The Tuesday boy problem.
riderofgiraffes
causing some division and argument. Originally posed
by Gary Foshee.
Suppose I have two children, and one of them is a boy
born on a Tuesday. What is the probability that both
my children are boys?
Don't be too quick to react. Just as with Monty Hall
some well-known and well-respected mathematicians are
quite definite, it's just that they don't all get the
same answer.
dannas
"riderofgiraffes" <mathforu...@solipsys.co.uk> wrote in message
news:560624655.228654.12749...@gallium.mathforum.org...
>A relative of the Monty Hall problem, and currently
> causing some division and argument. Originally posed
> by Gary Foshee.
>
> Suppose I have two children, and one of them is a boy
> born on a Tuesday. What is the probability that both
> my children are boys?
tuesday is immaterial.
being born is immaterial.
the paragraph below is immaterial.
danh...@yahoo.com
wrote:
I claim the correct answer is 1/3 (with the usual assumptions of
independence and P{boy}=1/2).
The formal way to get the answer is with Bayes' theorem. The easy way
is to start with the four a priori equally likely events (in birth
order)
BB BG GB GG
The last one is ruled out and the other 3 are still equally likely.
Note that if we're told he is the older child, the answer is 1/2
because the second event is also ruled out.
hagman
wrote:
All depends on how the information was obtained:
a)
"Dear Mr. X. I hear you have two children. Is at least one of them a
boy born on a Tuesday?" - "Yes" - "Thank you"
b)
"Dear Mr. X. I hear you have two children. Is at least one of them a
boy?" - "Yes" - "Would you please tell me the weekday of his (or a
randomly selected boy's) birth?" - "Tuesday" - "Thank you"
c)
"Dear Mr. X. I hear you have two children. Would you please tell me
the sex and weekday of one of them?" - "(flips coin as he loves them
both the same) I say: boy and Tuesday" - "Thank you"
d)
"Dear Mr. X. I hear you have two children. Is at least one of them
born on a Tuesday?" - "Yes" - "And what is it?" - "A boy" - "Thank
you"
Moreover, for a question answered "Yes" in a scenario above, let us
assume that the interviewer walks from door to door until he finds
someone answering "yes"
For a) the probability of "two boys" is 13/27 if I'm not mistaken.
For b) it is 1/3.
For c) it is 1/2
For d) it is 1/2
For the curious, all those numbers are of the form (2-p)/(4-p) with p=
1/7, p=1, p=0, respectively.
Especially with c) and d) one sees that the interviewer obtains no
relevant information that would influence the odds of a bet on "both
children have the same sex".
I suspect that there are still more interpretations possible beyond a)
and d).
hagman
W^3
<560624655.228654.12749...@gallium.mathforum.org>,
riderofgiraffes <mathforu...@solipsys.co.uk> wrote:
My interpretation is to consider the collection of 4-tuples (s1 d1 s2
d2), where s1, s2 lie in {b, g} (the sexes of the children), and d1,
d2 lie in {1, 2, ..., 7} (the days of the week these children are
born). All of these variables should be independent, so each 4-tuple
has probability 1/2*1/7*1/2*1/7.
Let E be the event {s1 = b, d1 = 3} U {s2 = b, d2 = 3}. The
probability we seek is P({s1 = b, s2 = b} intersect E)/P(E). I get
13/27 for this.
Rob Johnson
riderofgiraffes <mathforu...@solipsys.co.uk> wrote:
I believe that implicit in the question (whether actually true or
not) is that a boy is as likely as a girl and that no day of the
week is any more likely for a boy to be born than any other.
To make sure we have disjoint cases, let's divide the possibilities
of having two children, one of them a Tuesday Boy, as follows:
Girl & Tuesday Boy = 7/14 * 1/14 = 7/196
Tuesday Boy & Girl = 1/14 * 7/14 = 7/196
Non-Tuesday Boy & Tuesday Boy = 6/14 * 1/14 = 6/196
Tuesday Boy & Non-Tuesday Boy = 1/14 * 6/14 = 6/196
Tuesday Boy & Tuesday Boy = 1/14 * 1/14 = 1/196
Thus the probability of a Tuesday Boy is 27/196.
