Tuesday’s Child
I came across this little teaser this morning and thought I’d share it here.
I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?
Please select an answer from the possibilities listed in the poll below.
This is not a new problem and you can probably find the answer on the internet very quickly, but please try to work it out yourself before doing so. In other words, try thinking before you google! I’ll add a link to a discussion of this puzzle in due course..
UPDATE: Here’s the discussion that triggered this post. As you can see from the poll, most of you got it wrong!
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In the Dark 
July 5, 2013 at 11:38 am
I prefer the version that says that there are two children, one of whom is a girls called Agatha – what is the probablity that the second is a girl?
July 5, 2013 at 11:46 am
Or indeed there are two children, one of whom is a boy named Sue…
July 5, 2013 at 1:28 pm
I have heard this before, but went through the explicit calculation to see the ambiguity.
It seems to all hinge on whether I consider you a random person from the set of all people with two children, and then I learn you have a boy born on a Tuesday, or if I consider you a random person from the set of all persons with two children one of whom is born on a Tuesday.
July 5, 2013 at 3:58 pm
Is five a random number?
July 6, 2013 at 12:21 pm
This version doesn’t quite work. The statement can be interpreted as identifying the child, which reduces the chance that the other child is a boy born on that day. Even if it is probably not meant that way, the probably still affect the probability. It is not a problem when giving the name of the child as the name is chosen by the parent, not by chance.
July 6, 2013 at 2:43 pm
Are you saying that the problem is ill-specified, or that it is more difficult than it at first seems?
July 6, 2013 at 3:18 pm
The problem as given has an answer but it is a range rather than a number. If you want more precision than [6/13 , 0.5], more information would be needed.
July 6, 2013 at 5:04 pm
Then you are saying that it is ill-specified (taking that description to mean that there is no unique answer). I don’t understand where, so let me ask: Do you think that the following problem, which I believe is what Peter meant, is ill-specified?
“I have two children, one ONLY of whom is a son-born-on-a-Tuesday. What is the probability that I have two boys?”
July 8, 2013 at 2:05 pm
Peter, I find the discussion that you have linked to confusing and believe it is actually wrong. The sentence in it, “Everything depends… on why I decided to tell you about the Tuesday-birthday-boy” is better phrased as “your (conditioning) information is ambiguous.” If you have an unambiguous statement (ie, unambiguous conditioning information) then you can always assign a definite probability. I also think that the grid diagram showing the 100+ possibilities as squares is confusing. Its caption says that “it shows all the possible variations of birth day and gender for children in a two-child family”, but there are 14×14=196 such possibilities, and the diagram has twice that number of squares.
Let me rephrase the problem a bit legalistically to avoid ambiguity:
You overhear a woman say that she has two children, and that exactly one of them falls in the category “boy born on the first day of the week”. What probability do you assign that she has two boys? Make the ‘normal’ assumptions: that boys and girls are equally likely on every day, and that birth is equally likely on any day of the week (which is not true if doctors are inducing potentially problematic imminent births on a Friday so that they can have a quiet weekend); and that she has been pregnant twice, once with each child only (so that they are not twins).
I’d draw a diagram as follows. First, you need a label for the two children. (They are not bosons or fermions!) The obvious choice is younger and older (although anything will do, eg shorter and taller). Construct a 2×2 grid for gender, having the two possibilities (Boy or Girl) for the Younger child along the top and the two possibilities for the Elder down the side. Now subdivide each of the resulting four squares into a 7×7 grid for the days of the week. You end up with 4×49=196 small squares. There is a vertical line of 14 small squares corresponding to the younger child being a boy born on the first day of the week, and a horizontal line of 14 small squares corresponding to the elder child being a boy born on the first day of the week. All other 13×13=169 small squares correspond to neither child being a boy born on the first day of the week. Along those two lines, 13+13=26 small squares correspond to just one child being a boy born on the first day of the week, and one small square (where the two lines intersect) corresponds to both children being boys born on the first day of the week. Our conditioning information now restricts us to those 26 small squares. Of them, 6+6=12 correspond to two boys. So the answer is 12/26, or 6/13, which means that I’d have ticked Peter’s last box (“none of the above”). The answer of 13/27 given in the linked discussion corresponds to the problem with the phrase “AT LEAST one falls in the category…” but the normal meaning of the phrase “one falls in the category” is “EXACTLY one falls in the category”. That is correctly assumed in the discussion at the link, so I disagree with their answer.
You can do it instead by starting with a 7×7 grid for the days of the week, and subdividing each of the resulting 49 squares into a 2×2 grid for gender, giving the same 4×49=196 small squares but reordered. (You get the same answer, of course.) That is closer to the diagram in the link. But I have no idea why they drew not 196 but 392 small squares, ie twice as many.
