
Think you understand Monty Hall? Try the Tuesday boy problem. - ColinWright
http://scienceblogs.com/evolutionblog/2011/11/the_tuesday_birthday_problem.php
======
user24
Doesn't this rest on the simple ambiguity in the phrasing?

> I have two children and one is a son born on a Tuesday.

If by that is meant:

> I have two children. Here is some information about one of them: son, born
> on Tuesday.

Then the probability of the other child being a son is 1/2.

If on the other hand we mean:

> I have two children. One or more is a son. Exactly one of them was born on a
> Tuesday.

Then we get the 13/27 probability.

In fact it doesn't seem reasonable at all to assume that _only_ one was born
on Tuesday, while _at least_ one is a son. One single interpretation of 'one
of them' must be applied to _both_ the gender and day of birth. Otherwise
we're picking and choosing our interpretation on a whim.

edit: Colin appears* to think that what I've said here is incorrect, and I'd
like to know why. I'm not a maths/stats person at all so am very keen to be
re-educated on this matter.

* based on his now-deleted reply to ars which said "no, it doesn't, and no, you're not"

 _Important Edit Two:_

If I'm reading this right, I think the defender of the 13/27 solution would
say:

No. We don't discount the possibility of both being tuesday-boys (TB), we just
adjust the calculation so that it doesn't count (eldest=TB, youngest=TB) and
(youngest=TB, eldest=TB) as two separate possibilities.

To which I respond:

Right, so it's not down to ambiguity. But shouldn't you also discount every
other symmetrical pair such as (eldest=TB, youngest=WB) and (youngest=TB,
eldest=WB) and thus return the odds to 1/2? Or does that not return the odds
to 1/2?

~~~
ColinWright
No, you shouldn't "also discount every other symmetrical pair", for exactly
the same reason as there's a 1/36 chance of rolling double 6, but 2/36 chance
of rolling a six and a one. It's all to do with labellings, and it's the most
common source of error[1] in statistics.

[1] By "error" I mean calculations that then don't agree with the experimental
results.

~~~
user24
What about this:

I have two teenagers. One is a boy of 13.

Do we encounter a similar situation with regard to the odds of the second
teenager being a boy?

edit: I'm thinking the odds are exactly the same, 13/27, by coincidence, as
there are seven possible teen ages.

So then, what about this: One is a boy named George. Or One wears a black
shirt. Or One likes chocolate.

Doesn't this mean that the more information we gain about the boy, the less
likely it makes it that his sibling is a brother?

~~~
orangecat
_Doesn't this mean that the more information we gain about the boy, the less
likely it makes it that his sibling is a brother?_

More likely, but only if that information being true was a precondition for
knowing about the boy in the first place. Take the following scenarios,
assuming Alice knows Bob has exactly two children.

Alice: Do you have a son? Bob: Yes Alice: Pick one of your sons, and tell me
the day of the week he was born Bob: Sunday

Here the day of week provides no additional information because Bob will
always have an answer (like in Monty Hall, where Monty will always reveal a
losing door), so the probability that Bob has two boys is 1/3.

Alice: Do you have a son who was born on a Sunday? Bob: Yes

Here having a son isn't enough; he also has to satisfy a condition that occurs
with only 1/7 probability. Bob is more likely to be able to answer yes if he
has two sons and thus two chances to satisfy that condition.

------
reitzensteinm
I've been thinking about this for an hour, and I'm now convinced that the
author is wrong. The fact that we found out about one of the children _from
the father_ means that all probabilities are not equal, even though they're
treated here like they are.

The difference is between the information being offered, and determined
independantly. I'll do this with the boy/girl problem, for simplicities sake.

If we ask a man if he has at least one boy, and he says yes, we can work out
the chance the other child is a boy like so:

Assume all four possibilities are equally likely:

    
    
       BB
       BG
       GB
       GG
    

If we ask him if he has at least one boy, and he says yes, we effectively
filter off GG, which brings the list down to:

    
    
       BB
       BG
       GB
    

Therefore, the chance of the other child being a boy is 1/3. Pretty straight
forward.

However! Because the father offered the information on his own, it effectively
turns it into the author's other problem, where the older child is a boy, find
out the gender of the younger child.

The trick is that he's equally likely to give information about either of his
children, therefore there are eight possibilities:

He gives information about child A:

    
    
       BB - B
       BG - B
       GB - G
       GG - G
    

He gives information about child B:

    
    
       BB - B
       BG - G
       GB - B
       GG - G
    

There are still 3 possibilities, but BB is twice as likely as the others,
because if both children are a boy he's definitely going to reveal the gender
of one of them as a boy; whereas if one is a girl and one is a boy, there's
only a 50% chance he will.

    
    
       BB - 50%
       GB - 25%
       BG - 25%
    

So, the answer to this:

 _You meet a man on the street and he says, “I have two children and one is a
son born on a Tuesday.” What is the probability that the other child is also a
son?_

Is 1/2

The answer to this:

 _A man has two children, and one is a son born on a Tuesday. What is the
probability that the other child is also a son?_

Is 13/27

It's nitpicky, but I think the author should be very exact about this kind of
thing, since he's trying to clear things up.

~~~
NickPollard
The issue here is with assumptions - you have made a different set of
assumptions from the author, and hence are getting a different result.

A lot of people here are having similar issues, by misreading exactly what the
initial proposition means.

Your reasoning above relies on the 'likeliness' of a man giving you the
information, which is something that is not meant to be a part of the problem.
Although it is phrased as a man 'telling' you something, that statement is
really a metaphor for 'you determine the following piece of information, 100%
truthfully'.

In particular, your explanation assigns agency to the man - that if he does in
fact have a son, he may or may not choose to reveal the truth 'I have a son'.
However it makes no allowance for the man lying - so you are assuming if he
answers it will be truthfully, but you are allowing him the lie of omission.