The probability of a Tuesday Boy and two boys is 13/196.
Thus, given two children, one a Tuesday Boy, the probability of two
boys is 13/27.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
master1729
probability = 21/40
Tim Little
> Suppose I have two children, and one of them is a boy
> born on a Tuesday. What is the probability that both
> my children are boys?
There are numerous ways to interpret that information, giving
different answers. The question is ambiguous (as is the usual
statement of the Monty Hall problem).
For example:
"I have two children. One of them (coin flip) is a boy born on
Tuesday." In this case the event space has five independent uniform
variables: Sex1, Day1, Sex2, Day2, Flip. Of the 392 possibilities,
the given information excludes all but 28, 14 of which result in two
boys. So p = 1/2.
"I have two children. One of them is a boy born on (flip if there are
two boys) Tuesday." This time the coin flip is contingent: it is only
performed at all in 1/4 of the event space. The provided information
leads to p = 1/3.
"I have two children. (At least) one of them is a boy born on
Tuesday." In this case there was never a choice to be made and only 4
independent variables. The provided information rules out all but 27
of the equally-likely events, of which 13 have two boys. Then
p = 13/27.
There are even more possibilities, e.g. if you happen to be more
likely to inform me about the sex and day of birth of one of your
children than the other by some unknown margin, and so on.
- Tim
jmor...@idirect.com
I find it puzzling that some of the answers depend on the use of a
particular calendar...
Suppose the information was that there was a boy born on a
Tuesday...in July. Now do you use 7, or 12 to create your fractions?
Oh, and it was also a Tridi in Thermidor. The probabilities keeps
changing, and yet the two kids are still standing there, wondering how
likely they are...
Tim Little
> I find it puzzling that some of the answers depend on the use of a
> particular calendar...
That because some of the assumptions depend on the use of a particular
calendar...
> Suppose the information was that there was a boy born on a
> Tuesday...in July. Now do you use 7, or 12 to create your
> fractions?
Neither. In the first two of my sets of assumptions, the calendar is
irrelevant. In the third, you would additionally have to assume a
probability distribution for their ages and apply that to the number
of Tuesdays in July in those years. If the current date was not given
then I suppose it would be reasonable to assume a uniform distribution
over the whole Gregorian calendar cycle. There may or may not be a
1/84 chance of being born in a Tuesday in July over that distribution.
> Oh, and it was also a Tridi in Thermidor. The probabilities keeps
> changing, and yet the two kids are still standing there, wondering
> how likely they are...
Calculations applied to models always depend upon the assumptions of
the model. When a hypothetical problem omits most of the assumptions,
and there are many reaonable ones, the problem is ambiguous.
- Tim
Richard Tobin
Tim Little <t...@little-possums.net> wrote:
>If the current date was not given
>then I suppose it would be reasonable to assume a uniform distribution
>over the whole Gregorian calendar cycle. There may or may not be a
>1/84 chance of being born in a Tuesday in July over that distribution.
1722/146097
-- Richard
John L. Barber
During each iteration, I "construct" two random, independent "children". Each child has a sex randomly chosen as either 0 or 1, with equal probability. Each child also has a birth day randomly chosen as either 0, 1, 2, 3, 4, 5, or 6, with equal probability.
I define a "Tuesday Boy" as a child with sex = 0 and birth day = 0.
I keep track of Ntb, the total number of pairs so far which have included at least one Tuesday Boy, as well as Nbb_tb, the number of pairs containing at least one Tuesday Boy which also contain two boys.
I hope all of that is clear enough.
