Lewis Carroll's Pillow Problem

This problem is cited by M. Gardner in his Mathematical Circus and also Gardner's Workout.

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?

Below is a simulation^(*) device you may use to gain insight into the problem. To make a selection click on one of the radio buttons. Once you did the program will show your guess. It's either White or Black. The color of the second checker will appear in the second edit control. If you hit White and the other color is also White, you win. If the other color is Black, you lose. Hitting Black just does not count.

The simulation works the following way. First, I randomly select one of two positions. Next, a White/Black selection is made for this position and the remaining position is assigned White.

Two different solutions are given after the simulator device. Which is the right one?

Important note:

You run a simulation. During a simulation you are allowed to make as many selections as indicated in the "To Go" control before your first selection. Remember also that after each selection the device needs approximately 1 second to clear up controls. Please wait till it does. To start a new simulation please press the "Reset" button.

Lewis Carroll offers two solutions - one short and wrong (Proof 1 below), another right but long. (He of course new which is which.) Martin Gardner [Circus, p. 189] calls the second of Carroll's proofs "long-winged" and affers instead a shorter one (Proof 2 below) by one of his readers, Howard Ellis from Chicago.

Solution #1

As the state of the bag, after the operation, is necessarily identical with its state before it, the chance is just what it was, viz. 1/2.

Solution #2

Let B and W1 stand for the black or white counter that may be in the bag at the start and W2 for the added white counter. After removing white counter there are three equally likely states:

Inside bag	Outside bag
W1	W2
W2	W1
B	W2

In two of these states a white counter remains in the bag, and so the chance of drawing a white counter the second time is 2/3.

References

M. Gardner, Gardner's Workout, A K Peters, 2001, pp. 129-132
M. Gardner, Mathematical Circus, Vintage, 1981

Chris Conradi mailed me the following remark:

Carroll's Pillow Problem -- see the Probability problems -- says that we know a bag contains one counter, and it is either black or white. The solution presumes that the bag is equally likely to contain a black counter or a white counter, although neither Carroll nor Bogomolny makes that clear. It would be helpful to make that point in the statement of the problem. Otherwise, the answer cannot be determined from the statement of the problem.

This is just a particular instance of a more general problem. Suppose that the first counter is drawn at random from another bag containing d counters, n of which are white and d - n of which are black. Without revealing its color, this first counter is placed in a second, empty bag. Then a white counter is added to this bag. Now you draw a counter from this second bag and it turns out to be white. What is the probability that the remaining counter is white? The general answer is 2n/(n + d). I.e., if n = 1 and d = 2, the answer is 2/3, as given. But we can also determine the probability if n = 1 and d = 4 (2/5); if n = 3 and d = 4 (6/7); or if n = 5 and d = 18 (10/23).

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