A Twist on the Two Children Problem (Not a HW question)

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The discussion centers on the Two Children Problem, specifically the scenario where one child is a boy born on Tuesday, leading to a probability calculation of having two boys. Contrary to the intuitive answer of 1/2, the actual probability is 13/27. The reasoning involves analyzing the possible outcomes based on the days of the week and the genders of the children. The author expresses skepticism about the relevance of the birthday in determining the probability, suggesting that the method may be flawed due to its reliance on conditional probabilities.

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waterchan
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On this website (http://www.sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong), I recently came across a twist on the Two Children Problem. The problem is deceptively simple: "I have two children, one of whom is a boy born on Tuesday. What is the probability that I have two boys?"

Intuitively the answer would appear to be 1/2, but in fact it turns out that the probability is 13/27. I don't really understand the graphical solution presented on that website, so here's my own understanding of the "official" solution.

_____________________________________________________

Let's name the two children A and B. One of them is the Tuesday-born son. We don't know which one, so we consider the case that A is the Tuesday-born son. Then we consider the case that B is the Tuesday-born son.

If A is the Tuesday-born son, then B is either a boy born on any of the days (7 outcomes), or a girl born on any of the days (7 outcomes). 7 + 7 = 14 possible outcomes so far.

If A is not the Tuesday-born son, then B MUST be the Tuesday-born son. Then what is A? A can be a girl born on any of the days (7 outcomes). Or, A can also be a boy born on any of the days EXCEPT TUESDAY. Because otherwise, that would contradict the assumption for this case that A is not the Tuesday-born son. Thus, we remove the Tuesday possibility and are left with 7 - 1 = 6 outcomes.

The total number of outcomes is thus 7 + 7 + 7 + 6 = 27. The total number of desired outcomes, where both children are boys, is 7 (A is Tuesday boy, B is boy) + 6 (B must be Tuesday-boy and A is a boy) = 13 outcomes.

Thus, probability = 13/27.

______________________________________________________

Now, personally I suspect the solution is flawed because it considers the birthday which is arguably irrelevant, and applies conditional probability to unconditional outcomes. But I am unable to prove that it's flawed.

Another intuitive reason I believe the method is flawed is the following. Consider replacing the verb "have" in the question with any other verb. For example:

I kidnapped two children, one of whom is a boy born on Tuesday. What is the probability that I kidnapped two boys? (Does the birthday and sex of the kidnapped child affect anything?)

I knocked down two children, one of whom is a boy born on Tuesday. What is the probability that I knocked down two boys? (Does the birthday and sex of the knocked-down child affect anything?)

I ran over two children, one of whom is a boy born on Tuesday. What is the probability that I ran over two boys? (Does the birthday and sex of the injured child affect anything?)

I vaporized two children, one of whom is a boy born on Tuesday. What is the probability that I vaporized two boys? (Does the birthday and sex of the vaporized child affect anything?)

For all three of these questions, the exact same calculation can be applied to arrive at the exact same answer of 13/27 (which is slightly less than 1/2, the answer obtained if the information about being born on a Tuesday is disregarded). Doesn't it seem wrong that the sex and birthday of the kidnapped/knocked-down/ran-over /vaporized child has any affect on the other's sex and birthday?
 
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The article points out that the reason that the odds are skewed is because your inital selection criteria is biased - it is not a random sampling. Read the part about "the pair of puppies, at least one of which is male".
 

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