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sttscitrans@xxxxxxxxx wrote:
> Proginoskes wrote:
> > sttscitrans@xxxxxxxxx wrote:
> > > matt271829-news@xxxxxxxxxxx wrote:
> > > > Jim Dars wrote:
> > > > > <matt271829-news@xxxxxxxxxxx> wrote in message
> > > > > news:1146141536.984996.226130@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> > > > > >
> > > > > > Proginoskes wrote:
> > > > > > > Jim Dars wrote:
> > > > > > > > <matt271829-news@xxxxxxxxxxx> wrote in message
> > > > > > > > news:1146006008.923415.229970@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> > > > > > > > >
> > > > > > > > > matt271829-news@xxxxxxxxxxx wrote:
> > > > > > > > > > Is this meant to be the same problem as was discussed at
> > > > > > > > > >
> > > > > > > >
> > > > > http://groups.google.com/group/alt.math.recreational/browse_frm/thread/6a5d73da8194f46d,
> > > > > > > > > > or is the behaviour different on the first turn?
> > > > > > > > >
> > > > > > > > > Sorry, I meant different on the first few turns, not
> > > > > > > > > specifically on
> > > > > > > > > the first turn - where all are R anyway.
> > >
> > >
> > > > > > > > It's the same problem, if you recall not one one math.rec could
> > > > > > > > solve
> > > > Still no closed form solution though!
> > >
> > >
> > > Looking back at the thread, I see that I did in fact give
> > > a closed form solution in terms of a summation.
> >
> > In your post on March 6, 2005, you are summing over all matrices A with
> > a fixed set of properties:
> >
> > ----------------------------------------------------------------------
> > The following is an obvious generalisation and closed-form solution.
> > Given N balls and a sequence of colours c0,c1,c2,.....,cm
> >
> > Consider all nx(n-1) matrices A (postive integral entries) that meet
> > the following conditions
> > a_11 = N-1, other entries =0
> > a_21 +a_22 = N-1, a_22 >=1, other entries =0
> > a_N-1,1 + a_N-1,2 + ... a_N-1,N-1 = N-1,a_N-1,N-1 >=1
> > a_N,1 + a_N,2 + .... + a_N,N-1 = K
> > where k is the index of the colour of the last remaining ball.
> >
> > Let BN be the last remaining ball of colour ck
> > Let B1,B2,B3,..., B_N-1 be any permuation of the balls that disappear.
> >
> > Let R_j be the sum of the entries in the jth row
> >
> > In the sequence of ball selections, the first ball B1 will disappear at
> > R_1+1 in the sequence, the second ball at R_1+R_2 +2 and so on. The
> > probability that this sequence is chosen is then
> > 1/N^(R_1+1)*(N-1)^(R_2+1)* 2^(R_N-1+1) = p(R_1,R_2,...,R_N-1)
> >
> > Let P_j = (R_j!)/aj1!aj2!.....ajN!
> >
> > T = P_1*P_2*........P_N-1
> >
> > The probability that the last remaining ball has colour k then equals
> >
> > SUM (over A) N*(N-1)!*T*p(R_1,R_2,...,R_N-1)
> >
> > Say you had 4 balls and 3 colours. A typical matrix for k =2 would be
> >
> > 3 0 0 0
> > 0 3 0 0
> > 0 0 3 2
> >
> > The matrices A are obviously sums of column vectors
> > 1st column
> > (3,0,0)
> > 2nd column
> > (0,3,0)
> > (1,2,0)
> > (2,1,0)
> > 3rd column
> > (0,0,3)
> > (0,1,2)
> > (0,2,1)
> > (1,1,1)
> > 4th column
> > (0,0,2)
> > (0,1,1)
> > (0,2,0)
> > (1,1,0)
> > (2,0,0)
> > ----------------------------------------------------------------------
> >
> > And in your February 26, 2005, post, you give something closer to
> > what's viewed as a "closed form solution":
> >
> > ----------------------------------------------------------------------
> > Say you had 3 marbles, A,B,C and you wanted to calculate the
> > probability that C was the last marble remaining and white. This would
> > take 7 selections. C can't occur as XXXXXXC or it would be red.
> >
> > A typical sequence is then XXAXXXB and the associated probability is
> > 3^(-3)2^(-4) = (3,4)
> > You know that 2 A's must appear in "XX" and 2B's and a C in "XXX"
> >
> > The number of sequences of this form is therefore
> > (2!/2!0!0!)*(3!/0!2!1!) = 3
> >
> > The total probability for sequences of the form XXAXXXB and XXBXXXA is
> > then 2*3 = 6
> >
> > The number of sequences with probability (4,3) is
> > 2*{(3!/2!1!0!)*(2!/0!1!1!) +(3!/2!0!1!)*(2!/0!2!0!} =18
> >
> > Simarly for (5,2) and (6,1)
> >
> > The probability that C is the last remaing marble and white is then:
> > 6(3,4) +18(4,3) +36(5,2) + 60(6,1) = (2,3) +(2,2) +(3,0) +10(5,0)
> > = (5,3)(27 + 2*27 +9*8 + 10*8)
> > = 233(5,3)
> >
> > The probability that A or B or C is the last marble and white is
> > therefore 3*233(5,3) =233(4,3) = 233/648
> > ----------------------------------------------------------------------
> >
> > > How are you defining a "closed form solution" ?
> >
> > Probably the following way:
> >
> > (1) If R is a real number or a variable, R is a closed form solution.
> > (2) If R and S are closed form solutions, then R+S, R-S, R*S, R/S, and
> > R^S are all closed form solutions.
> > (3) If R is a closed form solution, then ln(R), sin(R), cos(R) are
> > closed form solutions.
> > (4) Any expression which can be obtained by a finite number of
> > applications of (1) through (3) is a closed form solution.
> >
> > This would usually exclude sigma (and product) notation,and "..."'s.
>
> I could understand the exclusion of infinite sums and products,
> but as there is nothing more complicated than finite sums and products
> of rationals involved, it seems paradoxical to include
> ln , sin, cos (infinite series).
>
> 1 + 2 + 3 + 4 = 10
>
> sum (m = 1 to 4) m = 10
>
> What is the essential difference ?
>
> Sum (m=1 to n) m = n(n+1)/2
>
> Are you saying this is not a closed form ?
OK, let's allow finite summation (sigma) notation. Is it then possible
to express your solution to the problem as a single formula that will
work for any N? (Or, actually, I guess you'd need three formulas, one
for each of R, W, B).
Sorry, I'm possibly being a bit lazy here but I've now forgotten most
of what was originally discussed about this problem, and it would take
a while now to go back and try to figure out exactly how your method
works.
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