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When intuition and math probably look wrong
A twist on the Two Children Problem shows how information can steer what looks probable
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WORKING IT OUT
This grid gives a visual way of calculating the answer to the question, “Among families that have two children, one of whom is a boy born on a Tuesday, how many have two boys?” It shows all the possible variations of birth day and gender for children in a two-child family. The families with the saturated colors are those which might qualify, while those that are shaded out are eliminated. The row gives the birth day of the older child and the column gives the birth day of the younger, while the color represents the sex of each child. If the older child is a boy born on Tuesday, then the family must lie on the Tuesday row and the box representing the older child must be blue. If the younger child is a boy born on Tuesday, the family must lie on the Tuesday column and the box representing the younger child must be blue. The combination of all those boxes show all the possible families that qualify. The total number of saturated boxes is 27. The number of saturated boxes in which both children are boys is 13. So the probability is 13/27.
Bill Casselman

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

Gary Foshee, a puzzle designer from Issaquah, Wash., posed this puzzle during his talk this past March at Gathering 4 Gardner, a convention of mathematicians, magicians and puzzle enthusiasts held biannually in Atlanta. The convention is inspired by Martin Gardner, the recreational mathematician, expositor and philosopher who died May 22 at age 95. Foshee’s riddle is a beautiful example of the kind of simple, surprising and sometimes controversial bits of mathematics that Gardner prized and shared with others.

“The first thing you think is ‘What has Tuesday got to do with it?’” said Foshee after posing his problem during his talk. “Well, it has everything to do with it.”

Even in that mathematician-filled audience, people laughed and shook their heads in astonishment.

When mathematician Keith Devlin of Stanford University later heard about the puzzle, he too initially thought the information about Tuesday should be irrelevant. But hearing that its provenance was the Gathering 4 Gardner conference, he studied it more carefully. He started first by recalling a simpler version of the question called the Two Children Problem, which Gardner himself posed in a Scientific American column in 1959. It leaves out the information about Tuesday entirely: Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?

Intuition would suggest that the answer should be 1/2, since the sex of one child is independent of the sex of the other. And indeed, had he been told which child was a boy (say, the younger one), this reasoning would be sufficient. But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth:

Boy, girl
Boy, boy
Girl, boy

Since one child is a boy, we know that girl, girl isn’t a possibility. Of the three approximately equally likely possibilities, one has two boys and two have a girl and a boy — so the probability of two boys is 1/3, not 1/2, Devlin concluded.

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VISUAL CALCULATION
This grid gives a visual way of calculating the answer to the question, “Among children with a single sibling, how often is that sibling of the same sex? And oh, by the way, mention the sex and birth day of the selected child along the way.” This is the interpretation that Peres argues is the most natural one in real life. Because the sex and birth day don't influence the selection of the family, any two-child family would qualify, and in half of them, both are boys or both are girls.
Bill Casselman

He used this same method on the Tuesday birthday puzzle, enumerating the equally likely possibilities for the sex and birth day of each child and then counting them up.

If the older child is a boy born on Tuesday, there are 14 equally likely possibilities for the sex and birth day of his younger sibling: a girl born on any of the seven days of the week or a boy born on any of the seven days of the week. (This analysis ignores minor differences like the fact that slightly more babies are born on weekdays than on weekend days.)

Now suppose that the older child isn’t a boy born on Tuesday. The younger child then must be, of course. Now we count up the possibilities for the sex and birth day of the older child. If she’s a girl, she might have been born on any day of the week, generating seven more possibilities. If he’s a boy, he could have been born any day except Tuesday. (Otherwise this case would already have been counted in the first scenario: the older child a boy born on Tuesday). This second scenario generates just six, rather than seven, more possibilities.

Since each of these cases is (approximately) equally likely, we can compute the probability by dividing the number of cases in which there are two boys by the total number of cases. The total number of cases is 27: 14 if the older child is a boy born on Tuesday and 13 if the older child isn’t. In 13 of those cases both children are boys (7 if the older child is a boy born on Tuesday and 6 if he isn’t), yielding a probability of 13/27.

Devlin was astonished by this answer. As a mathematician, he had long been familiar with the Two Children Problem and its answer of 1/3. “Knowing the birth day is a Tuesday may (and does) make a difference, but it surely cannot make much of a difference, right?” he wrote in his blog, Devlin’s Angle. “Wrong.” After all, 13/27 is far closer to 1/2 than 1/3.

So why does intuition seem to lead us so astray? Both the intuitive and the mathematically informed guesses are wrong. Are human brains just badly wired for computing probabilities?

Not so fast, says probabilist Yuval Peres of Microsoft Research. That naïve answer of 1/2? In real life, he says, that will usually be the most reasonable one.

Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2. The child’s sex and birthday are just information offered after the selection is made, which doesn’t affect the probability in the slightest.

Gardner himself tripped up on his simpler Two Children Problem. Initially, he gave the answer as 1/3, but he later realized that the problem is ambiguous in the same way that Peres argues that the Tuesday Birthday Problem is. Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests. On the other hand, suppose that you are looking for a male beagle puppy. You want a puppy that has been raised with a sibling for good socialization but you are afraid it will be hard to select just a single puppy from a large litter. So you find a breeder who has exactly two pups and call to confirm that at least one is male. Then the probability that the other is male is 1/3.

In the scenario of Mr. Smith, you’re randomly selecting a child from his two children and then noticing his sex. In the puppy scenario, you’re randomly selecting a two-puppy family with at least one male.

The remarkable thing that Foshee’s variation points out is that any piece of information that affects the selection will also affect the probability. If, for example, you selected a family at random among those with two kids, one of whom is a boy who plays the ukulele and wants to become a dancer, the ukulele-playing and dancing ambitions would affect the probabilities about the sex of his sibling.

Peres says that we shouldn’t despair about our probabilistic intuition, as long as we apply it to familiar situations. The difficulty of these problems is rooted in their artificiality: In real life, we almost always know why the information was selected, whereas these problems have been devised to eliminate that knowledge. “The intuition develops,” he points out, “to handle situations that actually occur.”

Still, Gardner’s initial overly narrow interpretation warns of the dangers of over-hasty analysis of probability questions — and shows the wonder that can come from them.

Comment

Martin Gardner. The Second Scientific American Book of Mathematical Puzzles and Diversions. University of Chicago Press, 1961.

More on the Two Children Problem:
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If you're looking for another mind-bending probability puzzle with a similar ambiguity, read about the Monty Hall problem: [Go to]

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• Both the arguments presented in the article are invalid.

The correct analysis was provided in the comment by B. Hackney.

The problem with the problem is that the language of the example makes it appear reasonable to introduce birth-order, when it is an entirely unrelated factor that has no bearing on the problem presented.

Consider this formulation of the problem:

I have two boxes. Each box contains a mouse. One male mouse was put into its box on Tuesday. What is the probability that both boxes contain male mice?

At this point, does it seem reasonable to perform the following analysis?

"Intuition would suggest that the answer should be 1/2, since the sex of one mouse is independent of the sex of the other. And indeed, had he been told which mouse was a male (say, the one in the box constructed first), this reasoning would be sufficient. But since the male could be in either the younger or the older box, the analysis is more subtle. Devlin started by listing the mice’s sexes in the order the boxes were constructed:

Male, Female
Male, Male
Female, Male

Since we know one mouse is male, we know Female, Female isn’t a possibility. ..."

Can you spot the problem at this point? While the introduction of birth-order appears to be relevant to the children, it actually has no bearing whatsoever. It does not seem reasonable to consider the order of constructing boxes to hold mice, so why should it seem reasonable to consider the birth-order of the children?

It's the use of words like "have two children" that suggest the children must have been born and thus must have a relative birth-order that must be taken into account. What if the words were "adopted two children" -- would it be obvious that the order of adoption is relevant? What about "kidnapped two children" -- does the order of the kidnapping affect anything?

On the other hand, the analysis that "Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy." is also incorrect. Firstly, because it is constructed upon the invalid premise that birth-order is relevant. Secondly, because it introduces further additional criteria that were also not present in the original problem, such as remaining silent in cases that were not part of the stated problem.

The correct analysis requires knowing what needs to be analyzed.
D.P.
Jul. 1, 2010 at 10:09am
• Another twist: determining the sex of a baby before birth is possible, with time for legal abortion. In the U.K., almost 25% of pregnancies end in abortion. Among younger women, the typical reasons can be guessed, but among older women, including those that already have at least one child, family composition comes into play. In the UK, there is not the general preference for boys; in Asian countries this preference leads to increasing likelihood that a born baby will be male if the preceding first, or first-and-second, children were female. In the U.K., the preference is for a family with mixed genders: a boy and a girl, a boy and two girls, two boys and a girl.

Legal sex-selection abortion makes it easier for families to suit their preferences. The way to detect this is by looking at likelihood of opposite-sex child at 2nd or 3rd child to mother, more challenging to analyze than simple birth ratio.