What is the algebra corresponding to this geometric picture? In more complicated problems in which you cannot draw a picture, you will need an algebraic way to proceed. Only Bayesians who understand the relation between probability and the Boolean calculus of propositions can do this. We want the probability of truth of the proposition E(B)Y(B), where E and Y denote Elder and Younger child, and B denotes Boy. (Write also E(3), etc, as the proposition that the elder child was born on the 3rd day of the week.) Our key conditioning information is that exactly one child is a boy born on the first day of the week, ie that the Boolean proposition E(B)E((1)~(Y(B)Y(1)) + ~(E(B)E(1))Y(B)Y(1) is true. This corresponds to the two ways in which exactly one child is a boy born on day-1: in the first way, the elder is and the younger isn’t, and vice-versa in the second way (exclusive OR). Here, the negation operator ~ operates only on what is immediately to its right (overlines are clearer but are not possible here). We also have conditioning information Z that everything else is as you expect.
You now have to use the sum and product rules, and de Morgan’s theorem working on the propositions, to get the answer in terms of probabilities of the ‘elementary’ propositions E(B), E(1) etc conditioned on Z alone. These are probabilities we know: p(E(B)|Z)=1/2, p(E(1)|Z)=1/7, etc. It is quite a lengthy manipulation because of the logical sum in the conditioning information, and this is what was meant in the link by the phrase “the difficulty of these problems is rooted in their artificiality”. The logical sum must be brought across the conditioning solidus using the product rule (and/or Bayes’ theorem). Once it is all done, the answer is 6/13.
I cannot overemphasise how important it is to understand that the arguments of probabilities are Boolean propositions, and how to manipulate the propositions and the probabilities using Boolean algebra and the sum and product rules. Unless you can do this, there are whole classes of problems that you will not be able to solve when they get a bit more complex than this one.
July 8, 2013 at 2:27 pm
I googled for 6/13 in relation to this problem and found this 3-year-old page:
http://www.freerepublic.com/focus/bloggers/2543558/posts
Here, ‘djf’ cracks it in very few words: “Whatever the other child is, it is NOT a son born on Tuesday… It could be a son born on one of the other six days of the week. Or it could be a daughter born on any of the seven days of the week. So that’s 13 possible outcomes. Six of those outcomes leads to two sons… the answer is 6/13.” And the next contributor (KarlInOhio) correctly says “13/27 if you mean at least one boy born on a Tuesday. 6/13 if you mean one and only one boy born on a Tuesday.”
July 8, 2013 at 2:44 pm
Anton,
I linked to the discussion because that’s what happened to spark off the post, not because it was very clear. Links are not necessarily endorsements!
However, I read the statement that “one is a son born on a Tuesday” as saying nothing about the other child at all, so I went for the answer 13/27…
Peter
July 8, 2013 at 3:27 pm
I’m not ranting at you Peter, I’m ranting at the material in the link!
July 12, 2013 at 10:15 am
“You overhear a woman say that she has two children, and that exactly one of them falls in the category “boy born on the first day of the week”. What probability do you assign that she has two boys?”
I assign a probability of 1/2 that she has 2 boys, on the basis that I’m equally likely to overhear the woman talking about her other child (the one not born on the 1st day of the week).
I guess the conditioning information in your rephrased problem is ambiguous too 🙂
July 8, 2013 at 2:36 pm
Here is Wikipedia on this stuff
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
The article includes the totally misleading comment that “The moral of the story is that these probabilities don’t just depend on the information we have in front of us, but on how we came by that information.” That is absolute nonsense. To definite conditioning information there always corresponds a definite probability assignment. Wikipedia should have said that the conditioning information is ambiguous (leading of course to differing probability assignments). This reminds me of some of the frequentist nonsense that arises in “optional stopping” problems in which the frequentist assignment depends on why you stopped gathering data where you did, but the Bayesian assignment doesn’t (as of course it shouldn’t – only on the data).
July 9, 2013 at 8:54 am
Yes, the hard part is turning what you have learned into binary propositions and making sure that you have not left anything relevant out. Once that is done, you should be able to assign a definite probability. Think of it as an interrogation in which the subject may answer only Yes or No, eg “Do you have two children?” If Yes, then “Were you pregnant twice, once with each only?” If Yes, then “Is any of them a boy who was born on a Tuesday?” But note that this is different from asking “Is any of them a boy?” followed if Yes by “Was he born on a Tuesday?” For if she has two boys then saying “he” is ambiguous to her. So it is not the same as seeing her with a boy on the street and asking her about the birth day of THAT boy. Good subtle stuff.
You can gain insight in the present case by considering not what day of the week a child was born, but whether a child was born at night or in the day. (Assume 12-hour days and nights.)
August 1, 2013 at 1:24 am
Actually, most got it right. Just like the Monty Hall Problem, the correct solution requires you to recognize that whoever gave you the information may have had a choice between two options.
While 13/27 of all fathers-of-two with a Tuesday Boy will have two boys, 1/2 of all such fathers who *tell* you about a Tuesday boy have two. The difference is because some fathers-of-two with a Tuesday Boy tell you they have a Thursday Boy, or a Friday girl, so you can only count half of the fathers who only have one Tuesday Boy.
September 12, 2013 at 4:40 pm
A side note … in optional stopping problems the amount of data gathered is also an observable. In order to assess its likelihood you need conditioning information, ie the probabilities of stopping at any point given the existing data. Bayesian statistics won’t give the right answer unless you know the stopping criteria.