Whilst there is nothing specific that rules out your interpretation, it is not
what is intended. Read it instead as:

\----

There exists a man, A.

A has exactly two children.

The statement 'A has at least one Son, B' is true

What is the chance that the statement 'The Non-B child of A, is a son' is
true?

\----

~~~
reitzensteinm
I agree that that is what the author intended to communicate. But I actually
think that's a bigger stretch than my own interpretation - it changes how the
information was determined, which has a definite impact on the outcome.

The reason why I posted was to suggest that the author should have worded it
the second way, i.e.

 _A man has two children, and one is a son born on a Tuesday. What is the
probability that the other child is also a son?_

Which leaves no doubt. I guess I didn't really make that clear enough with my
original post.

------
raldi
I found this post confusing and ambiguous, so I restated it in simpler terms,
with pictures:

[http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-
proble...](http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-problem-in-
under-300-words.html)

~~~
bittermang
Thank you for this. I slugged through the whole original article and felt like
I was being beat up with words.

However, I still fail to comprehend how the "at least one is a boy" quirk
maths out to a 1 in 3 chance that his second child is also a boy.

Taken literally, it does. I understand that, in a set of data, GB is different
from BG. But for the sake of our comparison, the order the children were born
in doesn't matter. We're seeking if the other child is a boy, or not.

In my mind, the bit about it being the younger or older sibling is irrelevant
information. We're comparing gender, not age. Regardless of if she is the
younger sister, or the older sister, she's still his sister, and therefore not
a boy.

I think the introduction of age is convoluting the issue, unnecessarily.

I now standby, ready to be proven wrong. I'd really like to wrap my head
around this one, but I must insist that the age information is irrelevant.

~~~
drblast
Probability is much less meaningful when you're not talking about large groups
of repeated trials.

In the context of the article, imagine that you didn't do this just one time,
but that you asked 100 fathers about the compostion of their children.

The question posed in the article is essentially, "Of those fathers who
responded 'I have one son,' (which is likely 75 of the 100), how likely is it
that they have another son, (which is likely 25 of that group of 75, or 1/3).

When the article talks about the father standing next to one of his children
at random and the probability of another son being 1/2 at that point, it helps
to imagine those same 100 fathers all standing next to their children. Of that
group, you're not eliminating the fathers standing next to girls based on the
way the situation is posed.

The English words used to describe each case make it much less clear which
group of 100 people we're talking about.

Also, when we talk about one father and not a group of fathers, the 1/3 or 1/2
number is much less meaningful. This is where insurance companies make their
money (ideally). It's impossible to predict whether a single person will die
in a car accident over their lifetime, and any number is essentially a guess.
But it's very easy to predict that, say, 1 in 50,000 people will.

------
nadam
As the currently top voted comment does not get it, I try to intuitively
explain the paradox.

No, it is not ambiguity of language. It says formally:

I have two children. There exists a child of mine who is (boy and born on
tuesday).

And yes, the probability of the other child being a boy is 13/27.

To understand this, try a more extreme case:

When a child is born we generate a random number: rnd(1billion)

Now the man says:

'I have two children. There exists one of them which is boy with generated
id=456765234'

As we can see, the probability of the other child being a boy approaches 1/2
as we increase the bound of our random number!

Why is this intuitively?

Because by using a very specific id we are closer and closer to specifying one
of the boys, as the probability of the id not being unique decreases. We are
closer and closer to saying this:

'Randomly pick one of my children. He is a boy. What is the probability of the
other being a boy?'

Yes, that would be 1/2.

Edit.: Seriously downvoting this??? For most commenters the correct result was
not intuitive. (It was not intuitive for me at first also). I tried to make
this intuitive to them. (a bit deeper than usual paradxes though) Or do you
still think the 13/27 result is not correct? In that case you are fighting
mathematical truth with the downvote button.

Edit: Thanks, corrected probability from 2 to 1/2

~~~
ars
Why can't the other boy also be born on Tuesday?

~~~
nadam
The other can also be born on Tuesday. (generated id = 1 in rnd(7)) And the
other can also have a generated id = 456783123 in rnd(1billion). But as the
bound of the rnd is bigger and bigger, the probability of that approaches to
zero. So as we know more and more information of one of the children, we are
closer and closer to simply uniquely identifying him, so the probability of
the other being a boy approaches 2.

~~~
ars
If both can be born on Tuesday then telling me one was born on Tuesday does
not uniquely identify him! It's not unique if both can do it.

And since it's not unique the rest of your analysis is based on a faulty
assumption.

(BTW I did not downmod you, in case you were wondering.)

~~~
nadam
It is about specifying a child, and then saying it is a boy.

The one extreme is: one of my children is a boy. in this case the other is a
boy with 1/3 probability.

The other extreme is: One of my children has a national unique id=... He is a
boy. The other is a boy with 1/2 probability.

And other cases are in between. The more information you provide on the first
child (the less chance there is that the other can have the same property, the
more close you are to 1/2.

Of course in the rnd(7) example we are in-between:

1/3 < 13/27 < 1/2

~~~
ars
"the less chance there is that the other can have the same property"

No! That is not true! The two events are independent, specifying information
on one has zero impact on the other.

~~~
iliis
They are independent, but you get information about both ("One of _them_ is a
boy."), therefore p=1/3.

If you say "the older is a boy" then you have a statement about one son and no
information whatsoever about the younger one, therefore p=1/2.

~~~
ars
"One of them is a boy." is a completely different scenario and no one is
arguing about it. (BG, GB, BB, GG - by saying one is a boy you exclude GG
leaving 1/3.)

