A Father has Two Children at Least One is Boy (born on a Tuesday)

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In summary, the conversation discusses a variation of the classic problem where a man has two children and at least one is a boy. The probability of having two boys is initially assumed to be 50%, but it is actually 1/3 when considering the gender combinations BB, BG, GB. However, when the condition of one child being born on a specific day is added, the probability changes to 13/27 due to the elimination of certain possibilities. This leads to a paradox where the probability can be calculated without knowing the day of birth, but the expected answer is 1/3. This paradox has been discussed in a blog post by Keith Devlin and on a physics forum.
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starfish99
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There is a variation of an old problem that is driving me nuts
The original problem is : a man has two children and at least one of them is a boy. What is the probability that he has two boys? You immediately think 50%, but you realize the gender combinations BB, BG,GB(where BB reresents Boy born first, Boy born second,etc.). The answer turns out to be 1/3. I understand this . No problem.
Suppose,however,you say :a man has two children at least one of which is a boy born on a Tuesday. What is the probability that he has two boys? Your fiirst instinct is to ask what does Tuesday have to do with anything. It turns out that it does change the answer.

(The analysis that follows is from Keith Devlin's Blog called Devlin's Angle titled Probablity Can Bite in the Mathematical Association of America"[URL

That makes the mathematics very different, as I'll now show. Instead of just the two genders, B and G, of the original puzzle, there are now 14 possibilities for each child:

B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su
G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su

When I tell you that one of my children is a boy born on a Tuesday, I eliminate a number of possible combinations, leaving the following:

First child B-Tu, second child: B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.
Second child B-Tu, first child: B-Mo, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.

Notice that the second row has one fewer members than the first, since the combination B-Tu + B-Tu already appears in the first row.

Altogether, there are 14 + 13 = 27 possibilities. Of these, how many give me two boys? Well, just count them. There are 7 in the first row, 6 in the second row, for a total of 13 in all. So 13 of the 27 possibilities give me two boys, giving that answer of 13/27.

Now the following scenario is what is driving me, Starfish 99,nuts.
Can we use the sudden shift in probability from 1/3 to 13/27 to create a paradox?
Say, for example there are eight fathers:Mr. Sunday, Mr Monday, Mr Tuesday, Mr Wednesday, Mr Thursday, Mr Friday, Mr. Saturday, and Mr Unknown.

All of the fathers have two children of which at least one is a boy and Mr. Sunday's boy is born on Sunday, Mr. Monday's boy is born on Monday, Mr Tuesday's boy is born on a Tuesday,etc.
I talk to Mr . Sunday who tells me he has two children of which at least one is a boy. I tell him that he has a 1/3 probability of having two boys . He then tells me the boy was born on Sunday. I immediately tell him the probability is 13/27 of having two boys.

I then talk to Mr. Monday who tells me he has two children of which at least one is a boy. I tell him that he has a 1/3 probability of having two boys. He then tells me the boy was born on a Monday, I again change my answer to the probability is 13/27 of having two boys.

I talk to Mr. Tuesday, Mr. Wednesday, Mr, Thursday, Mr. Friday and Mr. Saturday
and I get the same results.
Finally I talk to Mr Unknown.
He tells me he has two children of which at least one is a boy and he was born on a...
I say Stop ! Wait! Don't tell me!
I know from the previous seven fathers that the probabilty is 13/27 of having two boys for Misters Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday. There are only seven days of the week. So I am going to say that Mr . Unkown's probability of having two boys is 13/27. So did I (correctly?) answer the probability question without his telling me what day of the week his son was born? But the probability answer without the day of the week given is supposed to be 1/3. What is going on?
 
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1. What is the probability that the other child is a girl?

The probability that the other child is a girl is 50%, as the gender of one child does not affect the gender of the other child.

2. Why specify that one child is a boy and the other is born on a Tuesday?

This specification is used to add context and eliminate any ambiguity in the question. It is important to specify that one child is a boy and the other is born on a Tuesday to accurately calculate the probability.

3. How does the day of the week affect the probability?

The day of the week does not affect the probability, but it is used as a way to add specificity and make the question more interesting.

4. What if the question stated "at least one is a boy and the other is not a girl"?

The probability would still be 50%, as the statement "the other is not a girl" would include both boys and non-binary genders.

5. Can this question be applied to real-life situations?

Yes, this question can be applied to real-life situations where there are two children and the gender of one is known, but the gender of the other child is unknown. It can also be used as a mathematical problem to test and improve probability skills.

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