform
        
                           x1             xm
        	f(z) = p1 z   + ... + pm z
        
        Likewise let the second die have probabilities q1,...,qn
        of results y1,...,yn, represented by the z-transform
        
                           y1             yn
        	g(z) = q1 z   + ... + qn z
        
        The requirement is then that
        
                             2    3          12
        	f(z)g(z) = (z  + z  + ... + z  ) / 11
        
        So it looks like a factoring problem to me. :-)
        
        Dan

>>        The requirement is then that
>>        
>>                             2    3          12
>>        	  f(z)g(z) = (z  + z  + ... + z  ) / 11
        
                                      2      2      10
        The right hand side is (1/11)z (1+z+z +...+z  ).  If you
        assume factoring into sums of nonnegative integral powers
        of z, then the long term is
        
                         2        10                 2 pi i / 11
        	(z-w)(z-w )...(z-w  )		w = e
        
        This can be grouped into two polynomials in a number of
        ways.  However, you want the coefficients to be reals in
        the interval [0,1].  This requires that each linear term
        be grouped with its conjugate.  This already restricts
        you to dice with these many sides: 1 and 11; 3 and 9; 5 and 7.
        
        The 1 and 11 case has the easy solution: a one sided die
        with k on its "face" and an eleven sided die (equally
        probable sides) with 2-k thru 12-k on its faces.  I don't
        know if any of the 3 and 9 groupings or the 5 and 7
        groupings of factors yields all coefficients in [0,1].
        
        Again, there was an assumption here of nonnegative
        integral values for the faces.
        
        Dan

    Re .2:
    
    Ah, by "unfair", I just assumed the faces were non-standard.  I guess
    I'm not devious-minded enough.
    
    
    				-- edp

    A fair die can be made for any even number of values by constructing a
    prism with that number of sides and simply numbering them with the
    desired values.

    A fair die can be made for any odd number by constructing a prism with
    twice that number of sides and numbering them with the desired values,
    each value appearing twice.  (An even number is required to create an
    unambiguous "upper face").

    The prism should be sufficiently long relative to its cap diameter so
    that the probability of it landing on its end is negligable (as we
    normally neglect the probability of a coin landing on its edge --
    though it could -- and reportedly has -- happen).  Alternately one
    could taper the ends to a point making a kind of "spindle" shape.

				    Topher

    If we assume that the dice in question have six-faces numbered 1-6
    (but with unequal probabilities) and furthermore assume that the die
    have identical biases, then the answer is no, a "fair-pair" are not
    possible.

    I got this result fairly laboriously, with some help from maple.  Let
    p1, p2, p3, p4, p5, p6.  Then the following must be true:

	    p1� = 1/6

    since this is the only way to roll a 6.  p1 must, therefore be
    sqrt(6)/6.

    Since there are "two" ways to roll a 3 (1+2 and 2+1):

	    2*p1*p2 = 1/6

    Solving this for p2 using the previous value for p1, and using p1 as
    a shorthand for sqrt(6)/6, we find that:

	    p2 = (1/2)p1

    The next equation is:

	    2*p1*p3 + p2� = 1/6

    which gets us:

	    p3 = (3/8)p1

    And

	    2*p1*p4 + 2*p2*p3 = 1/6

    so

	    p4 = (5/16)p1

    But this runs us into a problem.  I could have gone thorough exactly
    the same chain of reasoning starting with

	    p6� = 1/6

    for 12.  The equations, and their solutions would have been the same
    except that p1 would become p6, p2 would become p5 and p3 would become
    p4 (and vice versa).  Therefore p3 has to equal p4, which it clearly
    does not, so we have a contradiction, so there is no consistent set
    of values for p1, p2, p3, p4, p5 and p6.  Also note that by that same
    symetry, p1+p2+p3 must equal 1/2, but they all contain a factor of p1,
    and hence a "bare" factor of sqrt(6).  They obviously cannot sum to
    1/2.

    Just to be on the safe side, I extended the reasoning to p6 to get,

	    p5 = (35/128)p1
	    p6 = (63/256)p1

    Obviously p6� does not equal p1� equals 1/6, nor does the sum of p1..p6
    equal 1, nor does the probability for, say, an 8 come out right.