In this case we are asking if the other is a boy NOT if the other was born on
Tuesday.

~~~
iliis
Ah, sorry, I misunderstood your question.

You are asking, why the information "one is born on Tuesday" says anything
about the gender or birthday of the other, right?

More specificaly, this:

> _Let us first assume that it is the older child who was a son born on a
> Tuesday. In this case the second child could be either of two sexes, and
> could have been born on any of seven days of the week, for a total of 14
> possibilities.

> Now let's suppose it is the younger child who was a son born on a Tuesday.
> Then the older child could, again, be either of two sexes and could have
> been born on any of seven days of the week, again providing 14
> possibilities. Added to our original 14 that would seem to give 28
> possibilities.

> But be careful! One possibility got counted twice. Specifically, the one
> where both children are boys born on Tuesdays. So really there are only 27
> possibilities. And since 13 of them involve the second child being a boy,
> the probability would be 13/27_.

Maybe it helps, if you first imagine _all_ 196 (2 * 2 * 7 * 7, think of four
7-by-7 tables [draw them, it helps a lot ;)]) possibilities. Eliminate 49 of
them where both are girls (one of the four tables) and you have 147
possibilities/cells left. Then, in each of the two BG and GB tables leave only
one row/column, eg. eliminating another 2 * 6 * 7=84 possibilities.

Now comes the interesting/tricky part: The last case/table, where both are
boys contains only 13 possible cases (instead of 2 * 7)! Imagine a 7-by-7
table, where each row and column corresponds to a day. Mark all the cells
which are not in a Tuesday-row or a Tuesday-column, eg. eliminate all
posibilities where none of the two boys are born on a Tuesday. This eliminates
another 49-13=36 cases.

So, we have a total of 196-49-84-36=27 cases, 13 of which "the other is a boy"
and 3/27=1/9 cases where "the other is born on a tuesday".

I hope this makes somewhat sense. Once I've drawn all the possibilities and
marked all the impossible ones, it became a lot clearer.

~~~
iliis
ADDENDUM: The tables:

    
    
           A: Boy                     A: Boy
           M T W T F S S              M T W T F S S
                               
     B M   . 1 . . . . .        B M   . 1 . . . . .
       T   1 1 1 1 1 1 1          T   . 1 . . . . .
     B W   . 1 . . . . .        G W   . 1 . . . . .
     o T   . 1 . . . . .        i T   . 1 . . . . .
     y F   . 1 . . . . .        r F   . 1 . . . . .
       S   . 1 . . . . .        l S   . 1 . . . . .
       S   . 1 . . . . .          S   . 1 . . . . .
      
      
           A: Girl                    A: Girl
           M T W T F S S              M T W T F S S
         
     B M   . . . . . . .        B M   . . . . . . .
       T   1 1 1 1 1 1 1          T   . . . . . . .
     B W   . . . . . . .        G W   . . . . . . .
     o T   . . . . . . .        i T   . . . . . . .
     y F   . . . . . . .        r F   . . . . . . .
       S   . . . . . . .        l S   . . . . . . .
       S   . . . . . . .          S   . . . . . . .
    
     total: 196
     of which are possible ('1'): 27
     of which 'the other is a boy': 13
       (all the ones from the upper left table)
     of which 'the other child is born on a tuesday': 3
       (the T/T-cell of each table)

------
Estragon
It's fun to reason these things out, but it's easier and faster to simulate,
at least at first.

    
    
      import random, time, collections, pprint
      
      sample = collections.defaultdict(float)
      
      now = time.time()
      
      known_child = ('M', 2)
      
      while sum(sample.values()) < 100000:
          genders = [random.choice('MF') for dummy in '..']
          birthtimes = [random.uniform(0, now) for dummy in '..']
          birthdays = [time.localtime(t).tm_wday for t in birthtimes]
          gendertimes = zip(genders, birthdays)
          if known_child in gendertimes:
              gendertimes.remove(known_child)
              sample[gendertimes[0]] += 1
              if (sum(sample.values()) % 10000) == 0:
                  print sum(sample.values())
      
      males,females = [sum(c for (g, t), c in sample.items() if g == gender)
                       for gender in 'MF']
      
      print males / (males + females)
    

Result when I ran this was .4816, approximately equal to 13/27.

------
extension
The more sons you have, the more likely that one of them is born on a tuesday.
Thus, if you take all the two-child families with at least one son, and
eliminate the families without a son born on tuesday, the two-boy families are
more likely to remain than the one-boy families, and you will end up with a
higher proportion of two-boy families than before.

Simple.

(edited)

~~~
gcp
_Simple._

And wrong. Note that the correct answer is 13/27 probability of a boy, which
is less, not more, than 50%.

~~~
extension
Sorry, I should have said a higher proportion of two-boy families, not a
higher proportion of boys. 13/27 is more than 1/3, which is the proportion of
two-boy families before you eliminate the families without tuesday boys.

------
yuvadam
I stopped caring about 'paradoxes' ever since I understood there's no
isomorphism between pure mathematical concepts and human language.

~~~
lurker17
The paradoxes are important in highlighting the ambiguities in our language.

The annoying aspect is when someone uses a paradox to show off how
mathematically clever they are, instead of to show how ambiguous language is.
As XKCD illustrates: <https://www.xkcd.com/169/>

------
ars
I hate this one because while it says:

    
    
      "I have two children and one is a son born on a Tuesday."
    

It actually means:

    
    
      "I have two children and only one of them is a son
       born on a Tuesday."
    

You are supposed to just assume this modification.

~~~
danparsonson
No I don't think that's true. Bear with me :-)

There are 14 possible permutations for a child

Mon-Sun Girl

Mon-Sun Boy

which for two children gives a total of 28:

Child 1 is Boy, born Mon-Sun

Child 1 is Girl, born Mon-Sun

Child 2 is Boy, born Mon-Sun

Child 2 is Girl, born Mon-Sun

Note that we haven't said which child is 1 and which is 2, and in fact we
don't know because we haven't been told - this is the important bit.

So, for our unknown child to be a boy, it must be one of the following 14
permutations:

unknown child is Child 1, boy, born Mon-Sun

unknown child is Child 2, boy, born Mon-Sun

but one of those permutations is already taken by our known Tuesday boy, so we
have to remove one of the Tuesday permutations leaving 13 possible outcomes
out of 27. Note that one of those 13 outcomes is 'unknown child is son born on
Tuesday' - it's perfectly valid to have both sons born on a Tuesday.