The comment form says links will be removed, so for the intersted, you will have to get hold of birth vital statistics and run the numbers for yourself.

Also, as more Asians enter into US, UK, and Canada, these trends are detectable among these Western countries when examined by Asian background. Almond and Eklund noted this based on USA 2000 census data, and when the USA 2010 data emerge, it will give the opportunity for re-analysis.

So, with one on four or five UK pregnancies resulting in abortion, there is bias that must be factored in. To a small degree, sex preference will throw another prarameter into the calculation of the likelihood of a boy following a boy.

In 1978, MacFarlane (BMJ) noted that births in UK are more likely to happen on Tuesday-Friday, as opposed to Sat Sun Mon. Current numbers would be needed to fine-tune calculations.

So, if I were a gambling man, if I knew a parent had one child, I would guess the next child would be opposite sex, and born on a Friday. Unless the parent were Asian background, then I would pick male. In the long-run, this would beat the odds, but it is overall a small factor, so the winnings would not be very big.
MeToo
Jul. 1, 2010 at 10:42am
• Math is never wrong when used correctly. The problem is that this article and many comments, are using math incorrectly.

The solution proposed, and supported by many, concludes that birth order matters but then neglects one possible combination.

First, let's assume birth order does NOT matter, here are the possibilities for the gender of a child:
1) Boy
2) Girl
so 50% chance of Boy.

If we assume birth order matters, then these are the possible combinations:
1) Boy(known), Boy
2) Boy(known), Girl
3) Boy, Boy(known)
4) Girl, Boy(known)
Still 50% chance of Boy.

All of this assumes that chance of Boy or Girl on any given try is 50%, like the flip of a coin. Any other answer is using math based on incorrect inputs.
AJ Mac
Jul. 1, 2010 at 11:41am
• D.P.: The solution beginning "Everything depends" in no way depends on birth order, which is, as you say, completely irrelevant. The point Peres is making---that the answer depends on how we came to learn the sex and birth day of one of the children---is not bringing in information not in the problem. Rather, it's underscoring the fact that we need information not in the problem to answer the question.

Recall the Monty Hall problem. In that problem we need to know how Monty chooses doors to open, because the probability the goat is behind the door you've selected depends on Monty's behavior. If Monty's behavior is not specified in a version of the puzzle, then there is no way to unambiguously assign probabilities.

Similarly, in the boy-girl problem we need to know the behavior of the "someone" telling us about one of their kids. Since the problem doesn't tell us that information, we can't answer. A reasonable assumption, though, is that we just randomly found out the sex and day of birth of one of the kids. So a strict answer might be, "There is not enough information to answer. If we assume boys and girls are equally likely and independent within and across families, and we assume that we discover the sex and birth day of a kid randomly, then the other kid is a boy 1/2 the time."

If you write down the math, you'll find that the 13/27 answer invokes an implicit and bizarre assumption that the someone in the puzzle will ALWAYS say "I have a boy on Tuesday" if at least one of their kids is a boy born on Tuesday. That's a strange assumption and certainly not specified in the riddle. Under this weird assumption, the answer to the riddle is equivalent to, "what fraction of two-child households with at least one boy born on Tuesday have two boys," which is 13/27. As written, the most reasonable answer is the quite intuitive 1/2, although strictly speaking there is not enough information to answer.

Incidentally, Peres's claim that the 13/27 answer only obtains if the person keeps quiet if they don't have a boy born on Tuesday is mistaken: if they don't have a boy born on Tuesday, they just need to say anything other than "I have a boy born on Tuesday." They could, for example, say they have a girl born on Friday.
Chris A
Jul. 1, 2010 at 1:22pm
• Chris A:

You say, "Rather, it's underscoring the fact that we need information not in the problem to answer the question."

I don't require any more information in order to answer the question: 1/2. In fact, what I needed was less than the information present in the problem. The problem statement did not require either the birth ordering or the birth date to be used.

If someone presents a problem:

"A diesel train leaves Madrid traveling at 40mph on Tuesday and travels east for 3 hours. How far has the train traveled?"

A reader is expected to discard the parts of the question that have no bearing on the mathematical question. The type of engine, the date, the city, the fact it was a train and not a car... this is all irrelevant. We do not need to know "more" -- the weight of a train, the time of day, the name of the train engineer.

If someone were to say "I need a map to answer that so I can determine what is east of Madrid.", they would be wrong. If they said "Trains can not accelerate to 40mph instantly, so I need to know about the weight of the train and the engine power.", they would be wrong.

Those are valid lines of reasoning, if one wanted to address different questions that were not actually asked. But they have as little relevance to the actual question as does the Tuesday and birth-order in this problem.
D.P.
Jul. 1, 2010 at 2:38pm
• D.P.: You do need to know the behavior of the person telling you about your kids to answer the question. This behavior is absolutely not an irrelevant detail, like the manufacturer of the train engine in your example.

Suppose we play a game. I flip two coins, look at them, and show you one. You have to guess the probability the other coin is heads. We play a round and I show you heads. What's the probability the coin I didn't show you is heads?

If I just randomly select a coin to show you, you should guess the other coin is heads with probability 1/2. But suppose that I always show you heads if I can, and only show you tails if I flipped two tails. Then you should conclude the other coin is heads with probability 1/3. Other behaviors give uncountable different probabilities. If you don't know how I choose coins to show you, you cannot unambiguously assign a probability.

Similarly, in the boy-girl problem, you do need to know the behavior of the person asking the question. Perhaps fathers with a boy and girl always talk about the girl and never mention anything about the boy, whereas fathers with two boys randomly choose a boy, and fathers with two girls randomly choose a girl. In this case, you should infer the other kid is a boy with certainty.

Having said that, the most reasonable assumption is surely that we've randomly discovered the sex and day of birth of one of his kids. That's a puzzle--appropriate assumption, like assuming coin flips are fair, or assuming boys and girls are equally likely and independent. Under that assumption the answer is 1/2, which is why Peres says 1/2 is the most "reasonable" answer, not unambiguously the answer---the question does not really give us all the information we need.

Mathematically, let X be the event "Mr Smith says he has a boy." For simplicity condition on learning the sex of one child and that boys and girls are equiprobable and independent. Use Bayes Rule and the law of total probability to find

Pr ( BB | X ) = 1 / ( 1 + 2Pr( X | M )),

where BB means the father has two boys and M means the father has a boy and a girl. P(X|M) is the probability a father with a boy and girl reports "I have a boy." The question does not tell us Pr(X|M), but we need to know it to answer. A reasonable assumption is that Pr(X|M)=1/2. The answer Foshee et al give implicitly makes the strange assumption that Pr(X|M)=1.0, that is, that a father with a boy and girl will always reveal he has a boy and never that he has a girl.
Chris A
Jul. 1, 2010 at 4:55pm
• On Jul. 1, 2010 at 9:34am, AK wrote in response to my earlier post:
'Certainty' might be a little strong given they could instead have two girls or one girl and a boy...

No, not with my postulate that parents want two boys, and will stop having children as soon as this happens. Then the only possible family constitutions are of the form, in birth order, (xG)B(yG)B, where (xG) denotes x consecutive girls, and x, y can take any of the values 0, 1, 2,.. The total family size is n=x+y+2. If I learn n=2, I can deduce x=y=0, so that the family must contain two boys and no girls.

Alternatively, we might postulate that parents stop having children as soon as they have one of each sex. Then knowing that a parent has exactly two children, and one is a boy, the other must be a girl.

The same general principle applies to any family planning strategy, even if we also allow some coin-tossing in the decision-making: the size n of a family is itself informative about its make-up, and in particular about the number of boys in it.
Philip Dawid
Jul. 1, 2010 at 6:10pm
• The reason why people find counterintuitive the correct results (1/3 and 13/27, respectively) is IMO the misguided focus on one particular family that the wording of the problem seems to imply. Indeed, when we look at one particular family, it's easy to ask ourselves: why would the sex of one kid have anything to do with the sex of the other kid? And stranger still: why would the day of week when one of the kids was born have anything to do with the sex of the other kid?

To come to terms with our intuition I think we need to remember that despite its wording, this problem is not at all about *one* family: as the word "probability" implies, the problem is about *all* the families that share a specified set of characteristics with the family described in the problem. So when we hear "I have two children, one of whom is a son (born on a Tuesday); what is the probability that I have two boys?", it helps our intuition a lot if we rephrase the problem in a way that eliminates the particular and highlights the overall context within which the probability is to be determined. We can do this in 2 steps. First we eliminate the specificity implied by the 1st person, obtaining: "If *someone* has 2 children, one of whom is a son (born on a Tuesday), what is the probability that *that person* has 2 boys?". Then we replace the singular by the plural: "Of all the people who have 2 children, of which one is a son (born on a Tuesday), how many people have 2 sons?"