    If there is a solution with regularly numbered, six-sided dice, the two
    dice must be biased differently.

				    Topher

    re .4,
    
    By symmetry, you can also assume xi = yi for all i.  That reduces 10
    variables to 5.
    
    Eugene

        re .9,
        
>> By symmetry, you can also assume xi = yi for all i.  That reduces 10
>> variables to 5.
        
        Actually, I used m x's and n y's, so they don't
        necessarily line up like that. :-)  But yes, to restrict
        to six-sided dice set m = n = 6, and to give both dice
        identical markings set xi = yi for all i.  Then to bias
        both dice identically set pi = qi for all i.
        
        Dan

    If we restrict the available symbol set for each die to a subset of the
    integers 1 to 6, then if we can do "it" with two dice of any number of
    sides (biased or not), we can do it with two biased six-sided dice.

    More -- if we can do it with no more than 6 arbitrary numbers per die,
    however many sides there are per die, then we can do it with two biased
    6-sided dice.

    Obvious once you see it.

    So the question is, are there any really different solutions with more
    than six symbols for at least one of the die?  Yes, trivially -- one
    die has 11 equiprobable faces numbered 2 through 12 and the other has
    any number of however biased faces all numbered 0. (You can avoid the
    0 if you want, by numbering the faces on the former 1 through 11 and
    the other 1 everywhere).

				    Topher

If a pair of (some-sided) dice exist that always roll integers, then
all the face values that have non-zero probabilities are integral,
or can be transformed to a functionally equivalent, integral pair of
dice, by increasing all face values on one die, and decreasing all
face values on the other die.

So we can stay in the domain of integral face values.

.5>	Again, there was an assumption here of nonnegative
.5>	integral values for the faces.

    Would anyone volunteer to solve those five equations to see if it is
    indeed possible?
    
    Eugene

        Which five equations?
        
        Dan

    After giving it more thoughts, I retract .9.  It is not necessary for
    xi = yi for all i.  So we are back to 10 equations and 10 unknowns...
    
    Eugene

        What precisely are you trying to prove/disprove?
        
        Dan

    re .16,
    
    Nothing just things like:
    
    x1*y1 = 1/11
    x1*y2 + x2*y1 = 1/11
    x1*y3 + x2*y2 + x3*y1 = 1/11
    ...
    
    and so on and so forth.
    
    Eugene
       

    re .17

    Isn't that 11 equations in 12 unknowns?

    Not only that, the extra constraints (sum of x's = sum of y's = 1) mean we
    have 13 equations in 12 unknowns, so a solution is unlikely.

    I think you can invoke some sort of symmetry argument so Topher's analysis
    in .8 becomes general.

    I had a quick look at the simpler case of unfair two-sided dice (ie pennies
    :-). The equations had no solution, so I tried minimising the sum:

    	(prob(2)-1/3)^2 + (prob(3)-1/3)^2 + (prob(4)-1/3)^2 

    	for x1 + x2 = y1 + y2 = 1

    The solution seemed to be that x1 = x2 = y1 = y2 = 1/2 (ie fair dice!).
    Maybe the equivalent is true for 6-dice also?
    

        .17
        
        .5 and .12 together show that can't happen with two
        six-sided dice with "probabilities" that are real and in
        [0,1].
        
        Dan

RE: .8 (me)

    I can't believe I did that!

    Note that for some stupid reason I assumed that the probability of each
    of the eleven equiprobable combinations was 1/6.  .5 shows that the
    calculation would fail anyway, but I thought that a direct
    demonstration of the failure of the "base case" (ordinary, normally
    numbered, identical biased dice) would add conviction to the rather
    more abstract argument .5.  Just doesn't seem worth going through
    that work with the right numbers though.  As for the 12 variable case
    -- I started it and it blows up into pretty big equations pretty
    quickly.  If it seemed likely to produce a solution, a numeric solution
    (as opposed to an analytic one) might be worthwhile -- but that will
    either fail or produce an unimprovable near miss, which won't tell us
    much at all.