Note also that we can't say which permutation we're removing until we know
which is Child 1 and which is Child 2, just that one of them is taken.

If the original statement were phrased as '... the older child is a son born
on a Tuesday' then you would have a constraint on which was Child 1 and which
was Child 2 and then the probability would be 1/2 as expected because you
would know up front that you were entirely discounting, say, Child 1, AND that
the removed Tuesday permutation also belonged to Child 1.

But I stand to be corrected!

~~~
ars
First, saying Mon-Sun is confusing almost everyone since most people start the
week on Sun, not Mon. (Even in Europe Christians start the week on Sunday.)

In any case, as I replied here <http://news.ycombinator.com/item?id=3290118>
you can not remove the duplication! It's two different situations, even though
they may appear the same.

~~~
danparsonson
ajanuary explains the duplicates problem nicely here:
<http://news.ycombinator.com/item?id=3290185> so I won't double up.

I've spent the last hour and a half wrestling with the same clash between
maths and intuition that you're experiencing so I understand your frustration
- 13/27 really is the right answer though.

~~~
ars
He explains it, and he's wrong.

Look, I understand your intuition wants you to remove duplicates "They are the
same!". But you should resist, because doing that gives incorrect results.

~~~
danparsonson
On the contrary - I'm saying that 'removing the duplicates' is _counter_
intuitive and that's why you're having trouble with it, the same as I did.

Besides: please re-read my original post - I didn't mention anything about
removing duplicates, and in fact chose to explain the problem in a different
way precisely because the duplicate removal thing was so difficult for me to
grasp sufficiently to be able to explain it.

Ask yourself this: where does the original 28 come from? One of those 28
permutations appears to be 'Child 1 is Tuesday boy' and 'Child 1 is Tuesday
girl'. If you can answer that then you should be able to understand the whole
thing.

(edited for clarity)

------
rubergly

        Let's try a simpler problem. Suppose we know that a 
        certain man has two children and we also know that the
        older one is a boy. In this case we would say that the
        probability that the other child is a boy is 1/2. After
        all, the sex of one child is independent of the sex of
        the other child. That the older child is a boy has no
        bearing on the sex of the younger child.
    
        Now suppose we know simply that a man has two children
        and that one of them is a son. This time we would reason
        that there is no possibility that the person has two
        girls. It follows that the sexes of his two children,
        ordered from oldest to youngest, are either BB, BG or GB.
        Since these cases are equally likely, and since only one
        of them involves having two boys, we would say the
        probability that the man has two boys is 1/3.
    

Maybe I'm missing something, but this seems fundamentally wrong. Pr(two boys |
older child = boy) is not equal to Pr(two boys | one+ child = boy)? Why is
time so special? Could we not order them based on their height, and assert
Pr(two boys | taller child = boy) = 1/2. Or, if there were a scale of
masculinity, order them based on that and assert Pr(two boys | manlier child =
boy) = 1/2, where manlier child = boy <==> one+ child = boy, so Pr(two boys |
manlier child = boy) = Pr(two boys | one+ child = boy) (contradiction).

~~~
incremental
For a randomly selected family with two children, there are four possible
boy/girl combinations: B B, B G, G B, G G

In the first case we are told that the older child is a boy. This leaves only
two cases: B B, B G Therefore, there is a 50% chance the second child is a
boy.

In the second case, we are told only that [at least] one child is a boy. This
leaves three possibilities: B B, B G, G B Therefore, the probability that both
children are boys is 1/3.

Enumerating possible states of the world like this is the fundamental insight
you need to have to be able to understand these types of problems - but it
does take a while to get used to!

~~~
rubergly
But you are assuming that each of those three possibilities has equal
probability; can you explain the rationale for that? (it is clear why BB, BG,
GB, and GG have equal probability in the unrestricted case, but less clear why
BB, BG, and GB have equal probability in this restricted case)