In this new light our intuition serves us well, as it no longer focuses on one particular pair of kids, but rather on how we select a certain subset of the hundreds of millions of people out there who have 2 kids. So when we impose the first condition -- "one of whom is a son" -- it's no longer about some kid A and kid B in family X; instead, it's about the fact that of those hundreds of millions of people having 2 kids, we select only those who have at least 1 son. And it becomes obvious that once we eliminated the 1/4 of families who had 2 daughters, what we're left with is 1/3 families with 2 sons and 2/3 families with a son and a daughter.

Same goes for the seemingly counterintuitive Tuesday condition, though maybe less obvious from a strictly quantitative standpoint: once we eliminated the people with 2 daughters, when we add the condition that we want families who have a son born on a Tuesday, something interesting happens: of the 1/3 of this intermediate set of families who have 2 sons, we'll keep more families (since 2 sons yield higher chances that one of the was born on a Tuesday), while of the other 2/3 we keep less families, so we end up with a final set of families of which on average, many more than 1/3 (actually almost 1/2, or more exactly 13/27) have 2 sons.

Finally, thinking of all this selection process on large numbers of families also makes clear why the actual condition doesn't matter much -- the son being born on a Tuesday, or playing ukulele: once we eliminate the 2-girl families, we always end up with about 1/3 families with 2 boys and 2/3 families with a boy and a girl, and no matter the subsequent condition(s) we impose on one boy, the 1/3 families having 2 boys will have almost twice the chances of one of their sons matching the condition, which causes almost twice as many (percentually) of these families -- with respect to those 2/3 having a boy and a girl -- to make it past the final selection.
Andrei Juan
Jul. 1, 2010 at 8:28pm
• Andrei--1/3, or 13/27, are not correct. I understand that the probability that there are two boys, given that at least one of the children is a boy born on a Tuesday, is 13/27. But that is not what we're asked to calculate in the problem. The problem asks us to calculate the probability Mr Smith has two boys given he tells us he has a boy born on Tuesday.

To see why the first calculation doesn't answer the problem, consider these two urn problems.

Problem 1: There are four urns. One urn is blue and contains two white balls. The other three urns are red; two of the red urns contain one black and one white ball, the last contains two black balls. You push the blue urn off a cliff and select a ball randomly from one of the red urns. Seeing the ball you've selected is black, what's the probability the other ball in that urn is black?

Problem 2: is the same as problem 1, except you're blindfolded and can't throw the blue urn off a cliff. You randomly select one of the four urns and randomly select a ball. Someone tells you it's black. What's the probability the other ball in that urn is black?

It is easy to prove that the answer to problem 1 is 1/3 and the answer to problem 2 is 1/2. Notice the boy-girl problem (in it's simpler form) is equivalent to problem 2, under the assumption not explicitly stated that children are randomly selected from households. That is so even though seeing a black ball in problem 2 allows you to infer you must have chosen a red urn---pulling a black ball from an urn is not the same information as seeing the urn is red.

Similarly, Mr Smith telling you "I have a boy" is not equivalent to simply eliminating two-girl households. Unless fathers with a son and daughter always report they have a son and never that they have a daughter, you should infer from the report "I have a boy" that it's relatively more likely that the father has two sons and not one son and one daughter.

Again, under puzzle-appropriate assumptions, it is possible to prove

P(BB|X) = 1/(1+2P(X|M)),

where X denotes the event "the father says he has a boy," BB means he has two boys, and M means he has a boy and a girl. If fathers randomly report the sex of one of their kids, P(X|M)=1/2, and the probability the other kid is a boy is 1/2.
Chris A
Jul. 2, 2010 at 12:07am
• Following up, here's a straightforward, non-mathematical explanation as to why 1/3 is not the correct answer to the boy-girl problem.

Suppose there are four houses. One house contains two girls, one two boys, and two contain a boy and a girl. Both of the following statements are true:

1. One-third of the houses with at least one boy contain two boys.

2. If you randomly select a child from a randomly selected house and see you've selected a boy, the probability that the other kid in that house is a boy is 1/2.

It is easy to see that statement 1 is true. To see that statement 2 is true, note there are a total of four boys in these houses. Two of the boys are in the house with two boys, and the other two are in the mixed sex houses. So the probability you've selected the house with two boys, given you randomly selected a boy, must be 1/2.

Notice that is so even though you can infer from the fact that you drew a boy that you could not have selected the house with two girls.

The only difference between statement 2 and the boy-girl problem is statement 2 makes it explicit that the child is randomly selected. Without that assumption, or some other assumption about how children are selected from households, the problem has no unambiguous answer.
Chris A
Jul. 2, 2010 at 12:55pm
• I believe everyone who says the answer is ambiguous is correct, including Peres (in the article), the author and, among other commenters, "bremen b". The correct answer to the problem depends upon how you interpret the information given.

To amplify on the second part of bremen b's comment, if you interpret the question as:

1) Take all the two-child families.
2) Limit this set to all those with at least one boy born on Tuesday
3) Randomly select one family from this subset; what are the odds of a two-boy family?

The answer to this formulation of the problem is indeed 13/27 as detailed in the article, although I find the associated graphic more confusing than clarifying.

I believe you can generalize this formulation to say that the odds of a two-boy family are equal to (2n-1)/(4n-1) where n represents the number of discrete states of the non-gender characteristic specified.

So for example, if the problem is given as "What are the odds of a two-boy family, if one of the children is a boy born between noon and midnight?" then the odds are (2*2-1)/(4*2-1) = 3/8.

(This of course assumes you interpret the problem as above and that the non-gender characteristic has n discrete values that are uniformly distributed and independent of the sex of the child.) This genralization is consistent with the notion that the more specific the information supplied (actually, under this formulation the information supplied is best thought of as "sample selection criteria") the closer the odds of a two-boy family approach 50%.

If, on the other hand, you interpret the question as:

1) Take all the two-child families with at least one boy
2) Randomly select one family.
3) Assume someone (the "revealer") privately inspects the children's birth certificates and he reveals that one child is a boy born on a Tuesday (or any day of the week). What are the odds that this is a two-boy family?

In this case, the correct answer is most likely 1/2.

I say "most likely" because you have to assume the revealer is acting randomly. If the revealer says to you, "By the way, one of the children is a child born on Tuesday, and I'll pay you a dollar if the family has two boys, but you pay me a dollar if it does not" then you'd better assume that he is not acting randomly and you will lose your bet (with probability=1).
Robert Heath
Jul. 2, 2010 at 1:57pm
• This article and the question it presents are more telling as a demonstration of the psychology of asking simple questions to people who expect to be asked complicated and tricky questions.

This question has been presented as a mathematical puzzle requiring "subtle" reasoning. That challenges the reader to reject the immediate and obvious answer -- which happens to be the correct answer -- and to attempt to find an answer which is less obvious.

******

Consider this restatement of the problem:

A robot with three arms is holding two bags. One bag was filled with an even number of blue marbles in April. What is the probability that the robot is holding an even number of marbles?

There is no mathematically interesting difference between this problem and the problem in the article. A number of nouns have been substituted and a number of additional facts have been added. But none of these change the calculation.

The reader might be tempted to consider questions such as the following:

...Which arm is holding which bag?
...What month was the other bag filled in?
...What color are the marbles in the other bag?
...In what order were the bags filled?
...Did the robot fill the bags itself?
...Do the bags contain anything other than marbles?
...Are the bags the same size?
...Why a robot?

These questions and many others are all utterly irrelevant to the question stated in the problem. If one chooses to, one can use them to formulate many new questions, but that would mean creating new problems, not answering the one that has been presented.

Chris A wrote: "Having said that, the most reasonable assumption is surely that we've randomly discovered the sex and day of birth of one of his kids. That's a puzzle--appropriate assumption, like assuming coin flips are fair, or assuming boys and girls are equally likely and independent. Under that assumption the answer is 1/2, which is why Peres says 1/2 is the most "reasonable" answer, not unambiguously the answer---the question does not really give us all the information we need."

This seems to be a defense of Peres deliberately miscomprehending the problem.

In the original problem, there are a number of assumptions the reader is _required_ to make in order for the problem to be considered a mathematically-addressable problem and not an open-ended invitation to speculate on the mysteries of knowledge. Specifically:

"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"

...There are only two children to be considered

The speaker does not explicitly say they have only two children -- it could be read as "at least two". If, for example, there are an infinite number of children, that would change the odds of having both only two boys and at least two boys.

...Children can only be either boys or girls.
...The odds of an given child being a boy are 1/2.
...There are seven days in a week and one of them is Tuesday

These assumptions are, as Chris says, like assuming a coin can only be heads or tails. They are conventional to the language and common to everyday experience.

At this point, one must consider, "Can the question be answered?" If the answer is that it can -- and clearly it can -- then there is no basis for claiming that "the question does not really give us all the information we need". If the question can be answered, then by what can we claim that needed information is not present?

Certainly, many things are not present. We do not know if the speaker is male or female. We do not know if the children speak French. We do not know if the children are young or if they are grown and have children of their own.