The results: After 1,138,989 randomly-selected pairs of children, I have
Ntb = 156638
and
Nbb_tb = 75459
So, the probability requested in the OP is approximately:
75459/156638 = 0.481741
(Note that 13/26 = 0.481481.)
jmor...@idirect.com
>>>> (Note that 13/26 = 0.481481.)<<<<<<
Let's not make things more confusing than they have to be :>
Ostap Bender
Here is more information for you: I have two children, and one of them
is a boy born on a Tuesday who has brown hair, reads comic books, and
is left-handed.
Is that going to change your probabilities?
Ostap Bender
Wow. That's nostalgic. I remember reading about this paradox in one of
the popular math books (Gardner?) in 7th grade. Only a year later did
I realize that the reason for this paradox is the way the storyteller
chose which child to tell us about.
Ostap Bender
How much did you pay for your calculator? You overpaid.
master1729
13/26 = 0.5 according to me.
john overpaid.
maybe we should abolish fractions now ? :p
John L. Barber
> >
> > How much did you pay for your calculator? You
> > overpaid.
>
> 13/26 = 0.5 according to me.
>
> john overpaid.
>
> maybe we should abolish fractions now ? :p
Dammit. I edited that post to read "13/27 = 0.481481" moments after I posted it.
How come no one can see the correction but me?
Rob Johnson
I don't know how mathforum works, but you cannot correct a message
that has been posted to a large number of NNTP servers; you have to
post a correction.
I took a look at mathforum and I see that the mathforum post has
been edited, but that only affects the mathforum post:
<http://mathforum.org/kb/message.jspa?messageID=7085258&tstart=0>
John L. Barber
> correct a message
> that has been posted to a large number of NNTP
> servers; you have to
> post a correction.
OK, I've done so.
Thanks very much!
John L. Barber
Of course I meant to type 13/27...
Jesse F. Hughes
> In article <871895705.265860.12754...@gallium.mathforum.org>,
> "John L. Barber" <jlba...@lanl.gov> wrote:
>>> > > (Note that 13/26 = 0.481481.)
>>> >
>>> > How much did you pay for your calculator? You
>>> > overpaid.
>>>
>>> 13/26 = 0.5 according to me.
>>>
>>> john overpaid.
>>>
>>> maybe we should abolish fractions now ? :p
>>
>>Dammit. I edited that post to read "13/27 = 0.481481" moments after I posted it.
>>
>>How come no one can see the correction but me?
>
> I don't know how mathforum works, but you cannot correct a message
> that has been posted to a large number of NNTP servers; you have to
> post a correction.
In fact, Usenet was built with the ability to supersede posts in order
to correct errors, by canceling the old post and replacing it with the
new. Because it was easy to forge cancels (and hence supersedes), few
servers honor cancel requests these days.
Of course, that probably has little to do with mathforum's edit
functionality. Who knows how they implemented that and whether they
bothered trying to propagate it to Usenet?
--
Jesse F. Hughes
"If the world weren't rather strange, by now I should at least be with
some research group talking about my number theory research."
-- James S. Harris learns the world is a funny place
Joshua Cranmer
> Of course, that probably has little to do with mathforum's edit
> functionality. Who knows how they implemented that and whether they
> bothered trying to propagate it to Usenet?
Doesn't look like it. My newsserver accepted cancels and Supersedes when
I last checked (I do recall this because there was a troll superseding
someone's posts with vulgar drivel), and I see a 13/26.
Or it could be that one of these servers doesn't propagate such messages:
pitt.edu
news.cse.ohio-state.edu
news.glorb.com
news2.glorb.com
news.glorb.com
tr22g12.aset.psu.edu
news.mathforum.org
--
Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth
Ostap Bender
Your mommy must be very proud of you!
>
> john overpaid.
Isn't this what I said?
> maybe we should abolish fractions now ? :p
Better abolish typos.
Ostap Bender
Have you ever read Andersen's The King's New Clothes?
Rob Johnson
Joshua Cranmer <Pidg...@verizon.invalid> wrote:
>On 06/02/2010 01:29 PM, Jesse F. Hughes wrote:
>> Of course, that probably has little to do with mathforum's edit
>> functionality. Who knows how they implemented that and whether they
>> bothered trying to propagate it to Usenet?