Besides, this is just a rephrasing of the original article's argument, and
doesn't counter mine at all. I am open to the possibility that there is a flaw
in my argument, but where is it?

~~~
incremental
So you say: "it is clear why BB, BG, GB, and GG have equal probability in the
unrestricted case"

The restricted case is just the unrestricted case + one additional bit of
information, that is, you're told that GG is not an option. This eliminates GG
from the unrestricted case, but says nothing more about the probabilities of
the other options. So the probabilities stay equal, although they now equal
1/3 each (if you eliminate options, the remaining options all become more
likely).

What you're missing is this: The statement "the older child is a boy" has more
information than the statement "one of the children is a boy". The first
statement allows you to eliminate two options (GB and GG), while the second
statement only allows you to eliminate one option (GG).

The "older" part is not fundamental to the problem. Equally, the statement
"the taller child is a boy" has more information than "one of the children is
a boy". The problem with this is that probabilities for height are not so
friendly like the 50/50 probabilities for birth order (e.g., boys are likely
to be taller than girls, older children are taller than younger, etc), which
introduces unnecessary complexities to a logic problem. So that's why birth
order is used for these types of puzzles.

------
jules
For the people who don't believe this, here's a Python program:

    
    
        x = [(c1,c2,d1,d2)
                for c1 in ["B","G"]
                for c2 in ["B","G"]
                for d1 in [0,1,2,3,4,5,6]
                for d2 in [0,1,2,3,4,5,6]
                if (c1 == "B" and d1 == 1) or (c2 == "B" and d2 == 1)]
    
        num = sum(1 for (c1,c2,d1,d2) in x if c1 == "B" and c2 == "B")
        denom = sum(1 for _ in x)
    
        print("%s/%s" % (num,denom))

------
_dps
If you generalize "born on Tuesday" to a generic attribute, with probability p
(equal for boys and girls) it becomes easier to see what's going on. For
brevity, let's say people with the attribute are positive and people without
are negative.

There are seven (ordered) possibilities involving at least one positive boy,
and we'll subdivide them into two subgroups: B+B- B-B+ B+G- G-B+ / B+B+ B+G+
G+B+. The important thing to note is that _within each group the outcomes are
equiprobable_.

If p is near 1, then the first group has almost zero probability and we have
~1/3 chance of two boys. If p is near 0 then the second group has almost zero
probability and we're left with ~2/4 chance of two boys.

Intuitively, if p is near 1 then the statement "I have at least one positive
boy" is almost equivalent to "Both my children are positive, and I have at
least one boy" (group 2, 33% chance). If p is near 0, then the statement is
almost equivalent to "I have exactly one positive boy" (group 1, 50% chance).

------
Raphael
I understand thanks to a comment I found here:
[http://helives.blogspot.com/2010/07/tuesday-child-
puzzle.htm...](http://helives.blogspot.com/2010/07/tuesday-child-puzzle.html)

====================================

Unfortunately, these questions you ask are ambiguous, and it is the failure to
recognize how they are ambiguous that causes the results to seem unexpected.
Consider two versions of what led up to the first statement:

Case #1: A father is chosen at random. He is given a slip of paper as he is
led onto a stage. The paper says "Pick one of your children. Tell the audience
the number of children you have, the chosen child's gender, and the day of the
week on which it was born."

Case #2: A father is chosen at random from all fathers who have two children,
including one boy born on a Tuesday. He is also ushered onto a stage and given
a slip of paper that instructs him to tell the audience the criteria used to
select him.

Now shift scenes. You are in the audience when a man is ushered onto the
stage. He looks at a slip of paper, thinks a moment, and says "I have two
children and one of them is a boy born on Tuesday." What is the probability
that he has two boys?

The answer to the question depends on which case applies to the man you
listened to. In Case #1, it is 1/2. In case #2, it is 13/27. Your simulation
only covered the second case. To get the first, after you have two children,
flip a coin to see which one the father will tell about. If it is not a
Tuesday Boy, don’t keep that trial even if the other child is a Tuesday Boy.
You will find that the 27 cases where you have a Tuesday Boy reduce to 14
(just over half, since one father didn’t need to flip the coin), the 13 where
you also have two boys reduces to 7, and the answer is exactly 1/2.

If you simulate the simpler problem, where you don’t worry about the day of
the week, the answers are 1/2 and 1/3 for the two cases, respectively. The
reason 13/27 seems unintuitive, is because the fact that a Tuesday Boy was
REQUIRED in the second case is not intuitively obvious from the statement "one
of them is a boy born on Tuesday." In fact, as you point out, the puzzle could
equally well be named after either of your two children, which is probably two
different names. You choose one, just like the father in case #1, so the
better answer to your question is 1/2, not 13/27. It is still ambiguous, but
there is no valid reason to assume that case #2 applies.

------
jfager
When in doubt, use brute force:

    
    
        1BT 2BM    1BM 2BT
        1BT 2BT    1BT 2BT <-- Counted twice
        1BT 2BW    1BW 2BT
        1BT 2BR    1BR 2BT
        1BT 2BF    1BF 2BT
        1BT 2BS    1BS 2BT
        1BT 2BU    1BU 2BT
        1BT 2GM    1GM 2BT
        1BT 2GT    1GT 2BT
        1BT 2GW    1GW 2BT
        1BT 2GR    1GR 2BT
        1BT 2GF    1GF 2BT
        1BT 2GS    1GS 2BT
        1BT 2GU    1GU 2BT
    

I think the issue people are having is that they want to distinguish between
the two instances of '1BT 2BT', but it doesn't work like that. Let's say I
numbered each of the possibilities written above 1 through 28. If I handed you
a piece of paper with '1BT 2BT' written on it, could you tell me which number
that corresponded to? No - it would be one of two numbers - so the numbering
is additional information not given in the problem statement.

~~~
zerostar07
Alternatively, suppose we label the combinations of child-day BM, GM, BT, GT
... BU, GU (14 items). We have a bucket with 2 of each of these combinations,
as there are 2 kids (total 28 items). The problem states that we have taken
one "BT" item out of the bucket (so 27 items left), and are asking how many
"B*" items are left in the bucket (13).

The confusion arises because one can consider the same problem with
replacement, putting back the BT item in the bucket before picking a second
time. It's something that's not clear if u are not thinking of enumeration.

~~~
DougBTX
What would the question have to be to make the following the answer?

 _Suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14
items). We have two buckets which both contain a copy of each combination
(total 28, 14 in each bucket). The problem states that we have taken one "BT"
item out of one bucket, and are asking how many "B_ " items are left in the
other bucket (7 out of 14). _

~~~
zerostar07
That would be two parents having one child each, one of whom says his only son
was born tuesday, and asking what 's the probability the other guy has a son.

------
bermanoid
Sometimes I hate these types of problems - though they're quite fun to argue
about - because though they seem to be simple, straightforward enumeration
problems, the "right" answer depends on assumptions about how the information
that you've got is obtained, and what the other possible pieces of information
you could have obtained are.

Here's the thing: nobody needs to see how you obtained your answer, we're all
smart enough to enumerate things, so don't bother; similarly, leave out the
equations, they're simple enough, and that's not where anyone falls over. If
you want to argue productively about this type of problem, you need to explain
_why_ you enumerate the possibilities the way you do, not _how_.

And actually, I think the article goes through this problem pretty well, as
mentioned there it hinges on two questions:

1) In this "random sampling", was "girl" a possible answer? Or did we restrict
the sample pool to only people with boys, and only let them reveal that they
had a boy?

2) Similarly, was any day other than "Tuesday" a possible answer?

