You have four balls in a bag.

probability/four.balls New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

You have four balls in a bag.  One is blue, one is black and two are
red.  Two balls are drawn out.  You are told that at least one of these
two balls is red.  What is the chance that both are red?

-- Paul Sloane <PSloane@MathSoft.com>

Show solution

*sigh*  The question is ill-posed, even if one assumes that the drawing
procedure is completely random and that the teller of the information
is always truthful.

If the person who told you this information is impartial (will always
say "at least one ball is red" and nothing else whenever this is in
fact the case), then the probability is 1 in 7.

If the person who told you says "at least one ball is red" when there
is exactly one red ball drawn out, and says something different when
two red balls are drawn out (e.g. "both balls are red"), then the
probability is 0.

If the person instead only says "at least one ball is red" when _both_
balls are red, and says something else when only one ball is red (e.g.
"exactly one ball is red"), then the probability is 1.

Of course any other value of probability is possible if this person
chooses to give random (but always truthful) pieces of information.
For example, if this person only examines the first ball that comes
out, and says "at least one ball is red" if this ball is red, and
offers no conclusion otherwise, then the probability is now 1 in 3.

Basically, information consists of more than just a statement which
happens to be true under the present circumstances.  To properly use
information to determine how it affects probabilities, you have to
know what statements would have been given under all other possible
circumstances that are under consideration.  Quite often you do not
know what information would have been given under those circumstances
(especially in those hateful problems where a fact is simply stated
(e.g. "You draw two balls.  At least one of the balls drawn is red.")
with no clue as to what facts would have been provided under which
circumstances), and so the problem becomes ill-posed.

-- Terry Tao <tao@grapes.princeton.edu> (rec.puzzles and sci.math,
   April 1995)

[ Editor: with an impartial teller the answer is 1 in 5, not 1 in 7:
of the 6 equally likely pairs, 5 contain a red ball and 1 has both
red. ]