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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

559.0. "What's the probability that the last ball is red?" by CLT::GILBERT (eager like a child) Fri Aug 08 1986 14:21

This problem is making the rounds on the USENET.  It's difficult,
but most of the inroads into solving it have so far been either
trivial or incorrect.

A box contains i red balls, j white balls, and k blue balls.  Two
random balls are removed, and one ball replaced.  If the removed
balls are the same color, a ball of the same color is replaced;
if the balls are of different colors, a ball of the third color
is replaced.

Eventually, just one ball will remain.  What is the probability
that it is red?

Let P(i,j,k) be the probability that the last ball is red.  We have
the following:

	P(1,0,0) = 1,  P(0,1,0) = 0,  P(0,0,1) = 0,
	P(i,j,k) = P(i,k,j),  and  P(i,i,i) = 1/3.
T.RTitleUserPersonal
Name		DateLines
559.1tends toward i=j=k...MODEL::YARBROUGHFri Aug 08 1986 16:035
    Looks to me as though the process tends to equalize the number of
    balls of each color, since e.g. i<j<k, prob(ii)<prob(jj)<prob(kk),
    and prob(ij)<prob(ik)<prob(jk). So the more common colors tend to
    be depleted faster. Assuming i, j, and k are large, prob(last=red)
    is 1/3.
559.2CLT::GILBERTeager like a childFri Aug 08 1986 23:1712
P(0,2,0) = 0		P(i,1,1) = 1 - P(i,2,0)
P(1,2,0) = 2/3
P(2,2,0) = 1/3
P(3,2,0) = 1/2
P(4,2,0) = 7/15
P(i,2,0) = (7i+5)/(14i+14), i >= 5

P(0,3,0) = 0
P(1,3,0) = 1/2
P(2,3,0) = 1/4
P(3,3,0) = 3/10
P(i,3,0) = (7i+20)/(28i+56), i >= 4
559.3Some results from Lambert Meertens in AmsterdamCLT::GILBERTeager like a childThu Aug 14 1986 23:21107
The notation p(ijk) is used for the probability that the last marble is red
if we started with i red, j blue, and k green marbles.  The following is a
very partial solution, concerning the cases where one of i, j and k is
large, compared to the other two.
 
I examined the behaviour, as a function of M, of p(Mab) (henceforth called
type I) and p(abM) (type II), for fixed values of a and b, not both zero.
The value of p(bMa) is of course the same as that of p(baM).  For most of
the experimental findings I present here I have no proofs.
 
Let d stand, throughout, for a+b-1 (so d >= 0).  It turns out that both
p(Mab) and p(abM) are rational functions in M for sufficiently large values
of M.  (A rational function is one that can be expressed as the quotient of
two polynomials with rational coefficients.)  Which rational function
depends on a and b and on the type.  However, the denominator depends only
on d (except that in two cases it happens to have a common factor with the
numerator); if we put the coefficient of the leading term equal to 1, it is
 
    (M+1)*(M+2)*...*(M+d).
 
For the numerators, I could not discern a pattern, except that each has the
same degree as the denominator, and the coefficient of the leading term
depends only on d.
 
The value of M from which onwards p(Mab) or p(abM) coincides with a
rational function depends only on d and the type (I or II).  Call it
M[I](d) or M[II](d).
 
It appears that p(baM) = p(abM) for M >= M[II](d).  Since
p(Mab)+p(abM)+p(bMa) = 1, it follows that for M >= M[II](d) we have
 
    p(Mab) = p(Mba) = 1-2*p(abM) = a rational function,
 
so M[I](d) <= M[II](d).
 
I found
 
              ( 2d   if d is even
    M[I](d) = (                   ,  M[II](d) = 2d+3.
              ( 2d+3 if d is odd
 
Since the limit of p(Mab) for M -> inf depends only on the ratio between
the leading coefficients in the numerator and denominator, which depends
only on d, that limit also depends only on d; it appears that we have
 
                          -d
                  1 - (-2)
    lim p(Mab) =  --------- .
    M->inf           3
 
For example, for p(M04) it is (1+1/8)/3 = 3/8.  From lim p(abM) = (1- lim
p(Mab))/2 we find the curious fact that this is equal to (2+(-2)^-d)/6 =
(1-(-2)^-(d+1))/3, so we get the same sequence of values as for lim p(Mab),
but shifted by one position.
 
Here are the numerators of p(Mab) for d <= 4.  I use a shorthand, in which,
e.g., (7, 27, 20) stands for 7*X^2+27*X+20.
 
d = 0
-----
 
M01: 0
 
d = 1
-----
 
M02: 1/14 * (7, 5)
M11: 1/14 * (7, 9)
 
d = 2*
-----
 
M03: 1/28 * (7, 27, 20)
M12: 1/28 * (7, 19, 12)
 
d = 3
-----
 
M04: 3/5096 * ( 637,  3458,  6071,  3370)    N.B. 3*637 = 1911
M13: 3/5096 * ( 637,  3822,  6851,  3546)
M22: 1/5096 * (1911, 11830, 22581, 13022)
 
d = 4
-----
 
M05: 5/10192 * ( 637,  6734,  24011,  34090, 16176)
M14: 1/10192 * (3185, 32214, 115063, 168546, 82512)
M23: 1/10192 * (3185, 31486, 108823, 153482, 72960)
 
*The numerators of M03 and M12 share a factor M+1 with the corresponding
denominators.
 
Another amazing result is that if you use yellow marbles instead of green
ones, the results appear to be the same!  This is useful for people who
have difficulties in distinguishing red and green.
 
-- 
 
Lambert Meertens, CWI, Amsterdam; [email protected]

Newsgroups: net.math,net.puzzle
Path: decwrl!amdcad!lll-crg!seismo!mcvax!lambert
Subject: Re: Probability problem that I'm too sleepy to think about.
Posted: 13 Aug 86 18:42:24 GMT
Organization: CWI, Amsterdam
Xref: decwrl net.math:3258 net.puzzle:1999
Apparently-To: rnews@mcvax