However, of all the things we do not know, one of the more unreasonable to introduce is that the speaker has mental states and may be lying about the facts presented. To suggest that the speaker has a psychology moves the problem out of mathematics. It essentially turns the question into:

I may be lying. I have two children. What are the odds I have two children?

******

A most telling passage in the article says:

"When mathematician Keith Devlin of Stanford University later heard about the puzzle, he too initially thought the information about Tuesday should be irrelevant. But hearing that its provenance was the Gathering 4 Gardner conference, he studied it more carefully."

Had this problem been presented in a grade-school text book and not a Gardner-related convention of mathematicians, how would people attempt to calculate it?
D.P.
Jul. 2, 2010 at 6:16pm
• D.P., how we came to learn the sex of one of Mr Smith's children is not an irrelevant detail, nor is it something commonly understood as a metaphor for some random event, like coin flips are understood to be a metaphor for binary, independent random variables.

Consider the trivial problem,

"If you flip a coin twice, what's the probability of getting two heads?"

A reasonable answer is "1/4." Even though coin flips are a common metaphor, "1/4, assuming a fair coin," is a better answer (and it would be strange if someone who gave that answer were accused of "deliberately misunderstanding the problem.") These answers don't change if it's revealed that the coins were once owned by Kevin Bacon, or that it's sunny outside, or that they're French coins. Those are irrelevant details.

In the boy-girl problem, the behavior of fathers is not such an irrelevant detail. The answer changes as we assume different things about how we came to learn the sex of one of the children. For example, if fathers with a boy and girl are more likely to talk about their son than their daughter, you should not conclude that a father with two kids who tells you about his son also has a daughter with probability 0.5.

For that reason, answering the boy-girl problem, "1/2, assuming the child whose sex we learn was randomly selected" is a better answer than "1/2," and clarifying what's being assumed about the unknown random process revealing the sex of one of the children does not imply a misunderstanding, quite the opposite. That would be true even if that process were a commonly understood metaphor, like coin flips, but how we learn the sex of one child in a two-child family is not a common metaphor.

Incidentally, your robot problem is not equivalent to the boy-girl problem, even ignoring the irrelevant details. There is really not enough information to answer your problem, nor can we fill in the missing information with some reasonable assumption based on common experience. I think "1/2" would be a poor answer to your problem. But "1/2, if we assume that that the any bag held by a robot is equally likely to contain an odd or even number of marbles" is a reasonable answer. So too is, "1.0, assuming the bags have the same capacity and both are full." Best is, "there is just not enough information to answer," because the missing information is not something we can fill in implicitly, like "fair" in front of "coin." We, or at least I, don't commonly encounter three-armed robots holding bags of marbles!

Chris A
Jul. 2, 2010 at 7:32pm
• The comments I've read overlook the fact that talking about the probability of something definite does not make sense (except for triviality). The guy has two boys, which do have a sex. So, the probability is 1 if both of them are boys and 0 otherwise.

If instead of the problem posed the guy says "Hey, I have a boy who is born on Tuesday, and my wife is pregnant, what's the probability that I'll have a girl? (assuming that the boy is my son :D )" The probability is surely 1/2 (just markovianity).

On the other hand, if I take a huge amount of people and say, "Get out of here unless you have two boys, one of which is born on Tuesday", and then I say "Raise your hand if you other son is a boy", then around 13/27 of the people there will raise their hand.

In the first case (the one that yields probability 1/2) we are computing the probability of the sex in a birth (restricted to births in a given family, which is pointless because of markovianity), while in the second case we are computing the probability of a set of familiar histories (restricted to some familiar histories, which does matter).
Sergio Giro
Jul. 3, 2010 at 8:44pm
• Carl Witthoft, You're a retard. There's only once Ace of Spades in a deck, but you can have two sons born on Tuesday. Well, obviously _you_ can't, because nobody would want to mate with you, but the ficticious Mr Smith can.

Jules Stoop, I don't think your stupidity has much to do with your lack of English comprehension. If one of the children is a boy, g/g isn't an option.

Jerry Vandesic, Yes.

Ralph Dratman, You;re a retard. The analysis also doesn't exclude the possibility that both boys are born on Tuesday.

Torbjørn Amundsen, I hope you were tired, otherwise you're a retard. "Boy, (boy mentioned)" and "(boy mentioned), boy" don't have the same probabilities as "Boy mentioned, girl" and "Girl, boy mentioned" - they are half as likely to occur.

Dace Hummel, Unlike Torbjørn, you definitely are a retard. The four scenarios B,B;B,G;G,B;G,G are just as likely (25% each). Given we know that one of the children is a boy, we eliminate G,G (applying this logic to a random sample of families with two children, this would occur 25% of the time). This leaves us with three possibilities each which had a 25% absolute probability, and now have a 33% relative probability.

Nathan Stanford, You're a retard. Just because there are two possibilities, does not mean each have equal weight. This is like saying that there are the same odds that #1 will come up in roulette as there are that #1 won't come up; or that there are the same odds you fathered your wife's children as there are that a random stranger did. Speaking of which, it's about time you got a paternity test.

Andrew Selkirk, You're a retard for the same reasons as Dace and Nathan. How's that job at McDonald's working out for you?

Chris A, You're a retard. The urn problem you mentioned takes order into account, the child problem in the article doesn't. The problem in this article would be the same as: "You randomly select an urn and draw both balls from that urn. At least one of the balls you drew was black. What's the probability the other ball you drew is black?"

Michael MacDonald, You're a retard for the same reason as Nathan. Roll two six-sided dice. Does 12 (6, 6) come up as 11 (6, 5 or 5, 6)? If you think so, please gamble with me.

I can't be bothered correcting the rest of you; evolution will thankfully take care of that.
Reow Reow
Jul. 4, 2010 at 10:35pm
• The birth order is only a means of labelling the kids. It doesn't matter as such.

Say we have kids A, B, and P(A=boy) = P(B=boy) = 1/2.

There are two different questions being considered here. One is, what is P(A=boy|B=boy) (or vice versa)? The answer is 1/2.

The other question however is, what is P(A=boy AND B=boy | A=boy OR B=boy)? The answer is 1/3.

This is really the most important point to make here, but for an explanation of the latter case, consider the set of all possible outcomes of A and B. A=boy partitions the set in half (P(A=boy) = 1/2). Considering that subset only: If we now look for the cases where also B=boy, that again halves the set (P(B=boy|A=boy) = 1/2). Similarly, if we start again with the whole set, B=boy halves it and the new set is halved again by A=boy.

Now consider the set A=boy OR B=boy. This is the union of the two respective sets. The subset A=boy AND B=boy is exactly the intersection of the two sets. The intersection makes up a half of each respective set as illustrated above, but this half is shared by both sets. Hence we get for each set a half that is not shared, and the half that is shared. Hence the latter, corresponding to A=boy AND B=boy, makes up a 1/3 of the relevant outcomes.
David R
Jul. 5, 2010 at 5:08am
• Reow, as David R says, birth order is irrelevant; I did not use birth order in my answer.

The problem as stated is,

"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"

We can just substitute words directly,

"I have [two white or black balls in an urn], one of which is [black]. What's the probability that I have two [black balls]?"

There is no "ordering" except in the sense that we learn the color of one of the balls (kids) and are asked to assign a probability distribution to the color (sex) of the other ball.

Suppose Mr Smith lives on a very long street with one pink house for every three blue houses. Every pink house contains two girls. One in three blue houses contains two boys, the other two contain a boy and a girl.

1. If Mr Smith tells you he has a boy, your most reasonable guess over the probability that he has two boys is 1/2.

2. If Mr Smith tells you that he lives in a blue house, your most reasonable guess over the probability that he has two boys is 1/3.

That is so despite the fact that the report you receive in statement 1 allows you to infer that Mr Smith lives in a blue house. These answers in no way depend on birth order. The phrasing of the puzzle at hand corresponds to the information in statement 1, not statement 2.

Incidentally, you mean "unconditional" and "conditional" probabilities, or perhaps "prior" and "posterior," but not "absolute" and "relative." Your reasoning that the conditional probabilities are equal because the unconditional probabilities are equal is mistaken.

Chris A
Jul. 5, 2010 at 1:33pm
• Chris A:

"Under this weird assumption, the answer to the riddle is equivalent to, 'what fraction of two-child households with at least one boy born on Tuesday have two boys,' which is 13/27. As written [...] there is not enough information to answer."

Agreed. The "two-child households" formulation of the problem is the only one that actually falls within the realm of probability theory. Probability, as a mathematical quantity, deals with a series of well-defined, repeatable experiments whose outcomes we plan to examine and count.

Meeting a father who makes incomplete assertions about the sexes of his children is neither well-defined nor repeatable.

