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The problem is ambiguous, due to under specification. That means that neither #1 nor #2 is "actually equivalent" to "at least one child is a boy," and more information is needed to construct a probability space. #1 is "When both genders are known, and boys are preferred in the description, at least one is a boy." The preference is what makes the answer 1/3, and assuming it adds information to the problem. #2 can be "When only one gender is known, and how we know it is uncorrelated with either possibility, at least one is a boy." But it can also be "When both genders are known, and the description reflects the probability of that gender being chosen at random from the two, at least one is a boy." In both cases, the answer is 1/2. But being under-specified does not mean the question can't be answered, it just requires applying a reasonable assumption instead of an unreasonable one. #1 is very unreasonable since it adds information, #2 is close, but #3 is best. And the proof is Bertrand's Box Paradox. That name does not properly refer to a probability problem, it applies to how to make this reasonable assumption. "Mr. Jones has exactly two children. I have written the gender, of at least one, inside this sealed envelope. What is the probability that both children have that gender?" If you were to open the envelope, and see the word "boy," the problem becomes the same as the one under discussion. If it can be answered, that answer is correct here as well. But it is an equivalent problem if you see the word "girl," and again the answer must be the same. If 1/3 is an acceptable answer, it means that 1/3 of all two-child families have two of the same gender, and 2/3 have mixed genders. But that is a contradiction. We know that the split is 1/2:1/2. So the assumption, that 1/3 is a reasonable answer, is disproven. Now, that does not mean that the information came to us via #2 or #3, it just means we can't assume that it was #1. Most often, the same logic is used for the Monty Hall Problem, it is just applied backwards. |