Ken's POTW

Balls and Urns
  1. 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?
  2. 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?
  3. 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.
  4. 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 3/22&29/94

Solutions were received from Phillippe Fondanaiche, Bert Sevenhant, Nick Baxter, Highlander, 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.

  1. 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) + Pr(B12/C4)*Pr(C4)
    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.

  2. 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 answers. 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.

  3. Place one white in one urn, and all the others in the other urn. This gives a 74/99 probability of picking white.

  4. 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.]

Mail to Ken