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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 ?
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