A lucid discussion of this and similar puzzles can be found in the Wikipedia article "Boy or Girl Paradox."
Ralph Dratman
Jul. 6, 2010 at 12:29am
• I want to add a point in defense of the position that 1/3 is the correct answer to the Gardner puzzle and 13/27 the correct answer to the nifty new Foshee puzzle. (I assume that in the formulation of the latter, “one of whom is a son born on a Tuesday” is short for “at least one of whom is a son born on a Tuesday.”) These answers can be reached by straightforward applications of Bayes’ Theorem, or by reasoning of the sort offered by Andrei and some others. Adam Augusta’s comments help to make the answers less counterintuitive.

Many discussants have disagreed. Most, led perhaps by Chris, think we need to ask what *prompts* the speaker to disclose what he does (i.e., that he has just two children, at least one of which is a son / a son born on a Tuesday). And when we do, they think, we’ll see that the correct answer to each puzzle is either “insufficient information” or, more commonly, “1/2.”

When addressing a concocted probability puzzle, however, we are expected to base our solution on just the information we are *given*. (Of course, with a less-than-ideally formulated puzzle we may have to make assumptions about what the concocter meant – e.g., that by “I have two children” he meant “I have just two children” –, as well as simplifying assumptions the concocter obviously intended for us to make – e.g., that the probability of a second child’s being male is .5, regardless of the gender of the first child.) In the Gardner and Foshee puzzles, we are told nothing about what prompts the speaker to disclose what he does. Accordingly, it is just the *truth* of what he discloses, and not the unspecified motive for the disclosure, on which we should (and can) base our solution. We can ignore the element of communication and take the Foshee puzzle to amount to this: Supposing that someone has just two children and that at least one of them is a son born on a Tuesday, what is the probability that both are sons? Chris, in one of his many incisive contributions (i.e., in his response to Andrei), acknowledges that the answer to *that* question is 13/27. Chris insists, however, that the puzzler is asking us to calculate the probability of the speaker’s having two sons given, not merely *what* the speaker has disclosed, but given also that the speaker has disclosed it. I simply see no warrant for this insistence.

Even if Chris should be right, however, many will still find the puzzles highly (and about equally) counterintuitive when they are construed as I suggest. And if they are finally satisfied that the correct answer to the simpler puzzle is 1/3, they will be surprised that the addition of “born on a Tuesday” not only changes the probability but pushes it all the way up to 13/27. The puzzles are instructive. Thinking hard about them can help us to improve our feel for probabilities.
Michael Burke
Jul. 7, 2010 at 12:53am
• A rectification of my earlier posting:

I want to add a point in defense of the position that 1/3 is the correct answer to the Gardner puzzle and 13/27 the correct answer to the nifty new Foshee puzzle. (I assume that in the formulation of the latter, “one of whom is a son born on a Tuesday” is short for “at least one of whom is a son born on a Tuesday.”) These answers can be reached by straightforward applications of Bayes’ Theorem, or by reasoning of the sort offered by Andrei and some others. Adam Augusta’s comments help to make the answers less counterintuitive.

Many discussants have disagreed. Most, led perhaps by Chris, think we need to ask what *prompts* the speaker to disclose what he does (i.e., that he has just two children, at least one of which is a son / a son born on a Tuesday). And when we do, they think, we’ll see that the correct answer to each puzzle is either “insufficient information” or, more commonly, “1/2.”

When addressing a concocted probability puzzle, however, we are expected to base our solution on just the information we are *given*. (Of course, with a less-than-ideally formulated puzzle we may have to make assumptions about what the concocter meant – e.g., that by “I have two children” he meant “I have just two children” –, as well as simplifying assumptions the concocter obviously intended for us to make – e.g., that the probability of a second child’s being male is .5, regardless of the gender of the first child.) In the Gardner and Foshee puzzles, we are told nothing about what prompts the speaker to disclose what he does. Accordingly, it is just the *truth* of what he discloses, and not the unspecified motive for the disclosure, on which we should (and can) base our solution. We can ignore the element of communication and take the Foshee puzzle to amount to this: Supposing that someone has just two children and that at least one of them is a son born on a Tuesday, what is the probability that both are sons? Chris, in one of his many incisive contributions (i.e., in his response to Andrei), acknowledges that the answer to *that* question is 13/27. Chris insists, however, that the puzzler is asking us to calculate the probability of the speaker’s having two sons given, not merely *what* the speaker has disclosed, but given also that the speaker has disclosed it. I simply see no warrant for this insistence.

Even if Chris should be right, however, many will find the 1/3 and 13/27 answers highly (and about equally) counterintuitive even when the puzzles are construed as I suggest. And if they are finally satisfied that 1/3 is the correct answer to the simpler puzzle, they will be surprised that the addition of “born on a Tuesday” not only changes the probability but pushes it all the way up to 13/27. The puzzles are instructive. Thinking hard about them can help us to improve our feel for probabilities.
Michael Burke
Jul. 7, 2010 at 2:25am
• My objection to treating the information "Smith tells you he has a boy" as equivalent to "Smith does not have two girls" could be put: if you tell me that two-thirds of two child houses with at least one boy also contain one girl, I'll shrug. If you tell me that if I meet some random guy with two kids and he reveals that one of his kids is a boy, I should infer that there's a 1/3 chance he has two boys, I'll insist otherwise. It's not kosher to tell laymen that their intuition that the answer is 1/2 is wrong, and that they'd understand that if they only understood the mysteries of probability theory. In fact, their intuition is completely correct (albeit likely based on an unstated, but reasonable, assumption) whereas the answer 1/3 is based on probability calculations which are simply wrong, or most charitably, based on unstated and bizarre assumptions.

I don't agree we can just condition on the *truth* of what we're told. We need information not given to us in the problem: if Mr Smith has a boy and girl, might he tell us about his girl instead of his boy? The answer Foshee gives implicitly assumes that fathers with a boy and girl always reveal they have at least one boy and never that they have at least one girl. It is this assumption which leads him to the counterintuitive answer 1/3. Under the more reasonable assumption that such a father is no more or less likely to mention his male child than his female child, the answer is 1/2. Without some assumption over how fathers reveal information about their children, we cannot provide a numerical answer.

Bayes' Rule does not (generally) tell us the answer to the simple question is 1/3. If we assume that we are being told the truth, and that we always discover the gender of one child, and that boys and girls are equally likely and independent within households, then Bayes' Rule tells us

Pr(BB|SHHAB) = 1/[ 1 + 2Pr(SHHAB|M)],

where SHHAB denotes the event "says he has a boy," BB means the father has two boys, and M means the father has a boy and girl. In deriving that expression we assume that Pr(SHHAB|GG)=0, that is, we impose the condition that Mr Smith is telling us the *truth*. That is not enough to get us to a numerical answer.

Foshee implicitly assumes that Pr(SHHAB|M)=1, that is, that fathers' utterances are like boy detectors: if a father has a boy, he will always say so. That's a silly assumption.

In the version in which we're also told day of the week, Bayes tells us

Pr(BB|SHHATB) = A / [ A + (2/7)Pr(SHHATB|M, TB) ],

where

A = [ 1 - (6/7)^2 ]Pr(SHHATB| BB, TB)

and SHHATB is the event "says he has a Tuesday boy" and "TB" means the father does have at least one boy born on a Tuesday. Pr(SHHATB|M, TB) is the fraction of fathers who have a boy born on a Tuesday and girl who utter "I have a boy born on Tuesday." Pr(SHHATB|BB, TB) is the fraction of fathers who have two boys, at least one of whom was born on a Tuesday, who utter "I have a boy born on a Tuesday."

If you impose Pr(SHHATB|BB, TB)=Pr(SHHATB|M, TB)=1, you will find that the expression evaluates to 13/27. That is, to arrive at the answer 13/27, we need to assume that fathers' utterances are "Tuesday boy detection mechanisms": if the father has a boy born on Tuesday, he will ALWAYS reveal that he has a boy born on Tuesday. There do not exist, for example, any fathers with a boy born on Tuesday and a girl born on Wednesday who say, "My daughter was born on a Wednesday." That's a silly assumption, too.

I think every single wrong answer I've seen, including those from experts who relied on their intuition rather than cranking through the math, is based on methods that look like: start with a sample space with equiprobable outcomes; treat the information "says he has a boy" as eliminating certain possibilities, then divide the number of remaining outcomes in which there's a boy or a Tuesday boy by the total number of remaining outcomes. That's what the grid at the top of the page does, for example. That method is generally wrong. It's generally wrong because it is not generally the case that all of the conditional probabilities which are not zero are all equal.

Again, I would prefer to phrase this as urn problem, and I don't think many people who understood their sophomore stats classes would get it wrong if phrased as such:

"There are four urns, one with two black balls, one with two white balls, and two with one black and one white ball. You [randomly] select a ball from a [randomly] selected urn. You see it's black. What's the probability you chose the urn with two black balls?"

This question is exactly the same as Foshee's if I leave out the "[randomly]": that, or equivalent, information is what is needed to answer Foshee's problem. When `randomly' is specified, a WRONG answer is, "1/3, because seeing a black ball means that I could not have selected the urn with two white balls. There are three remaining urns, one of which is the urn with two black balls, so 1/3." The correct answer, as Bayes or a little thought will tell you, is 1/2. Notice that is so despite the fact that it is correct to infer that, given you see a black ball, you could not have chosen the urn with two white balls.