				    Topher

re: .5
	             2    3          12                 2      2      10
	f(z)g(z) = (z  + z  + ... + z  ) / 11 = (1/11) z (1+z+z +...+z  )
        
>	This can be grouped into two polynomials in a number of
>	ways.  However, you want the coefficients to be reals in
>	the interval [0,1].  This requires that each linear term
>	be grouped with its conjugate.  This already restricts
>	you to dice with these many sides: 1 and 11; 3 and 9; 5 and 7.

I tried all the (3,9)-face possibilities, and all the (5,7)-face possibilities.
In each case, one of the die faces had a negative probability.  C'est domage.

Suppose that ten equiprobable consectutive numbers were desired (instead of 11).
Would each linear term still necessarily be grouped with its conjugate?
How about nine numbers, or twelve?

>> This requires that each linear term be grouped with its conjugate.
        
        Ummm, I had thought that the condition that the
        coefficients be all real required this, but this isn't
        "obvious" to me any more. :-)
        
        I did check out the case of f and g each having five of
        the ten linear terms (this corresponds to both dice
        having six sides).  The only combinations that gave a
        real product-of-constant-terms had a non-real
        sum-of-constant-terms.
        
        In fact, taking f to have the linear terms with exponent k
        (of the primitive root of unity e^(2 pi sqrt(-1) / 11))
        then the only cases where both f and g have both the
        product-of-constant-terms and sum-of-constant-terms real
        and of absolute value <= 1 are:
        
        k's for f               k's for g
        ---------               ---------
        none                    all
        3, 8                    1, 2, 4, 5, 6, 7, 9, 10
        2, 4, 7, 9              1, 3, 5, 6, 8, 10
        2, 3, 4, 7, 8, 9        1, 5, 6, 10
        
        and their "complements".  I didn't multiply out the
        entire polynomial in these cases to see if they would
        work (although "none" is the case of a one sided die and
        an eleven sided die with equally likely sides, which is a
        solution).  They deal with respectively: 1, 3, 5, and 7
        sided die [paired with an 11, 9, 7, and 5 sided die,
        respectively].  So there is empirically no solution with
        two six sided dice.
        
        And the exhaustive enumeration shows that "each linear
        term must be grouped with its conjugate" is correct :-)
        (in the sense: in order to have a chance for both all
        real coefficients *and* all coefficients in [0,1]), as
        the cases listed above all group terms with their
        conjugates and they are the only cases not yet ruled out.
        
.21
>>I tried all the (3,9)-face possibilities, and all the (5,7)-face possibilities.
>>In each case, one of the die faces had a negative probability.  C'est domage.
        
        The three cases I didn't rule out, were ruled out in the
        previous reply.  So only the one sided die (with value,
        say, 1) and the eleven sided die (with equally likely
        values 1,...,11) yield a solution.
        
        Dan
        

Let w(x) = cos(x) + i*sin(x).

Then (z-w(a))(z-w(b)) has real coefficients iff

  sin(a) + sin(b) = 0, and
  sin(a+b) = 0.

And (z-w(a))(z-w(b))(z-w(c)) has real coefficients iff

  sin(a) + sin(b) + sin(c) = 0,
  sin(a+b) + sin(b+c) + sin(c+a) = 0, and
  sin(a+b+c) = 0.

And (z-w(a))(z-w(b))(z-w(c))(z-w(d)) has real coefficients iff

  sin(a) + sin(b) + sin(c) + sin(d) = 0,
  sin(a+b) + sin(b+c) + sin(c+d) + sin(d+a) + sin(a+c) + sin(b+d) = 0,
  sin(a+b+c) + sin(b+c+d) + sin(c+d+a) + sin(d+a+b) = 0, and
  sin(a+b+c+d) = 0.

Of course.  Now does this help at all?

    How about a pair of dice, say d(1) and d(2) with the faces numbered:
    
    	d(1):	1,3,5,7,9,11
    and d(2):	0,0,0,1,1,1
    
    Since all outcomes from 1 through 12 are equiprobable, just ignore the
    case where d(1)=1 and d(2)=0, i.e. roll again. You then have a number
    generator where 2 through 12 are equiprobable outcomes!
    
    James.