Any mathematicians here?

[danmand]

Martin Gardner, who inspired the gathering we report on in this article, died on Saturday. New Scientist consultant Jeff Hecht has written an assessment of his career on our CultureLab blog.

Gallery: Mathemagical visions: a Gathering for Gardner album

Gary Foshee, a collector and designer of puzzles from Issaquah near Seattle walked to the lectern to present his talk. It consisted of the following three sentences: "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

The event was the Gathering for Gardner earlier this year, a convention held every two years in Atlanta, Georgia, uniting mathematicians, magicians and puzzle enthusiasts. The audience was silent as they pondered the question.

"The first thing you think is 'What has Tuesday got to do with it?'" said Foshee, deadpan. "Well, it has everything to do with it." And then he stepped down from the stage.

The gathering is the world's premier celebration of recreational mathematics. Foshee's "boy born on a Tuesday" problem is a gem of the genre: easy to state, understandable to the layperson, yet with a completely counter-intuitive answer that can leave you with a smile on your face for days. If you have two children, and one is a boy, then the probability of having two boys is significantly different if you supply the extra information that the boy was born on a Tuesday. Don't believe me? We'll get to the answer later.

As a melting pot of outside-the-box abstract thinking, this gathering is one of a kind. Attendees were invited to make the world's first snub dodecahedron out of balloons, shown how to solve the Rubik's cube while blindfolded and given tips on how to place a lemon under a handkerchief without anyone knowing. The 300 guests included magicians, origamists, artists, maze designers, puzzle writers, toy inventors and cognitive psychologists, as well as some of the world's most gifted mathematicians

 

[danmand]

Trouble with boys
The first thing to remember about probability questions is that everyone finds them mind-bending, even mathematicians. The next step is to try to answer a similar but simpler question so that we can isolate what the question is really asking.

So, consider this preliminary question: "I have two children. One of them is a boy. What is the probability I have two boys?"

This is a much easier question, though a controversial one as I later discovered. After the gathering ended, Foshee's Tuesday boy problem became a hotly discussed topic on blogs around the world. The main bone of contention was how to properly interpret the question. The way Foshee meant it is, of all the families with one boy and exactly one other child, what proportion of those families have two boys?

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.


Now we can repeat this technique for the original question. Let's list the equally likely possibilities of children, together with the days of the week they are born in. Let's call a boy born on a Tuesday a BTu. Our possible situations are:


■When the first child is a BTu and the second is a girl born on any day of the week: there are seven different possibilities.
■When the first child is a girl born on any day of the week and the second is a BTu: again, there are seven different possibilities.
■When the first child is a BTu and the second is a boy born on any day of the week: again there are seven different possibilities.
■Finally, there is the situation in which the first child is a boy born on any day of the week and the second child is a BTu – and this is where it gets interesting. There are seven different possibilities here too, but one of them – when both boys are born on a Tuesday – has already been counted when we considered the first to be a BTu and the second on any day of the week. So, since we are counting equally likely possibilities, we can only find an extra six possibilities here.

Summing up the totals, there are 7 + 7 + 7 + 6 = 27 different equally likely combinations of children with specified gender and birth day, and 13 of these combinations are two boys. So the answer is 13/27, which is very different from 1/3.

It seems remarkable that the probability of having two boys changes from 1/3 to 13/27 when the birth day of one boy is stated – yet it does, and it's quite a generous difference at that. In fact, if you repeat the question but specify a trait rarer than 1/7 (the chance of being born on a Tuesday), the closer the probability will approach 1/2.

Which is surprising, weird… and, to recreational mathematicians at least, delightfully entertaining

 

[Pilotas]

Gender is decided at conception. Birth day/date is irrelevant and therefor not be included in any calculations.

I agree though, these present fun problems.

 

[Mencken]

Gender is decided at conception. Birth day/date is irrelevant and therefor not be included in any calculations.

I agree though, these present fun problems.

And that misses the point.

The math as presented appears sound. Very interesting.

 

[Anynym]

The fact that one child is a boy is actually irrelevant: the question merely asks what the odds are that the *other* child is a boy, which is approximately 1:2 (or approximately 50% probability), same as for any child in a random sample.

The gender of one child has no bearing on the gender of the other. The combinatorics of gender between two children in a random sample are merely a diversion. The proof of that fact is that one of the children is already identified as a boy: the "GG" combination does not exist in the enumeration; the remaining members of the set are double-counting based on the order of birth, which wasn't part of the question.

 

[danmand]

The fact that one child is a boy is actually irrelevant: the question merely asks what the odds are that the *other* child is a boy, which is approximately 1:2 (or approximately 50% probability), same as for any child in a random sample.