>
>Doesn't look like it. My newsserver accepted cancels and Supersedes when
>I last checked (I do recall this because there was a troll superseding
>someone's posts with vulgar drivel), and I see a 13/26.
>
>Or it could be that one of these servers doesn't propagate such messages:
>pitt.edu
>news.cse.ohio-state.edu
>news.glorb.com
>news2.glorb.com
>news.glorb.com
>tr22g12.aset.psu.edu
>news.mathforum.org
I tried Supersedes several years ago and it failed to do what it was
supposed to. I had assumed that it had been phased out due to the
dangers it posed, but perhaps it was simply that the server I used
didn't support it.
Bastian Erdnuess
> I'd just like to mention that I've just done a small simulation of
> this process. [...]
> I keep track of Ntb, the total number of pairs so far which have
> included at least one Tuesday Boy, as well as Nbb_tb, the number of
> pairs containing at least one Tuesday Boy which also contain two boys.
How do you see from the OP that you really get informed about the
Tuesday Boy in anycase where there is a Tuesday Boy in the family?
Why should it be supposed to not happen that you have a pair say a
Sunday�Girl and a Tuesday Boy but you get only informed about that there
is a Sunday Girl in the family?
A similar question (with insufficient information) is the following:
"You throw two dice. Ana has a look at them and tells you that there is
a five within the dice. How big is the probability that you have thrown
doublets?"
Cheers,
Bastian
John L. Barber
>
> > I'd just like to mention that I've just done a
> small simulation of
> > this process. [...]
>
> > I keep track of Ntb, the total number of pairs so
> far which have
> > included at least one Tuesday Boy, as well as
> Nbb_tb, the number of
> > pairs containing at least one Tuesday Boy which
> also contain two boys.
>
> How do you see from the OP that you really get
> informed about the
> Tuesday Boy in anycase where there is a Tuesday Boy
> in the family?
It's not explicitly stated.
However, this type of problem is of a certain class of problems (the archetype being the Monty Hall problem) in which the same types of assumptions are always made.
The opening poster knew that the people on a forum like this one, being math-game-oriented folks, would be familiar with this kind of problem, in which the "you always get informed" assumption is always implicit.
In other words, it is assumed that the readers know the basic ground rules that invariably accompany such puzzles.
Tim Little
> Here is more information for you: I have two children, and one of them
> is a boy born on a Tuesday who has brown hair, reads comic books, and
> is left-handed.
>
> Is that going to change your probabilities?
For the three examples of fairly reasonable interpretations I posted
previously in this subthread: no, no, and yes respectively.
What do you expect with an ambiguous question?
- Tim
Ostap Bender
What do **I** expect? I expect most people here to be familiar with
the classic Martin Gardner paradox which I first read in 7th grade and
solved in 8th grade:
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
Boy or Girl paradox
The Boy or Girl paradox surrounds a well-known set of questions in
probability theory which are also known as The Two Child Problem[1],
Mr. Smith's Children[2] and the Mrs. Smith Problem. The initial
formulation of the question dates back to at least 1959, when Martin
Gardner published one of the earliest variants of the paradox in
Scientific American. Titled The Two Children Problem, he phrased the
paradox as follows:
* Mr. Jones has two children. The older child is a girl. What is
the probability that both children are girls?
* Mr. Smith has two children. At least one of them is a boy. What
is the probability that both children are boys?
Gardner initially gave the answers 1/2 and 1/3, respectively; but
later acknowledged[1] that the second question was ambiguous. Its
answer could be 1/2, depending on how you found out that one child was
a boy.
---------
Thus, the answer can be either 1/2 or 1/3, depending on how the answer
was selected. But under all reasonable interpretations, it cannot be
anything else, and it doesn't depend on the birth day, nor on left-
handedness, nor on any other properties of the reported boy.