In order to arrive at the answer 1/3, we have to assume that balls are not randomly selected. We have to assume that if there's a black ball in the urn, a black ball will always be selected.
Chris A
Jul. 7, 2010 at 6:28pm
• Actually, this article is one of the better presentations I have seen, and the points are all valid within the caveats applied. The proper solution does indeed depend on why a boy, or a Tuesday boy, was mentioned. I just wish the article had stressed one point a little better.

I don't think anybody ever suggested that birth order is, in and of itself, relevant. It isn't. The solution only uses order to establish a set of N equally-probable family types. To make it clear that as many families have a boy and a girl, as have two of the same gender. That is, it determines a proper Sample Space, which is not only relevant but critical.

What is used to provide that the order is unimportant, as long as it is independent of gender. I myself prefer to alphabetize the children's names, than to use age. And you don't need to utilize order at all, as long as you make a proper Sample Space. An equivalent solution can be given with the Sample Space of the three family types (0Boys, 1Boy, and 2Boys). But it is harder to represent it as a counting problem that way, since the probabilities are 1/4, 1/2, and 1/4, respectively.

Where I wish Rehmeyer would have gone further is when she said "The remarkable thing that Foshee's variation points out is that any piece of information that affects the selection will also affect the probability." I find this misleading, in the exact same way the original question is. It isn't the information itself that affects the probability. It is that, in Foshee's solution, the information is a prerequisite for selection; while the best interpretation of the problem statement is that it is an observation after selection.

When you insist on picking only families that include a boy with some property that has probability P, a "1Boy" family matches with probability P, but a "2Boys" family matches with probability (2P-P^2). Since "1Boy" is twice as likely initially, the proportion of families, from the two categories, that match the prerequisite is 2P:2P-P^2. That is equal to 2:2-P. Explained this way, it is obvious why the resultant selection becomes more likely to have two boys as P goes from 1 to 0. The proportion starts at 2:1, but ends at 1:1.

If, instead of choosing a family to match the information, you pick a family first and match information of the form "One is a " or "One is a born on a " to a random child in the family, then either probability is 1/2. Foshee's solution is unintuitive, because this is the more natural way to read either statement.

It's just that puzzle solvers (like Foshee) and pure mathematicians (like Keith Devlin) have misinterpreted what "it is given that..." means. All it means is that the information is accepted as true, not that it was required to be true. This difference doesn't affect other fields of mathematics, since they are never interested in what could have happened, but didn't. Probability is. Once they insisted the answer to the simpler problem is 1/3 based on that misinterpretation, they have to do the same with the complex form.
Jeff Jo
Jul. 14, 2010 at 5:09pm
• The day of the week/year/decade/century are irrelevant as all kids/coins have a 'birthdate'. The 'Tuesday' addition is a misdirection.

Line/Family 1st/Older 2nd/Younger Probability
A/Smith's T*Sue h*ron 1:4 (25%)
B/Jone's T*Sue t*tia 1:4 (25%)
C/White's H*Bob h*ron 1:4 (25%)
D/Black's H*Bob t*tia 1:4 (25%)

The rules:
1) In the 1st/Older column, the coin/child has a 50/50 chance of being either T*Sue or H*Bob.
2) In the 2nd/Younger column, the coin/child has a 50/50 chance of being either t*tia or h*ron.
3) The coin/child of the '1st/Older' column are not interchangeable with the coin/child of the '2nd/Younger' column, ever.
4) 'T*Sue' and 't*tia' are Tail/Girls
5) 'H*Bob' and 'h*ron' are Heads/Boys
6) 'T*Sue' and 'H*Bob' are older/1st (the range can be from a millisecond to years)
7) 'h*ron' and 't*tia' are younger/2nd (the range can be from a millisecond to years)
8) You actually never see the coin/child as you are blind.
9) You will meet up with one (and only one) of the above families, you have no idea which one though.

Your cane accidentally hits the father of one of the families. You both strike up a conversation, eventually he says:

"I have two children, at least one of whom is a son. What is the probability both children are boys?"

What say you?
Chucky
Jul. 20, 2010 at 10:59pm
• "Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests."
Actually this is wrong.
It's 1/3
Of the three possibilities, the two boys that can be the son are different from each other and cannot be interchanged.
1)Rob, Sue
2)Rob, Tim
3)Lisa, Tim
Obviously the 4)Lisa, Sue possibility is eliminated due to no boys.
The three remaining are all equally possible as one has no idea if the son introduced is Tim or Rob.
So of the 3 choices above, which are the Smith children?
The answer is 1/3, not 1/2.
Chucky
Jul. 22, 2010 at 4:47am
• I had a hard time believing this, so ran a simulation. You have to start with a universe of two child families, then pick out the "boyTuesay" families and check the distribution. It's true, I got 13/27. Here's some reasoning that explains the result:

We have to introduce BoyNotTuesday so that boyTuesday+boyTuesday doesn't get counted twice:

P(Girl+boyTuesday, boyTuesday+Girl, boyTuesday+Boy, and BoyNotTuesday+boyTuesday) =
P(boyTuesdayEitherChild) =
.5*1/7 + 1/7*.5 + 1/7*.5 + (.5*(6/7))*(1/7) = 1.5/7 + 3/49 = 13.5/49 % (almost 2/7)

P(boyTuesday+Boy, and BoyNotTuesday+boyTuesday) = P(boyOtherChild) =
.5*1/7 + (.5*(6/7))*(1/7) = .5/7 + 3/49 = 6.5/49 (almost 1/7)

P(boyOtherChild | boyTuesdayEitherChild) =6.5/13.5 = 13/27

PS - Why aren't the comments in time order?
Susan T
Aug. 12, 2010 at 6:42am

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xue you
Aug. 14, 2010 at 9:20am
• Yeah, this is the Bertrand Paradox (the straight kids example). The week stuff is adding complexity but of course it adds ambiguity by not stating the sample space correctly (similar with the Monty Hall problem). It is a shame people (mathematicians) waste their time to talk about these old problems and not advance their research.

Look these up (Bertrand paradox and Monty Hall) since I can't post links.
Ionut F
Nov. 19, 2010 at 2:47pm
• On 7/22/10 4:47 AM, Chucky said: "'Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests.' Actually this is wrong."

No; actually, it is right. Counting cases and throwing out the "impossible," as you did, is not always the right thing to do. You want to sum up the probabilities that each case produces the result in the problem statement, not just the counts. It works out to the same thing if every case you count has the same probability, but that isn’t true in the problem you stated.

If your Mr. Smith has two boys, there is a 100% chance the child you meet with him is a boy. If he has two girls, it is a 0% chance (so you don’t "eliminate" cases, you add 0% for those cases). But if he has 1 of each, there is a 50% chance. So the answer is not (1)/(1+1+1)=1/3, it is (100%)/(100%+50%+50%+0%)=1/2.

In fact, the answer to any version of the Two Child Problem is 1/2 unless the problem provides a reason why you couldn't know about either gender. Susan T's simulation will get that answer if she "picks out" not just the families that do not have a boyTuesday, but also those that do but the father chooses to tell us about his other child who is not a boyTuesday.

And Ionut F is quite right, this is a variation of Bertrand's Box Paradox (he has two named for him, and this is the "Box" one). Here's better way to express it: You have four poker chips, identical except for a dot painted in the center of each side. One chip has two blue dots (for boys), one has two pink dots, and two have one dot of each color. Scramble them in a hat and draw out one in such a way that only one side is ever visible. It has a blue dot. What is the probability the dot on the other side is blue?

It is tempting to count the chips to get the answer: three have a blue dot, and could be the one you see; one does not so it can't; and only one of those three has two blue dots. That's what Chucky did above, and it's wrong. There are eight ways you could draw a chip out this way, four of those ways show a blue dot, and two of those have a blue dot on the other side. You could also make 196 chips that have all of the possible combinations of the days of the week, painted in blue or pink, on them. If the one you draw has a blue "Tuesday," the chances are still 1/2 that the other side is painted in blue.

People have difficulty equating Bertrand's Box Paradox to the Two Child Paradox (or the Monty Hall Problem, or the Three Prisoners Problem, which are more direct comparisons) because the Box Paradox divides each case physically into two sub-cases. In some of them, the two sub-cases look the same, but because the difference is physical it is clear it exists. All the others have the same division into sub-cases, but (1) it is accomplished by a mental decision of a participant in the problem, and (2) it only applies to the cases where the sub-cases are different. These factors make it difficult to see it is there, but it is.