The gender of one child has no bearing on the gender of the other. The combinatorics of gender between two children in a random sample are merely a diversion. The proof of that fact is that one of the children is already identified as a boy: the "GG" combination does not exist in the enumeration; the remaining members of the set are double-counting based on the order of birth, which wasn't part of the question.

No, this is a variation of the "let's make a deal" problem. There are 4 combinations, bb gg bg gb. As one is a boy, there are only 3 left, bb bg and gb. Probablity 1/3.

 

[red]

thats aw sum

 

[great bear]

thats aw sum

And that my friend is a dum sum reply.

 

[red]

And that my friend is a dum sum reply.

at least it adds to the thread

 

[1HandInMyPocket]

The fact that one child is a boy is actually irrelevant: the question merely asks what the odds are that the *other* child is a boy, which is approximately 1:2 (or approximately 50% probability), same as for any child in a random sample.

The gender of one child has no bearing on the gender of the other. The combinatorics of gender between two children in a random sample are merely a diversion. The proof of that fact is that one of the children is already identified as a boy: the "GG" combination does not exist in the enumeration; the remaining members of the set are double-counting based on the order of birth, which wasn't part of the question.

I guess the expand on dammand's explanation is; one boy born on Tuesday and one girl born on Wednesday = BG combination. A girl born on Monday, and one boy born on Tuesday gives the GB combination. If the puzzle was worded given one is a boy what is the probablity that the the second child is also a boy, then the answer would be 50-50.

 

[oldjones]

…so that we can isolate what the question is really asking

And therein are lie all the problems of misunderstanding in the world.

E.g.: Can offshore drilling be safe?

 

[Anynym]

No, this is a variation of the "let's make a deal" problem. There are 4 combinations, bb gg bg gb. As one is a boy, there are only 3 left, bb bg and gb. Probablity 1/3.

No. I already explained the reason why not.

And no, it is not a variation on the "Let's Make A Deal" problem, for the same reasons. But this is why people (including, obviously, yourself) do get confused.

 

[Anynym]

BG and GB are the same as the problem never states if the boy was the older or younger sibling......right?

Correct.

 

[benstt]

Correct.

It's the other way around. If they tell you the order (ie the boy is the older), then some combinations are eliminated, and the chances revert to 50-50 for the gender of the other child.

The first article explains it very well.

So, consider this preliminary question: "I have two children. One of them is a boy. What is the probability I have two boys?"

...

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.

Think of this as a conditional probability problem. Given that one of the two children is a boy, what is the chance that both are boys?

Normally, you can separate out conditional probabilities cleanly if the events are independent. In this case, you can't, as you don't know what order the given boy is in, you must consider the whole set of possible events, and remove those that don't match the given condition.

How about another example to make it a bit more obvious. I have three children. Given one of them is a boy, what is the chance that the other two are also boys?

The full set of possible kids are:
{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} - eight possibilities before conditions are set

Given that one is a boy, you remove the GGG possibility only:
{BBB, BBG, BGB, BGG, GBB, GBG, GGB} - seven possibilities left after conditions set

How many have three boys? (Ie two more boys given one is a boy?) One of the seven {BBB} fits that description.

So, the chances are 1/7 if I get all this right.

In your independent events interpretation, you'd come up with 1/4, way different. (1/4 = 1/2 x 1/2)

I'm not a mathematician, I only play one on TV.

 

[danmand]

There is a difference between asking

1. I have two children. the first born is a boy. what is the probability that the next born will be a boy? answer: 1/2 ( B or G)

2. I have two children. One of them is a boy. What is the probability of me having two boys? Answer 1/3 (BG, GB or BB)



BUT, you seem all to have been stuck on the first part of the problem. The real interesting math comes when you answer the second question as quoted above: "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

Let's list the equally likely possibilities of children, together with the days of the week they are born in. Let's call a boy born on a Tuesday a BTu. Our possible situations are:

■When the first child is a BTu and the second is a girl born on any day of the week: there are seven different possibilities.
■When the first child is a girl born on any day of the week and the second is a BTu: again, there are seven different possibilities.
■When the first child is a BTu and the second is a boy born on any day of the week: again there are seven different possibilities.
■Finally, there is the situation in which the first child is a boy born on any day of the week and the second child is a BTu – and this is where it gets interesting. There are seven different possibilities here too, but one of them – when both boys are born on a Tuesday – has already been counted when we considered the first to be a BTu and the second on any day of the week. So, since we are counting equally likely possibilities, we can only find an extra six possibilities here.

Summing up the totals, there are 7 + 7 + 7 + 6 = 27 different equally likely combinations of children with specified gender and birth day, and 13 of these combinations are two boys. So the answer is 13/27, which is very different from 1/3.

