Lewis Carroll’s fuzzy counters

This site is mostly in Finnish, but I would like to get wider feedback on one posting. I suspect that I am wrong, but have not been able to prove it so far. The Finnish version is here.

A local book publisher, Terra Cognita, publishes weekly classic puzzles.  Last week’s problem has stuck in my mind. It is the fifth puzzle in  Lewis Carroll’s Pillow Problems. It is framed in this way“A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?”

I will analyze the solution  given by Terra Cognita. We need to count the possible scenarios. Call the counter in the bag B and the new white counter A. Counter B may be either black (marked BB) or white (marked BW). Counter A is always white, so it is marked AW.

There are then four scenarios.

S1. AW is drawn,  BW is left in  the bag

S2.  AW is drawn, BB is left in the bag.

S3. BW is drawn,  AW is left in  the bag

S4.  BB is drawn, AW is left in the bag.

Since we know that the drawn counter is white, S4 cannot be true. There are then three scenarios, two of which (S1 and S3) leave a white counter in the bag. There is thus a 2/3 probability that the counter left in the bag is white. This is the classical solution.

ButI believe the classical answer is elegant, but wrong. It is exactly true only if all scenarios are equally probable. They may not be. We need to count conditional probabilities.

P(S1) = P(A drawn)* P(B is white) = 0.5* P(BW)

P(S2)= P(A drawn)* P(B is black) = 0.5* P(BB) = 0.5* (1-P(BW))

P(S3)= P(B drawn) * P(B is white) = 0.5* P(BW)

P(S4) = P(B drawn) * P(B is black), mutta P(S4) is known to be zero.

The unknown counter is thus white with probability

P(XW)= (P(S1)+ P(S3)) / (P(S1)+P(S2)+P(S3)) = P(BW) / (0.5+0.5*P(BW)) = 2*P(BW) / (1+P(BW)).   [Equation 1]

If B has equal probability of being black and white, P(BW)=0.5 and P(XW) = 2/3 as above.

What is the problem?

According to the problem setting, we know nothing at all about the probability of B being white or black. We can consider the problem so that there is a barrel with say 1000 counters. Some are white, some black.  The experimenter chooses a counter B randomly and places it in the bag. He then places a white counter A in the bag and draws one out. If the counter happens to be black (Scenario 4) a new choice of B is made and the process repeated until the first drawn counter is white.

P(BW) then depends on the number of white counters in the barrel. If there are 500 white counters, P(BW) = 500/1000 = 1/2 and P(XW)=2/3 as above.

If there are 900 white counters, P(BW)=900./1000 ja P(XW)= 95%. In fact, all the counters may be white, in which case P(XW)=100%.

At the other extreme, if there is only one white counter,  P(BW)=1/1000 ja P(XW)=0.2%. In this case, the experiment may need to be repeated multiple times. If the experimenter happens to draw B first, it is almost  certainly black. On the other hand, he has a 50% chance of drawing A first, so that the experiment will not need to be repeated very many times.

In fact the experiment works even if there are no white counters in the barrel at all. The experimenter will eventually draw A, and the experiment is over. This “eventually” can actually be calculated quite accurately: the probability that he draws B n times in a row is 1-(0.5)^n. Thus, the experiment has a 99.9% probability of terminating within 10 draws. Even so, the counter in the bag is always black, with 100% probability, relentlessly.

Do we actually know anything at all?

Not really. Based on one experiment, we can say nothing about the contents of the barrel, and the contents of the barrel fully determine the probability that the counter left in the bag is white. We can, of course, calculate the value of Function 1 for different proportions of white counters.

The distribution has some interesting characteristics. Its median is 2/3, which is the classical solution. However, its mean value is only 0.61. Neither number really tells us very much. We are left with darkness, fuzziness, and uncertainty.

We can eliminate some of the darkness by repeating the experiment many times, and calculating how often the counter drawn from the bag is white. This allows P(BW) to be estimated. (The details will be left as an exercise to the reader).

The fuzziness and uncertainty remain, however. Even in the best case, the probability that we can derive is only a statistical estimate. The puzzle has no solution in a closed form.

The truly interesting question is: is my analysis correct? So far, I have found no errors in it. At best, I can state that is is wrong with some probability, but I have no idea what that probability is.

Other mathematicish postings, a few in English: WeirdMath