Jeff Jo
Jan. 14, 2011 at 12:15pm
• "Why does every one assume that, if I have two children, each combination BB, BG, GB, GG is equally likely? Even when all births are equally likely to be B or G this is typically false, since account must be taken of how parents choose to stop having children. As an extreme example, suppose every family goes on procreating until they have two boys (and any number of girls). Then knowing that a couple has exactly two children, I can infer with certainty that they have two boys. Philip Dawid"

'Certainty' might be a little strong given they could instead have two girls or one girl and a boy...
A K
Jul. 1, 2010 at 9:34am
• Except that the percentage of getting a boy or girl isn't 50/50. For every 100 girls born, there's about 105 baby boys. The numbers even out by age 20, due to less physical maturity in infant boys, and greater risk-taking in adolescent boys, among other factors. So the chance of getting 2 boys is greater than 50/50.

Further, if the parent bothers to mention that one is born on a Tuesday, normal human interaction would seem to forbid that both were born on a Tuesday. The birth date is less essential information than the gender in a conversation (I've had hundreds about birth gender, and only a handful about the day of the week) so there's a reason it's mentioned. Therefore it's probably more exclusionary than the gender--either the other's a Girl born on a Tuesday, or the other son is not born on a Tuesday.

Also, people are more likely to mention the quantity of one gender of child as a complete quantity...i.e. "I have two sons" or "I have 5 daughters" rather than "I have one son...and one son...and one son...and one son."

So my guess would be that the chances are pretty high that there's only one son, or else the other would be mentioned in the conversation, unless the discussion is about twins, as in "I have one son born on a Tuesday, but his twin, he was born on Monday (or Wednesday)" and half the sentence was left out.

I would discount the possibility of a girl twin in that case, because the normal usage would be "My son was born on Tuesday..." not "I have a son born on Tuesday"

So, either 95% chance there's only one son, UNLESS there's twins, and there's the midnight split, in which case it's 95% chance they're both boys.

I would leave the 5% chance of the other, because occasionally the topic comes up from a day-of-the-week perspective instead of a "what gender child is it" perspective, in which case there's a 95% chance only one child was born on a Tuesday, but the gender is back to the 100/105 split of girls/boys.
Elizabeth Lines
Jun. 30, 2010 at 12:59am
• i chose to call one child Alex and the other Bailey, and then enumerate all permutations that satisfy the stipulations of the puzzle by twiddling through them in a "kinda like when you're writing binary numbers out in long form" way:

Alex (boy, tuesday) Bailey (boy, monday)
Alex (boy, tuesday) Bailey (boy, tuesday)
Alex (boy, tuesday) Bailey (boy, wednesday)
Alex (boy, tuesday) Bailey (boy, thursday)
Alex (boy, tuesday) Bailey (boy, friday)
Alex (boy, tuesday) Bailey (boy, saturday)
Alex (boy, tuesday) Bailey (boy, sunday)
Alex (boy, monday) Bailey (boy, tuesday)
Alex (boy, tuesday) Bailey (boy, tuesday)
Alex (boy, wednesday) Bailey (boy, tuesday)
Alex (boy, thursday) Bailey (boy, tuesday)
Alex (boy, friday) Bailey (boy, tuesday)
Alex (boy, saturday) Bailey (boy, tuesday)
Alex (boy, sunday) Bailey (boy, tuesday)
Alex (girl, tuesday) Bailey (boy, monday)
Alex (girl, tuesday) Bailey (boy, tuesday)
Alex (girl, tuesday) Bailey (boy, wednesday)
Alex (girl, tuesday) Bailey (boy, thursday)
Alex (girl, tuesday) Bailey (boy, friday)
Alex (girl, tuesday) Bailey (boy, saturday)
Alex (girl, tuesday) Bailey (boy, sunday)
Alex (girl, monday) Bailey (boy, tuesday)
Alex (girl, tuesday) Bailey (boy, tuesday)
Alex (girl, wednesday) Bailey (boy, tuesday)
Alex (girl, thursday) Bailey (boy, tuesday)
Alex (girl, friday) Bailey (boy, tuesday)
Alex (girl, saturday) Bailey (boy, tuesday)
Alex (girl, sunday) Bailey (boy, tuesday)
Alex (boy, tuesday) Bailey (girl, monday)
Alex (boy, tuesday) Bailey (girl, tuesday)
Alex (boy, tuesday) Bailey (girl, wednesday)
Alex (boy, tuesday) Bailey (girl, thursday)
Alex (boy, tuesday) Bailey (girl, friday)
Alex (boy, tuesday) Bailey (girl, saturday)
Alex (boy, tuesday) Bailey (girl, sunday)
Alex (boy, monday) Bailey (girl, tuesday)
Alex (boy, tuesday) Bailey (girl, tuesday)
Alex (boy, wednesday) Bailey (girl, tuesday)
Alex (boy, thursday) Bailey (girl, tuesday)
Alex (boy, friday) Bailey (girl, tuesday)
Alex (boy, saturday) Bailey (girl, tuesday)
Alex (boy, sunday) Bailey (girl, tuesday)

of that list, there are exactly three duplicates (each of these following appears twice in the permutation set) in which the permutations of resultant children are literally identical.

'Alex (boy, tuesday) Bailey (boy, tuesday)' appears twice,
'Alex (boy, tuesday) Bailey (girl, tuesday)' appears twice,
'Alex (girl, tuesday) Bailey (boy, tuesday)' appears twice, the rest appear once each.

it seems intuitive to me to count them as only one possible outcome each.

so originally i had 42 permutations, now i'm down to 39. of those 39, 27 contain '(boy, tuesday)', which is a puzzle stipulation:

Alex (girl, monday) Bailey (boy, tuesday)
Alex (girl, tuesday) Bailey (boy, tuesday)
Alex (girl, wednesday) Bailey (boy, tuesday)
Alex (girl, thursday) Bailey (boy, tuesday)
Alex (girl, friday) Bailey (boy, tuesday)
Alex (girl, saturday) Bailey (boy, tuesday)
Alex (girl, sunday) Bailey (boy, tuesday)
Alex (boy, monday) Bailey (boy, tuesday)
Alex (boy, tuesday) Bailey (girl, monday)
Alex (boy, tuesday) Bailey (girl, tuesday)
Alex (boy, tuesday) Bailey (girl, wednesday)
Alex (boy, tuesday) Bailey (girl, thursday)
Alex (boy, tuesday) Bailey (girl, friday)
Alex (boy, tuesday) Bailey (girl, saturday)
Alex (boy, tuesday) Bailey (girl, sunday)
Alex (boy, tuesday) Bailey (boy, monday)
Alex (boy, tuesday) Bailey (boy, tuesday)
Alex (boy, tuesday) Bailey (boy, wednesday)
Alex (boy, tuesday) Bailey (boy, thursday)
Alex (boy, tuesday) Bailey (boy, friday)
Alex (boy, tuesday) Bailey (boy, saturday)
Alex (boy, tuesday) Bailey (boy, sunday)
Alex (boy, wednesday) Bailey (boy, tuesday)
Alex (boy, thursday) Bailey (boy, tuesday)
Alex (boy, friday) Bailey (boy, tuesday)
Alex (boy, saturday) Bailey (boy, tuesday)
Alex (boy, sunday) Bailey (boy, tuesday)

so i was thus able to understand the 27 everyone's talking about.

the puzzle question asks for the probability of having two boys. of the remaining 27 lines (which enumerate the entire scope of the given possibility set), 13 specify that both Alex and Bailey are boys.

i (finally) agree with 13/27.
jared spiegel
Jun. 29, 2010 at 11:15pm
• The article is correct, given the constraints, if you allow yourself to think logically for a second. Let's do a thought experiment on the simpler 2 child problem. A census in our heads (you could use STATA or SQL but it's easier and quicker to do it in your head). Let's assume we do a census on a population of 10,000 families with two children. You would expect a demographic something like this
Fistborn, Secondborn, incidences.
Boy, Girl x 2500
Boy, Boy x 2500
Girl, Boy x 2500
Girl, Girl x 2500
Total, 10,000

Apply the constraint that one must be a boy and we have this subset.
Boy, Girl x 2500
Boy, Boy x 2500
Girl, Boy x 2500
Total possibilities of 7,500.
Notice each outcome is equally probable.
Number of incidences where they are both boys is 2,500
Therefore P = 2500/7500 = 1/3
Married to Christ
Jun. 29, 2010 at 10:52pm
• Introducing birth order in the analysis when it was not part of the problem is the source of the apparent anomaly, but it's really simple when considered.
B Hackney
Jun. 29, 2010 at 10:20pm
• that should read " ... statistical analysis isn't obvious."
Greg Valcourt
Jun. 29, 2010 at 8:31pm
• I see exactly what your saying in your second post, Micheal. That perspective is correct. However, the title of this column is "When intuition and math probably look wrong", not "An introduction to how statistical isn't obvious."
Greg Valcourt
Jun. 29, 2010 at 8:30pm
• I recant.