Once people agree what the "obvious" answers to these questions are,
disagreements tend to evaporate very quickly, and the enumerations/equations
solve themselves. Unfortunately these assumptions are almost never spelled out
in the problem, which is why these damn problems keep confusing people...

------
ghshephard
"You meet a man on the street and he says, “I have two children and one is a
son born on a Tuesday.” What is the probability that the other child is also a
son?"

The fact that one of the sons is born on a Tuesday, is blonde, has freckles,
got an A+ on his math paper is irrelevant to the gender of the second son. The
author's entire reasoning misses the entire point of the Monty Haul paradox
where Monty Haul specifically reveals a door known _not_ to have the Goat,
thereby immediately modifying the probabilities of the remaining door.

In this "Tuesday Boy Problem" - no such selection takes place. There is no
paradox. There is a 50% chance that the other child is a boy and a 1/7 chance
that they were born on a tuesday (or a monday, wednesday, etc...).

Note - The wikipedia article on this topic does a much better job discussing
the ambiguity involved in asking the question -
<http://en.wikipedia.org/wiki/Boy_or_Girl_paradox>.

The sad part of this, is that if you had _selected_ a family in which at least
one son was born on a tuesday, then you would have modified the probabilities
of the other child being born a son - but no such selection was done here,
therefore no probabilistic impact on the chance of the other child being a
son.

~~~
_dps
There is a shorthand and implied understanding in mathematical word problems
(else they would be formulas!). As soon as you ask "what is the probability"
you are implying either a sampling/generative process, or a subjective (e.g.
Bayesian) probability framework. I think it's clear this is the former case.

The generative model implied for each child is: pick uniformly from boy/girl,
and pick uniformly from Mon-Sun. The generative model for the father is:
generate two children.

This generative process has a well-defined outcome distribution and it is
completely reasonable to ask "what is the probability of generating an outcome
with two boys, given that you generated an outcome with a Tuesday boy?"

------
bpm140
Wait. What? The author says there can only be one BB but that there can be
both GB and BG?

What he's doing here is saying that birth order functionally doesn't matter if
the sibling is a boy, but it _does_ matter if it's a girl. How is this
correct?

If you keep comparing apples to apple you get:

    
    
      Older Boy / Boy
      Boy / Younger Boy
      Older Girl / Boy
      Boy / Younger Girl
    

And we're back to a 50% chance that the other child is a boy.

~~~
suivix
Since he has two children, you know first off that these are all equal chance
historically:

BB BG GB GG

Now since you know the child is a boy, it eliminates the fourth option. So now
we have these possibilities:

BB BG GB

In only one of these is the other child a boy, so the chance is 1/3.

~~~
bpm140
This explanation makes a ton of sense. I get it now. Thank you.