It seems remarkable that the probability of having two boys changes from 1/3 to 13/27 when the birth day of one boy is stated – yet it does, and it's quite a generous difference at that. In fact, if you repeat the question but specify a trait rarer than 1/7 (the chance of being born on a Tuesday), the closer the probability will approach 1/2.

 

[aznguy99]

Trouble with boys
The first thing to remember about probability questions is that everyone finds them mind-bending, even mathematicians. The next step is to try to answer a similar but simpler question so that we can isolate what the question is really asking.

So, consider this preliminary question: "I have two children. One of them is a boy. What is the probability I have two boys?"

This is a much easier question, though a controversial one as I later discovered. After the gathering ended, Foshee's Tuesday boy problem became a hotly discussed topic on blogs around the world. The main bone of contention was how to properly interpret the question. The way Foshee meant it is, of all the families with one boy and exactly one other child, what proportion of those families have two boys?

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.


Now we can repeat this technique for the original question. Let's list the equally likely possibilities of children, together with the days of the week they are born in. Let's call a boy born on a Tuesday a BTu. Our possible situations are:


■When the first child is a BTu and the second is a girl born on any day of the week: there are seven different possibilities.
■When the first child is a girl born on any day of the week and the second is a BTu: again, there are seven different possibilities.
■When the first child is a BTu and the second is a boy born on any day of the week: again there are seven different possibilities.
■Finally, there is the situation in which the first child is a boy born on any day of the week and the second child is a BTu – and this is where it gets interesting. There are seven different possibilities here too, but one of them – when both boys are born on a Tuesday – has already been counted when we considered the first to be a BTu and the second on any day of the week. So, since we are counting equally likely possibilities, we can only find an extra six possibilities here.

Summing up the totals, there are 7 + 7 + 7 + 6 = 27 different equally likely combinations of children with specified gender and birth day, and 13 of these combinations are two boys. So the answer is 13/27, which is very different from 1/3.

It seems remarkable that the probability of having two boys changes from 1/3 to 13/27 when the birth day of one boy is stated – yet it does, and it's quite a generous difference at that. In fact, if you repeat the question but specify a trait rarer than 1/7 (the chance of being born on a Tuesday), the closer the probability will approach 1/2.

Which is surprising, weird… and, to recreational mathematicians at least, delightfully entertaining

Your logic does seem to make sense but somehow there is doublecounting here:
the last 6 possibilities should not be included, or else it is doublecounting

then you will get 7/21 = 1/3 same as previous example.

looking at the problem differently using your 1st example:
if we start stating order of child, then the outcomes
B1G2, G1B2, B1B2, B2B1, G1G2, G2G1 (using the same logic you will double count these BB and GG)
and since we know G1G2 and G2G1 is not a possibility,
we have 2/4 = 1/2 which is flawed.

 

[Cassini]

Okay Danmand, I'll bite. "I have two children. One is a boy born on Tuesday. What is the probability I have two boys?"

What exactly is your correct answer to this question?

 

[Anynym]

There is a difference between asking

1. I have two children. the first born is a boy. what is the probability that the next born will be a boy? answer: 1/2 ( B or G)

2. I have two children. One of them is a boy. What is the probability of me having two boys? Answer 1/3 (BG, GB or BB)

No. Repeating yourself does not change the mathematics involved.

And the very idea that the probability "changes" if you state what day of the week should be a clue that you've got something very wrong.

As has been clearly stated above, your enumeration of "BG" and "GB" are identical under (2): the enumeration does not change the odds, which are 1:2.

 

[pe1984]

No. Repeating yourself does not change the mathematics involved.

And the very idea that the probability "changes" if you state what day of the week should be a clue that you've got something very wrong.

As has been clearly stated above, your enumeration of "BG" and "GB" are identical under (2): the enumeration does not change the odds, which are 1:2.

I didn't read the whole solution of the puzzle but as someone doing his PhD in pure mathematics, I believe that those two particular statements of danmand are absolutely true, by the axioms of probability. Perhaps they don't make sense to some of us but any ways they are mathematically correct. These are simple exercises on so called conditional probability. You can read the basics on wikipedia.
Peace

 

[danmand]

I didn't read the whole solution of the puzzle but as someone doing his PhD in pure mathematics, I believe that those two particular statements of danmand are absolutely true, by the axioms of probability. Perhaps they don't make sense to some of us but any ways they are mathematically correct. These are simple exercises on so called conditional probability. You can read the basics on wikipedia.
Peace

I do not take credit for the "calculations" of probability. I merely posted the article about the meeting of recreational mathematics. The proofs have been hotly debated on the web ever since, but competent mathematics like yourself have confimed the correctness.

When I ran my own little research center some years ago, all 5 of us spent 2 days on a similar problem in probability, which was a variation of the "lets make a deal" problem, about a bag with red and white marbles, where the probability changes after you have taken 1 of the marbles.