Balls and Urns
An urn contains two balls, each black or white with even odds.
You pick a ball from the urn. It is black. You replace the ball.
You pick another ball from the urn. What are the odds that
it is also black?
Do the odds change if prior to picking the first ball in problem 1,
you are told that one of the balls is black, then again your first
pick is black?
There are two empty urns in a room. You have 50 white balls
and 50 black balls. After you place the balls in the urns,
a random ball will be picked from a random urn.
Distribute the balls (all of them) into the urns to maximize
the chance of picking a white ball.
You have one white ball and one black ball to place in an urn. What
picking method would you propose to create a probability of 1/3?
That is, if I am picking between A and B based on the outcome of your
method, I will pick A an average of 1 time out of 3. Multiple picks
are allowed (and probably necessary.) This is similar to asking how
to simulate a probability of 1/3 with only a fair coin.
Source: Internet newsgroup rec.puzzles, used as Ken's Puzzle of the Day
Solutions were received from
and David Shield.
The solutions were varied and much email was exchanged
about the correct analysis.
In the end, a little computer simulation pointed toward the answers below.
- 3/4. Phillippe's solution explains Bayes' Theorem and shows an
excellent use of it:
There are 4 possible
configurations to fill the urn,each of them with the same probability 1/4:
C1 = BB, C2 = BW, C3 = WB, C4 = WW
Let us call : Bi (or Wi) = a black (or white) ball is taken from the urn at
the i-th picking.
Obviously Pr(B1) = Pr(W1) = 1/2. Let us call B12 = B1 and B2.
According to Bayes' theorem, we have:
Pr(B12) = Pr(B12/C1)*Pr(C1) + Pr(B12/C2)*Pr(C2) + Pr(B12/C3)*Pr(C3) +
So Pr(B12) = 1*1/4 + 1/2*1/2*1/4 + 1/2*1/2*1/4 + 0*0*1/4 = 3/8
The requested probability is Pr(B12/B1) = Pr(B12)/Pr(B1) = (3/8) / (1/2) = 3/4.
- 3/4. This answer surprised me, but simulation verified it.
I extend Phillipe's solution from above. Learning that one
black ball exists removes the C4 configuration from the possibilities:
C1 = BB, C2 = BW, C3 = WB
Pr(B1) = 1*1/3 + 1/2*1/3 + 1/2*1/3 = 2/3
Pr(B12) = 1*1/3 + 1/2*1/2*1/3 + 1/2*1/2*1/3 = 1/2
Pr(B12/B1) = Pr(B12)/Pr(B1) = (1/2) / (2/3) = 3/4.
I think the above solution is the one that best matches the problem.
Several people read the problem differently and came up with different
One solver suggested that "hearing that one ball is black" is the
same as "picking a black". The resulting answer shows this is
not the case since the probability is different:
Again we'd start with the initial four configurations, and
Pr(B123) = 1*1/4 + 1/2*1/2*1/2*1/4 + 1/2*1/2*1/2*1/4 = 5/16.
The requested probability is Pr(B123/B12) = (5/16)/(3/8) = 5/6.
- Place one white in one urn, and all the others in the other
urn. This gives a 74/99 probability of picking white.
- Place a white and black in the urn. Draw one ball, put that
ball back in the urn, and draw again. If both draws were the white
ball, start over. Otherwise pick A if both draws were the black ball.
I recently read of a related puzzle from Sam Loyd in
The Greatest Puzzles of All Time,
(Matthew J. Costello, Prentice Hall Press, 1988, 3-10).
Here is that puzzle and its solution:
A bag contains two counters, as to which nothing is known except
that each is either black or white. Ascertain their colors
without taking them out of the bag.
Solution: We know that if a bag contained three counters, two
being black and one white, the chance of drawing a black one would be 2/3;
and that any other state of things would not give this chance.
Now the chances that the given bag contains BB, BW, or WW are,
respectively, 1/4, 1/2, 1/4.
Add a black counter. Then the chances that it contains BBB, BWB, or WWB
are, as before, 1/4, 1/2, 1/4. Hence the chance of now drawing a black one
= (1/4)*1 + (1/2)(2/3) + (1/4)(1/3) = 2/3.
Hence the bag now contains BWB (since any other state of things
would not give this chance.) Hence before the black counter was
added, it contained BW, i.e., one black counter and one white counter.
[Another solver suggested to just look in the bag!]
[Another (from the "A Set of Stamps" puzzle)
has suggested that if logic exists to show there are
two whites in the bag, that same logic can be used to show there
are two blacks in the bag, so if it is provable
there must be one white and one black.]
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