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A bag contains 2 counters, as to which nothing is known except that each is either black or white. Ascertain their colours without taking them out of the bag.

Carroll's solution: One is black, and the other is white.

Lewis Carroll's explanation:

We know that, if a bag contained $3$ counters, two being black and one white, the chance of drawing a black one would be $\frac{2}{3}$; and that any other state of things would not give this chance.
Now the chances, that the given bag contains $(\alpha)\;BB$, $(\beta)\;BW$, $(\gamma)\;WW$, are respectively $\frac{1}{4}$, $\frac{1}{2}$, $\frac{1}{4}$.
Add a black counter.
Then, the chances that it contains $(\alpha)\;BBB$, $(\beta)\;BBW$, $(\gamma)\;BWW$, are, as before, $\frac{1}{4}$, $\frac{1}{2}$, $\frac{1}{4}$.
Hence the chances of now drawing a black one,
$$= \frac{1}{4} \cdot 1 + \frac{1}{2} \cdot \frac{2}{3} + \frac{1}{4} \cdot \frac{1}{3} = \frac{2}{3}.$$ Hence the bag now contains $BBW$ (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.
Q.E.F.

Can you explain this explanation?

I don't completely understand the explanation to begin with. It seems like there are elements of inverse reasoning, everything he says is correct but he is basically assuming what he intends to prove. He is assuming one white, one black, then adding one black yields the $\frac{2}{3}$. From there he goes back to state the premise as proof.

Can anyone thoroughly analyze and determine if this solution contains any fallacies/slight of hand that may trick the reader?

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    $\begingroup$ Hehe...this is classical good'ol Lewis Carrol's logical stuff, like some of the stuff found in his books. You could change the setup and take instead two persons in a room, of which we only know that each is either a man or a woman... $\endgroup$ – DonAntonio Mar 17 '14 at 3:38
  • $\begingroup$ What does QEF mean? $\endgroup$ – Ben Grossmann Mar 17 '14 at 3:40
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    $\begingroup$ @Omnomnomnom Quod erat faciendum, "which had to be done". en.wikipedia.org/wiki/Q.E.D.#QEF $\endgroup$ – user856 Mar 17 '14 at 3:45
  • $\begingroup$ @Rahul ah, thank you! $\endgroup$ – Ben Grossmann Mar 17 '14 at 3:47
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    $\begingroup$ After reading this question, I have to fight the urge to put two black backgammon stones in a bag. $\endgroup$ – rumtscho Mar 18 '14 at 17:04
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There is a reason this is the last of Lewis Carroll's Pillow Problems. It is a mathematical joke from the author of Alice in Wonderland.

The error (and Lewis Carroll knew it) is the phrase

We know ... that any other state of things would not give this chance

since he then immediately gives an example of another case which gives the same chance. Indeed any position where the probability of three blacks is equal to the probability of two whites and a black would also give the same combined chance.

There is no need to add the third black counter: it simply confuses the reader, in order to distract from the logical error. Lewis Carroll could equally have written something like:

We know that, if a bag contained $2$ counters, one being black and one white, the chance of drawing a black one would be $\frac12$; and that any other state of things would not give this chance.

Now the chances, that the given bag contains (α) BB, (β) BW, (γ) WW, are respectively $\frac14$, $\frac12$, $\frac14$.

Hence the chance, of now drawing the black one, $=\frac14 \cdot 1 +\frac12 \cdot \frac12 + \frac14 \cdot 0 = \frac12.$

Hence the bag contains BW (since any other state of things would not give this chance).

If he had written that, it would be more immediately obvious that this was faulty logic with an assertion followed by a counterexample followed by faulty use of the assertion.

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  • $\begingroup$ "any position where the probability of three blacks is equal to the probability of two whites and a black would also give the same combined chance." I'm not quite sure what you mean here, but as stated this isn't correct; what if the probability of, say, three whites outweighs both of the stated probabilities? I think perhaps you meant any position "involving at least one black," or something similar. $\endgroup$ – Kyle Strand Mar 17 '14 at 16:58
  • $\begingroup$ @Kyle Strand: Three whites has a zero probability as the original question started with two counters and then added a black one. $\endgroup$ – Henry Mar 17 '14 at 20:27
  • $\begingroup$ Hence "involving at least one black." When you say "any position," it's unclear what you mean; clearly you're trying to talk about something more general than the problem as originally specified, but it's not obvious that you're still including the "add one black piece" constraint. $\endgroup$ – Kyle Strand Mar 17 '14 at 22:41
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Well, of course the reasoning is flawed, since it's certainly possible to have a bag with two counters of the same color in it!

The facts that are correct are:

The probability of drawing a black counter from a fixed bag with 3 counters is 2/3 iff the bag contains two black counters.

By adding a black counter to a randomly generated 2-counter bag, the probability of drawing a black from the resulting bag is 2/3.

The conclusion that this means the resulting bag in the latter case therefore contains 2 black counters and 1 white counter is what is flawed, because the bag itself is not fixed; the probability is being calculated over a variable number of possibilities for the bag.