~~~
suivix
Also I don't think the exact day of birth being Tuesday makes one bit of
difference to the gender. For example, if you were to say the known boy was
born crying, and the chance of this is 70%, it doesn't affect the gender of
the other children. It's just useless trivia.

~~~
jules
That's not true. It does affect the probability of the other child being a boy
just like being born on Tuesday does.

------
brutooths

      A small simulation in Python
    
      from random import randint
      oneBoy=0
      twoBoys=0
    
      # 0 = boy, 1 = girl
      for i in range(10000):
          s1 = randint(0,1)
          s2 = randint(0,1)
          if s1 * s2 == 0:  # Here is at least one boy
              oneBoy = oneBoy+1
              if s1 == s2:  # Here are two boys
                  twoBoys = twoBoys+1
      print("Two Boys")
      print(float(twoBoys)/float(oneBoy))
    
      # As expected the result is near 1/3.
    
      oneBoyT=0
      twoBoysT=0
      #days 0=Monday, 1=Twesday ...
      # (0,1) means (sex,day) = a boy was born one Twesday
    
      for i in range(1000000):
          s1 = randint(0,1)
          d1 = randint(0,6)
          s2 = randint(0,1)
          d2 = randint (0,6)
          if (s1,d1)==(0,1) or (s2,d2)==(0,1):
              oneBoyT = oneBoyT+1
              if s1 == s2:
                  twoBoysT = twoBoysT +1
      print("Two Boys in Twuesday")
      print(float(twoBoysT)/float(oneBoyT))
    
      # The simulation gives a result near 1/3, 
      # this is a hint to prove that 13/27 is incorrect
      # under the assumption that the population sex and
      # day are independent, and with probability 1/2 and 1/7
    
      # if you want to convince me otherwise, I'll be 
      # glad to see the code to generate the population
      # and the estimated probability.

------
adam-a
> I have two children. Child A is a boy, born on a Tuesday, what is the
> probability that child B is a boy?

We have no knowledge of child B so the probability is even for it being a boy
or a girl.

> I have two children, at least one of them is a boy who is born on a Tuesday.
> What is the probability that I have two boys?

We have incomplete knowledge of both children, we are constraining a
probability space in 2 dimensions (i.e. the 2 children) rather than refining
it down to 1 dimension as before. In this case the probability is 13/27 as
these tables (<http://news.ycombinator.com/item?id=3290349>) nicely show.

When the father says "one is a son born on a tuesday" this can be read as him
selecting a child and then asking you about the second or as giving you some
information that can apply to either child and then asking you about both.

In my opinion the language used leads much more easily to the first
interpretation, this is the reason surely for the confusion.

------
drblast
The crux of the issue is much better illustrated by the simplified problem
where if a man tells you one of his kids is a boy, the other child is a boy in
1/3 cases. That's intuitive looking at the possibilities:

BG GB BB

Where this analysis gets confusing is when you say that if you meet the man
with one of his kids at random and it's a boy, the probability changes to 1/2.
The reason is in that case one of the possible outcomes was indeed GG, even if
the child you met was a boy. The difference is how much knowledge you had
about the situation prior to making your measurement of the outcome.

If that doesn't make your head spin even a little you're not really human.

I was able to make peace with these problems when I read about the Two
Envelopes Paradox and realized that it's not really possible to pick a random
integer, because by picking a random integer you're essentially limiting the
domain of integers you've picked from to a finite one. If you can wrap your
head around that it might help you save your sanity.

------
hwiechers
I finally understand it.

Here are my events (taking 2 children as given):

    
    
      B = I have a boy
      BT = I have a boy and he was born on Tuesday
      2B = I have 2 boys
      B+G = I have a boy and a firl
    

Using Bayes' Theorem:

    
    
      Pr(2B|BT) = Pr(BT|2B)Pr(2B)/c
      Pr(B+G|BT) = Pr(BT|B+G)Pr(B+G)/c
      where c = Pr(2B|BT) + Pr(B+G|BT)
    
      Pr(BT|2B) = 1/7 + 1/7 - 1/49 = 13/49
      Pr(2B) = 1/4
      Pr(BT|B+G) = 1/7
      PR(B+G) = 1/2
    

So

    
    
      c = 13/49 * 1/4 + 1/7 * 1/2 = 27/49
      Pr(2B|BT and 2C) = (13/49 * 1/4) / (27/49) = 13/27
      Pr(B+G| BT and 2C) = (1/7 * 1/2) / (27/49) = 14/27
    

They main reason Pr(2B|BT) is around 50% is that Pr(2B|BT) is proportional to
Pr(BT|2B) while Pr(B+G|BT) is proportional to Pr(BT|B+G) and Pr(BT|2B) is
about 2 * Pr(BT|B+G).

It's much more likely you have a boy born on Tuesday given that you have 2
boys (rather than you have a girl and a boy) so it's more likely that you have
2 boys given that we know you have a boy born on Tuesday.

------
tzs
P(two boys, at least one on tuesday) = 1/4 (1 - 6/7 6/7). The 1/4 is the
probability of two boys, and the 1 - 6/7 6/7 is the probability at least one
is born on tuesday.

P(one boy, one girl, boy on tuesday) = 1/2 * 1/7. The 1/2 is the probability
of having one boy, one girl, and the 1/7 is the probability that the boy was
born on tuesday.

The man can make his statement if either of the above is true, and they are
disjoint, so their sum gives the probability that he has two children, one a
boy born on tuesday.

The ratio of the first to the second is the probability that both are boys.
Plug in the numbers and you indeed get the 13/27 the article claims.

------
james1071
p(2s\T)=p(2s & t)/p(T)

p(2s & T)=1/4 _(13/49)

p(T)=1/4_(13/49)+1/2*(1/7)

p(2s\T)=(13/49)/(13/49+14/49)

p(2s\T)=13/27

------
wylie
To answer the question posed by the title: no, I didn't think I understood
Monty Hall, but thanks to this article I'm pretty sure I do now. This is a
good explanation and analysis.

------
erikb
If you get that far down in the comments, you should also read why I think all
solutions given here and by the author are wrong:
<http://news.ycombinator.com/item?id=3291009>

------
raldi
This post would have been much better if the author had used more precise
language. I still don't know whether the assertion being made by the father is
that _at least_ one of his children is a son born on Tuesday, or _exactly_ one
is.

------
madamepsychosis
Scott Aaronson has a really good article about the anthtopic principle and
conditional probability.

<http://www.scottaaronson.com/democritus/lec17.html>

------
invertd
How would the solution change if the problem was stated as follows: 'You meet
a man on the street and he says, “I have two children and one is a son born in
March.” What is the probability that the other child is also a son?'

~~~
raldi
Or, "...one is a son born in an even-numbered month"

------
hermannj314
I have two children. One of my children is a son born on a Tuesday. Of course,
my wife is still pregnant with the 2nd, but she will be thrilled to know there
is a better than average chance she is having a girl!

------
vjk2005
"If you surveyed all such families, you would find that roughly 13/27 of them
have two boys." _roughly_ 13/27? How rough are we talking, maybe it's roughly
13/26 instead.

~~~
hexagonc
The degree of "roughness" depends on the number of families surveyed. The more
there are the closer to 13/27 you'll get.

------
twelvechairs
> Now suppose we know simply that a man has two children and that one of them
> is a son. [snip] It follows that the sexes of his two children, ordered from
> oldest to youngest, are either BB, BG or GB. Since these cases are equally
> likely, and since only one of them involves having two boys, we would say
> the probability that the man has two boys is 1/3.

Errr. except these cases aren't equally likely - because the boy we know about
could be either of the boys in the BB scenario, but only one either of the
other two. So the probability is still 1/2.

~~~
incremental
Since the constraint simply says that at least one child is a boy, we don't
need to distinguish between two types of BB.

This is a curious misconception, by the way. The typical failure mode I see on
these questions is people having difficulty accepting BG and GB as different
possibilities.

~~~
twelvechairs
If we are determined to look at order of birth (ie. have a BG and GB) then we
should consider all cases by order of birth, so where 'B' represents the boy
we know and 'b' or 'g' represents a child we dont, and the first character
represents the first child, and the second the second childe - we have:

Bb, bB, Bg, gB

50%.

~~~
incremental
Imagine a similar problem but with red and blue poker chips. Say, for example,
that I have a bag and I pull out two chips, one at a time.

In this problem, would you still try to distinguish the two identical red
poker chips using your logic?

~~~
twelvechairs
Okay. To respond to both of you - I have put this in programming just to
check, and I think you are taking a different inference from the question than
I am.

Under your inference, the man wouldnt have mentioned anything unless he had at
least one male child. (in which case you can say the GG scenario is gone, but
GB BG and BB are equally probable)

Under my inference, the man just told me the sex of one child at random....
(in which case BB is twice as probable as GB or BG - where he could have
equally said 'i have at least one girl')

That sound reasonable to you? I have it in ruby form if you are interested :)

The blogpost linked to by someone above
(<http://blog.tanyakhovanova.com/?p=221>) uses this explanation, which is very
different (to me) than the one in the main link, and I can see why this gets
to 1/3:

"A father of two children is picked at random. If he has two daughters he is
sent home and another one picked at random until a father is found who has at
least one son."

