[Guest]

This is one of Lewis Carroll's problems. This is not a trick question in any way, nor is it a lateral thinking type of riddle. You will arrive at the answer with mathematical and logical reasoning only.

A bag contains two marbles, as to which nothing is known except that each is either black or white. Ascertain their colors without taking them out of the bag or looking into the bag.

Again, this is not a lateral thinking riddle. So no taking pictures of inside the bag, no cutting the bag open, nothing like that.

If an answer hasn't been found, I'll post an answer in a couple days, because even the solution deserves discussion!

[Guest]

I would guess one black and one white. That gives me a 50% chance to be correct and I will always have at least one correct. And maybe, by telling me they are either white or black means that both colors are in the bag?

[Guest]

There is a 1/3 probability that the two marbles are either 1) both white, 2) both black, or 3) white and black pair.

Another way to look at it involves considering the probability of each marble: 1/2 . And then one takes the two together to yield 1/4 for the four scenarios, of which one is equivalent to another (namely, when the marbles are a black and white pair). From this it may be inferred that there is a 1/2 chance that the marbles are a black and white pair, whilst the other two possibilities are only 1/4 each.

So on second thought, my final answer is in the above paragraph: they are most likely a black and white pair.

[Guest]

Omnihistor, you've started on the right track. The answer, though, is definite; there is no "most likely." So what happens if you

put a black marble into the bag?

[curr3nt]

This is a trick question.

You can't use probability to prove the color of the marbles.

Just because the probability of pulling a black marble from BBB, BBW or BWW is 2/3 doesn't mean it HAS to be BBW since it's probability of pulling a black marble is 2/3.

Of the BBB, BBW and BWW, two are false and are altering the overall probability.

Edited March 31, 2011 by curr3nt

[Guest]

Omnihistor, you've started on the right track. The answer, though, is definite; there is no "most likely." So what happens if you

put a black marble into the bag?

is the same. Probabilistically, there is no effect that a white or black marble would have on result if one were added to the hypothetical scenario (that is, inserting a given into an unknown has no effect on the unknown component).

I suppose there could be a "definite" answer behind the marbles being a black and white pair, but I do not see how that would logically follow given that there is no empirical support to justify this view.

Edited March 31, 2011 by omnihistor

[Guest]

The three possibilities in the bag are BB, BW, and WW, having probabilities, respectively, of 1/4, 1/2, 1/4.

Add a black marble to the bag.

Now, the probabilities of BBB, BWB, and WWB (I just added a B to the three above) are still 1/4, 1/2, 1/4. Presumably this is correct, since adding a marble of a known color doesn't change the probabilities at all.

Now, what is the probability of pulling a black marble from the bag?

(Prob. of pulling B from BBB)*1/4 + (Prob. of pulling B from BWB)*1/2 + (Prob. of pulling B from WWB)*1/4

= 1*(1/4) + (2/3)*(1/2) + (1/3)*(1/4) = 2/3

But the ONLY way to have a 2/3 probability of pulling black from a bag with three marbles is if the marbles are BBW. Therefore, the bag with three marbles had BBW, and the original bag contained Black and White.

I know, I know, this sounds insane. curr3nt, could you explain some more where you think the flaw is, if there is one?

[curr3nt]

Now, what is the probability of pulling a black marble from the bag?

(Prob. of pulling B from BBB)*1/4 + (Prob. of pulling B from BWB)*1/2 + (Prob. of pulling B from WWB)*1/4

= 1*(1/4) + (2/3)*(1/2) + (1/3)*(1/4) = 2/3

You are calculating the probability of the bag as though it is BBB, BBW and BWW. However, we know two of those are cases are false since the bag can not be all three.

Someone knowing the state of the marbles would calculate the probabilities differently. It would either be 1, 2/3 or 1/3.

Since it is possible to have BB or WW in the bag at the start it should show that a probability of 2/3 with unknown states and a probability of 2/3 with known states are not equal.

The puzzle solution is based upon a trick getting the reader to believe that a 2/3 with three possible combinations BBB, BBW or BWW is equal to the 2/3 from a known BBW.

Adding a black marble is just a distraction. It is the same as saying that since the probability of drawing a black marble from BB, BW and WW is 1/2 then it must be BW since that is also 1/2.

Not sure if it can be explained better... I'm not the best at probability.

[Guest]

You changed the question. You went from what is in the bag to "Now, what is the probability of pulling a black marble from the bag?" and that is where your 2/3 came from.

The three possibilities in the bag are BB, BW, and WW, having probabilities, respectively, of 1/4, 1/2, 1/4.

Add a black marble to the bag.

Now, the probabilities of BBB, BWB, and WWB (I just added a B to the three above) are still 1/4, 1/2, 1/4. Presumably this is correct, since adding a marble of a known color doesn't change the probabilities at all.

Now, what is the probability of pulling a black marble from the bag?

(Prob. of pulling B from BBB)*1/4 + (Prob. of pulling B from BWB)*1/2 + (Prob. of pulling B from WWB)*1/4

= 1*(1/4) + (2/3)*(1/2) + (1/3)*(1/4) = 2/3

But the ONLY way to have a 2/3 probability of pulling black from a bag with three marbles is if the marbles are BBW. Therefore, the bag with three marbles had BBW, and the original bag contained Black and White.

I know, I know, this sounds insane. curr3nt, could you explain some more where you think the flaw is, if there is one?