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    $\begingroup$ +1 for certainly possible to have a bag with two counters of the same color. $\endgroup$ – Guy Mar 17 '14 at 5:54
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    $\begingroup$ Lewis Carroll is famous for being exactly the thing that plagues the internet: a troll. (The difference is Carroll had a large amount of class that typical internet trolls lack...) $\endgroup$ – corsiKa Mar 17 '14 at 18:34
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Essentially he argues (incorrectly) that since

$P(Y|X=x) = 2/3 \Leftrightarrow x=x_1$

then

$P(Y) = 2/3 \Rightarrow X = x_1$

Here

$Y$ = "drawing a black counter"

$X$ = "the three counters in the bag"

$x$ = "some particular state of counters"

$x_1$ = "two black counters, one white counter"

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    $\begingroup$ +1 for pointing out that he's using a statement about a conditional probability to draw a conclusion about an unconditional probability. $\endgroup$ – Kyle Strand Mar 17 '14 at 17:26
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The probability of drawing a black counter from a bag containing one white, and two black counters $= 2/3$.

The probability of drawing a black counter from a bag containing one counter that is known to be black, and two counters that have a 50/50 chance of being either black or white $= \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} = 2/3$.

Does this mean that these bags are necessarily the same? Of course not. You just happen to have the same probability of drawing a black counter from either one of them.

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  • $\begingroup$ Addressing the calculation of the respective probabilities is important and helpful. $\endgroup$ – grayQuant Mar 17 '14 at 14:24
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Lewis Carroll could have shortend his argument to the following.

Since we know nothing about the counters except that they are either black or white, the probability, if we would take out one counter, that it would be black, is $\frac12$. But knowing that probability, we can conclude (without actually taking out anything) that half the counters in the bag are black. Since there are two counters, one must be black and the other white.

But if he had done so, the flaw in the argument would have have been all too evident. (If you still cannot see it, consider that everything up to the last sentence also works for a bag with an odd number of counters, for instance just one.) A conjuror's trick consists for a large part of supplying distracting but irrelevant details.

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The problem lies in the difference between a probability to draw a certain counter given a set-up and get a certain counter given a certain set-up probability. You can see the fallacy more clearly if you take a counter out of the equation (and the bag); the logic (or illogic) is the same for any number of counters.

If we have two counters, they can be of the same colour (with the probability for drawing that colour being 100%, and for the other colour 0%), or they can be of different colours, in which case the drawing probability is 50% for each colour. Thus, the probability for drawing a token of a certain colour out of two counters can only be 0%, 50%, or 100%, and no state of things would give any other chance.

Now let us say we have one counter to start with. If we assume a random process (which is, by the way, not stated in the description), it has a 50/50 chance of being black or white. If we add a black counter, we have a 50/50 chance of BB or BW. In the first case, the probability of black is 100%, in the second case, it is 50%. Thus, the overall probability of drawing a black counter is 75% - a chance which, as we just stated, would not be given by any state of things.

So - the problem is that what we have here is, so to speak, not a state of things, but rather a state of probabilities, which is a juxtaposition of possible situations and thus not bound by the same limitations.

However, the number Lewis Carroll gets at the end does have a meaning, just not quite a strong one as he claims. It is the single most probable state of things. When, as in my example, there is no possible state corresponding to the probability, it means that there are multiple most probable situations.

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You cannot understand something that which is not to be understood. It is impossible to ascertain the colors, as stated in the problem by "nothing is known". Note that Carroll himself says the probability of having black and white in the bag is 1/2. Then he says it definitely is so. I think the slip is in the "every other state of affairs" line. He uses this line to talk about a state of affairs meaning a specific collection of 3 counters in a bag, but also when referring to a collection of such bags. How he gets 2/3 also is from arbitrarily adding a black counter.

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    $\begingroup$ Just a comment: I don't believe that there is a "slip" in this argument. I'm sure that Carroll knew perfectly well the argument was wrong, and was only publishing it as a joke. $\endgroup$ – David Mar 17 '14 at 5:44
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    $\begingroup$ Slip in that he's giving us the slip so to speak. $\endgroup$ – Jacob Wakem Mar 17 '14 at 5:45
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Interestingly, if the problem was rephrased, it could be answered.

A bag contains 2 counters, as to which nothing is known except that if we draw a counter at random, it could be either black or white (i.e. there is a non-zero probability that it is black, and a non-zero probability that it is white). Ascertain their colours without taking them out of the bag.

Carroll's solution: One is black, and the other is white.

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Lewis Caroll tells the content of a bag can be characterized by the probabilities to pick a black counter. Knowing that probability is the same as knowing the proportion of the number of black and white conters. This is reasonable, assuming the content is a fixed configuration.

But when he analyzes the case of 2 counters plus a black one, the content of the bag is a combination of different configurations with a probability attached. Here you are computing an average of different cases, which happens to equal one of the cases. That doesn't mean that the one case is certain.

You can apply the same reasoning to this paradox:
1. Consider a bag with 2 black or white counters. It countains BB, BW or WW. If the probability of drawing a black counter is exactly 1/2, that means that the bag contains exactly one black and one white counter.
2. Consider a bag of which you know it contains 2 identical counters. These can be either BB or WW. If you pull a counter, the probability of pulling a black counter is 1/2. From that you can conclude that the bag contains a black and a white counter.

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