~~~
incremental
"Under your inference, the man wouldnt have mentioned anything unless he had
at least one male child. (in which case you can say the GG scenario is gone,
but GB BG and BB are equally probable)"

No. I'm just assuming he has two children, and randomly mentions something
about one of them. The GG scenario is only eliminated _after_ he makes his
statement, because we then know he has at least one boy.

~~~
twelvechairs
Ah, then I do think you have an error of logic.

Put it this way. before he says it, we have GG, GB, BG, BB

after he says "I have a child that is [MALE OR FEMALE]" we have (where the
capital letter is the child whose sex has been mentioned, and the lowercase
letter is the other child):

Gg, gG, Gb, gB, Bg, bG, Bb, bB

So if he has said the sex is male, then we have four combinations left:

gB, Bg, Bb, bB.

Understand that we go to more scenarios (8) based on which child is mentioned,
before we go to fewer. Actually the order of the children is something you can
and should ignore, however as you are holding on to it, I show it this way....

~~~
incremental
I'm afraid this is wrong - you shouldn't distinguish between Bb and bB. In
this problem, they are not different states, so counting them messes up your
probability calculation.

Someone had a nice link higher up:
[http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-
proble...](http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-problem-in-
under-300-words.html)

------
IanDrake
The last Mythbusters episode put Monty Hall to the test and passed. I have to
say it was the first time I truly understood why Monty Hall works.

------
Rhapso
I think you a few words. I have only 1 son who was born on a Tuesday. Basing
this on ambiguity of phrasing is not novel, or interesting.

------
ashishb4u
I have a son born on xDay. What is the probability that my next kid will be a
son?

I dont get Monty Hall :(

~~~
pkteison
The Tuesday Birthday problem is related to you not having info about the order
leading to extra possibilities that you don't intuitively consider. Like in
the simple example, if one son is a boy, probability of other being a boy is
1/3rd because the info you have the sets available could be BB, BG or GB. But
if the younger son is a boy, then the GB possibility is removed, so it's 1/2
on the older son - BB or BG.

In your case, your younger kid is a son born on xDay. If you assume you will
have another kid, chances on being a son are still 50%. It only gets
interesting if either kid could have been a son.

In Monty Hall, the info that you don't intuitively consider is that Monty is
providing additional information. You think of it as 50/50 because Monty ruled
out a goat and car has to be behind one of two remaining doors, but actually
the way it works is you chose a door that 1/3rd had a car, leaving 2/3rds
chance of car on the other two doors. Then Monty eliminated one of those two
doors, still leaving 2/3rds chance of car on the remaining door. So you should
switch to that door. In Monty Hall, first you divide the set into a 1/3rd
chance group of 1 door and a 2/3rd chance group of 2 doors, then Monty makes
the second group a 2/3rd chance group of just 1 door.

------
shasta
Spoiler: Tuesday was the name of his horse.

------
its_so_on
ANYONE WHO SAYS THE ANSWER IS OTHER THAN 1/2:

Let's simplify the question:

If a man says "I have two children, one was born on a Tuesday," what is the
probability that they are both born on a Tuesday?

Is the answer to this 0 or 1/7 in your opinion? (Or something different).

~~~
danparsonson
You can follow a similar process to work out the probability as for the
original question. For any two children, the possibilities are:

Child 1 born Mon, Tue, Wed, Thu, Fri, Sat, Sun

Child 2 born Mon, Tue, Wed, Thu, Fri, Sat, Sun

We know that one of the children was born on a Tuesday, but we haven't said
whether it's Child 1 or Child 2 (and that lack of information is key to
understanding the original problem), so our unknown child could also be either
Child 1 or Child 2.

With that in mind, 'unknown child' has 14 possible options except that we know
one of those options (our known Tuesday child) is already taken - we don't
know whether it's Child 1 or Child 2 but that doesn't matter, we just know
that one of them is already taken. That leaves 13 possibilities for unknown
child, only one of which is 'born on Tuesday' (since we've removed the other
'born on Tuesday' option), so I would say the answer is 1/13.

~~~
its_so_on
This is ridiculous. Let's do some queries on the world:

From all women: (some 3 billion)

the ones who have exactly two living children: x matches

the ones in which it is true for them to say "One of my children was born on
Tuesday.". (this is true or false for every mother that comprises x. y mothers
remain (can answer "true" for that question")

the ones that can say "both my children were born on Tuesday": z matches

are you telling me that sizeof z is 1/14th the sizeof y???

(because for me it is either 1/7 that size or exactly 0 if the correct meaning
of "one of" is "exactly one of and not both of")

~~~
its_so_on
Wait, I think I get it. (Sorry, too late to update).

This is indeed ingenious. The key is the step from x to y. The women in x who
can say "one of my children was born on Tuesday" are the ones who can say
"EITHER my first OR my second child was born on Tuesday"; the initial
constraint (when you meet the person) is thus: "women who can say I have
exactly 2 children and it is true to say EITHER my first OR my second child
was born on a Tuesday"; obviously this is far more than 1/7th. Then the
additional constraint "BOTH of them were" is a smaller addition than 1/7th.

It's not 1/7th the first time and 1/7th the second time, because the question
wasn't "of women who can say they have exactly two children, the number who
can say 'my elder child was born on a Tuesday'" and then "of these the women
who can say 'my younger child was also born on Tuesday". This would indeed be
1/7th each time.

instead y is "EITHER my elder OR my younger child was born on Tuesday". This
obviously results in a set that is greater than 1/7th of all women with two
children, and, consequently, it is no surprise that there is a correspondingly
smaller than 1/7th possibility that both were.

~~~
danparsonson
Bingo :-) I was just racking my brain for a different way to explain it but
luckily you got there first!

It's a real mind-bender.