Bastian Erdnuess
>> John L. Barber wrote:
>>
>> > I'd just like to mention that I've just done a
>> > small simulation of this process. [...]
>>
>> > I keep track of Ntb, the total number of pairs
>> > so far which have included at least one Tuesday
>> > Boy, as well as Nbb_tb, the number of pairs
>> > containing at least one Tuesday Boy which also
>> > contain two boys.
>>
>> How do you see from the OP that you really get
>> informed about the Tuesday Boy in anycase where
>> there is a Tuesday Boy in the family?
>
> It's not explicitly stated.
>
> However, this type of problem is of a certain class
> of problems (the archetype being the Monty Hall
> problem) in which the same types of assumptions are
> always made.
In the Monty Hall problem it is explicitly stated how
the quiz master chooses the first door to open. And
that's important!
> The opening poster knew that the people on a forum
> like this one, being math-game-oriented folks,
> would be familiar with this kind of problem, in
> which the "you always get informed" assumption is
> always implicit.
I think you missed the point about the "always get
informed" thing. I think it is easier to work out in
the other question I posted.
You could also have a look at hangman's posting and
ask yourself what makes you beleve that you are in
case (a).
> In other words, it is assumed that the readers know
> the basic ground rules that invariably accompany
> such puzzles.
And that's why he writes:
"Don't be too quick to react. Just as with
Monty Hall some well-known and well-respected
mathematicians are quite definite, it's just
that they don't all get the same answer." ?
Btw. even Martin Gardner (author of the Mr. Smith
problem, rip) agreed later that his question in the
original form was not unambiguous[1]. For the very
same reason Gary Foshee's question is not either.
Cheers,
Bastian
[1] see e.g.
<http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Ambiguous_problem_statements>
Ostap Bender
Wait! I didn't give it much thought, did I? So, your point is that
there can be more than two ways to interpret this? You are probably
right. I was wrong. Sorry.
But why would you say that it can depend on whether he is left-
handed, but doesn't depend on whether his hair is brown, or that he
reads comic books?
What if I told you that
Pr{hair=Brown} = 30%, Pr{hair=notBrown} = 70%
Pr{readsComix=True}=50%, Pr{readsComix=False}=50%
Pr{handedness=Left} = 10%, Pr{handedness=Right} = 90%
and all these r.v's are independent of each other and of the gender?
What makes the effect of handedness different from hair color?
Ostap Bender
wrote:
> A relative of the Monty Hall problem, and currently
> causing some division and argument. Originally posed
> by Gary Foshee.
>
> Suppose I have two children, and one of them is a boy
> born on a Tuesday. What is the probability that both
> my children are boys?
The answer depends on your priors. For example when you say "one of
them is a boy born on a Tuesday", you could mean different things:
1. At least one of them is a boy born on a Tuesday
2. At least one of them is a boy, and he was born on a Tuesday
3. The first child that I lay my eyes on, is a a boy born on a Tuesday
Etc.
I think the answers are 13/27, 1/3, and 1/2 respectively.
Bastian Erdnuess
>> > > Here is more information for you: I have two children, and one of them
>> > > is a boy born on a Tuesday who has brown hair, reads comic books, and
>> > > is left-handed.
>>
>> > > Is that going to change your probabilities?
>>
>> > For the three examples of fairly reasonable interpretations I posted
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>> > previously in this subthread: no, no, and yes respectively.
> But why would you say that it can depend on whether he is left-
> handed, but doesn't depend on whether his hair is brown, or that he
> reads comic books?
It's not what he is saying. The "no, no, yes" does not refer to "brown,
comic, left" but to the "three examples of fairly reasonable
interpretations" he postend in his OP.
Cheers,
Bastian
Tim Little
> But why would you say that it can depend on whether he is left-
> handed, but doesn't depend on whether his hair is brown, or that he
> reads comic books?
My apologies for being ambiguous myself! I meant that in the first
two interpretations, none of the three additional pieces of
information would affect the probability. In the third
interpretation, any or all of them would affect the probability.
- Tim