The precise wording of the problem matters (as the article points out). If the focus is on the parent, then one third of the FATHERS with a son have two sons. 13/49ths of those fathers with two sons have one (or both) born on Tuesday. 7/49ths of the fathers with one son have a son born on Tuesday. Thus, the 13/27 [13/(13+7+7) as the one-son group is twice the size as the two-son group] number is also correct.
Frederick Michael
Jun. 29, 2010 at 7:41pm
• I agree with all the previous posters that said the article is flawed and the probability is 1/2. However, there is a more convincing explanation.

Consider the population of all 2 child families. Half of the boys are in two-boy families. Half of the boys born on Tuesday are also in two-boy families.

If this isn't obvious, let's do some actual numbers. Let's start without the Tuesday goof. Assume a population of 700 two child families with a distribution of children exactly according to the probabilities. There will be 700 boys and 700 girls.

175 families will have 2 boys, 175 have 2 girls, 350 with one of each -- 175 with an older boy and a younger girl & 175 the other way.

The 175 2-boy families have 350 boys. That's half of the total. So, if you pick a random boy, you have a 50-50 chance of it being from a 2-boy family. Get it?

Now add the Tuesday bit. 100 of the 700 boys are born on a Tuesday. 50 of them are in 2-boy families. Do I really have to keep going?

Golly.
Frederick Michael
Jun. 29, 2010 at 7:20pm
• When considering just the order the children were born in, I get 2/4. Here's how to do it properly: Label the child we know to be a boy as A. The child without a known gender is B.

Boy-A Boy-B
Boy-A Girl-B
Boy-B Boy-A
Girl-B Boy-A

2/4 = 1/2 as initial intuition expects. Math works as long as you break down the problem correctly, if you don't, you get funny answers like 1/3.

Now let's do the day of the week thing with the boy being born first on Tuesday:

Boy-A Girl-B-S
Boy-A Girl-B-M
Boy-A Girl-B-Tu
Boy-A Girl-B-W
Boy-A Girl-B-Th
Boy-A Girl-B-Fr
Boy-A Girl-B-Sat
Boy-A Boy-B-S
Boy-A Boy-B-M
Boy-A Boy-B-Tu
Boy-A Boy-B-W
Boy-A Boy-B-Th
Boy-A Boy-B-Fr
Boy-A Boy-B-Sat

Chances are 7/14. Filp around who was born first and add it in: 14/28. Even if we're not allowed to use Tuesdays for child B, we get 12/24 which is 1/2 as expected.
Greg Valcourt
Jun. 29, 2010 at 7:07pm
• The odds are 50%- regardless of day of the week or dancing ambitions, or ukuleles or anything else.

Torbjørn and Cletis are correct.

Since you don't know whether the son he is referring to is his oldest child or youngest child you must consider all of the options..

1)- the boy mentioned is the oldest and the youngest is a girl
2)- the boy mentioned is the youngest and the oldest is a girl
3)- the boy mentioned is the oldest and the youngest is a boy as well
4)- the boy mentioned is the youngest and the oldest is a boy as well
jon s
Jun. 29, 2010 at 6:22pm
• The question is ambiguous. It can read both ways:

"I have two children, one of whom is a son born on a Tuesday. What is the probability that the other child was born a boy?" - then the answer is 1/2

"Out of all families with two children, one of whom is a son born on a Tuesday. What is the probability that a family has 2 boys?" - then the answer is 13/27.

bremen b
Jun. 29, 2010 at 5:25pm
• Disclaimer: IANAS (I Am Not A Statistician), but it seems to me that this is a case of applying conditional probability where it does not belong.

I think the answer to the issue with this puzzle can be found in the silliness of the question "Given a two kid family where both kids are boys, what is the chance that at least one of them is a boy?", the answer to which unfortunately becomes a relevant part of the calculation if one wants to treat the puzzle as a problem of conditional probability. But let's begin by noting that the problem, as posed, doesn't actually force us to use conditional probability:

"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"

This question is insidious, because it invites the reader to apply conditional probabilistics against better judgement.

There are two ways to interpret the question: As a problem in conditional probabilistics - or not. That is to say, you can read it as "GIVEN that I have two children, one of whom is a son born on a Tuesday, what is the probability that I have two boys?" where the case could be made that the answer is 1/3, or as "I have two children, one of whom is a son born on a Tuesday. Regardless, what is the probability that I have two boys?" (in which case, incidentally, the answer becomes 1/4, not 1/2 as the article suggests: "so the probability of two boys is 1/3, not 1/2, Devlin concluded.").

The problem poser never explicitly states the word "given", so the blame can't be placed there but must instead be placed on anyone who "falls for it" and interprets the question as if there were somehow a relevant correlation present making the first statement a "given" to be considered in a conditional probability calculation that will derive the answer to the question in the second sentence. (Of course there could be real life correlations, but it is later in the article clarified that we are asked to not consider those). Precisely the same issue concerns the simpler two child problem, and as the math and logic is easier there I will use it instead of the Tuesday version to make my point.

"Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?"

The answer to the two child problem only becomes 1/3 instead of 1/4 if we treat the question as a conditional probability problem, which is a curious choice and one which also forces us to answer the embarrasingly silly question "Given a two kid family where both kids are boys, what is the chance that at least one of them is a boy".

The reason is that if we do want to treat this as a conditional probability problem, we have to apply Bayes' theorem:

P[A|B] = P[B|A] * P[A] / P[B]

That is, the probability of something "A" GIVEN something "B" equals the probability of B GIVEN A times the probability of A in general divided by the probability of B in general.

So in our case:

A = Both kids are boys
B = At least one kid is a boy

To resolve P[A|B], the probability that both kids are boys given that at least one of them is, we need the values for P[A], P[B] and P[B|A]. Well, P[A] is the probability in general (that is, without using conditional probability) that both kids will be boys which is 25%, P[B] is the probability in general that at least one kid is a boy which is 75% and P[A|B] is the unfortunately silly question of what is the probability that at least one kid will be a boy given that both kids are boys, which is 100%. This gives us 1 * 0.25 / 0.75 = 1/3 which is the "correct" answer according to the article but which should make anyone suspicious...how come our calculation has to contain such a silly part? Is the answer correct if part of the calculation to arrive at it is nonsensical?

Conditional probabilistics deals with the question "What is the probability of something GIVEN something else" under the assumption that the existence of the "something else" might somehow affect the probability of the first "something". For example, given that it is winter, the probability for snow might be up. Or given that you are by the equator, the chance might be down again. However, in this puzzle we are asked to disconsider all such real correlations that may in fact influence the probability of two boys in the family given that there is at least one boy.

For some reason the only influence that the "given" condition apparently may have on our calculation is to narrow down the dataset from which we must draw our conclusions. Reducing the available data this way could of course be valuable if it were the result of following a real, meaningful correlation but if it isn't, we're merely deciding not to access all data and risk reducing the quality of our results.

What happens is shown in a clear way in the earlier comment by signature "Married to Christ": suddenly you compare the 2,500 families with two boys to the 7,500 families with at least one boy, giving a 1/3 chance of having two boys for the two-kid family with at least one boy, instead of the unconditional approach of comparing the 2,500 families with two boys to all 10,000 families with two kids in the population to see that the general chance of having both kids be boys is in fact 25% which of course matches the general 50/50 expectation to have a boy (and, mind, is a necessary fact to solve the conditional equation as well).

I think that what this example does it to highlight the problem of applying conditional probability when not warranted - it might seem innocent, as in "if there is no correlation, shouldn't conditional probabilistics give the same answer as unconditional?", but this example shows that this is not always the case.

I guess the moral is: if we let meaningless information influence our decisions, they become worse.
mhelander
Jun. 30, 2010 at 1:34am
• Watch the file number 3 in the top figure, there is one posibility that is not mention in the tuesday column: the ones mention are (suppose John is the name of the child born on tuesday). Girl - John ; John - Girl ; John - Boy, but there is another possibility: BOY-JOHN that is not mention, count again and the probability is 1/2 as everybody suspect.
Javo
Jul. 1, 2010 at 7:48am
• Ok, is my understanding of this problem wrong? I only did the 2 children thing, but this is the pseudocode I came up with:

Pseudocode:

i = random(2); //random number between 1 and 2

if (i==1)
boy=older;

if (i=2)
boy=younger;

i = random(2);

if (i==1)
otherchild=girl;

if (i==2)
otherchild==boy;

if( boy==older || boy ==younger && otherchild == boy )
x= twosons;
else
x= !twosons;

As far as I can tell its a total representation of the problem, and its going to be 50%.
Michael MacDonald
Jul. 1, 2010 at 7:26am
• The deal is that 12.5% of the time it can be BG, 12.5% of the time it can be GB, 50% of the time it can be BB. They are trying to make it look like BG and GB each have the same probability as BB. Which they don't.
Michael MacDonald
Jul. 1, 2010 at 7:37am
• Ack, that if(i=2) should be if(i==2).
Michael MacDonald
Jul. 1, 2010 at 7:34am
• Why does every one assume that, if I have two children, each combination BB, BG, GB, GG is equally likely? Even when all births are equally likely to be B or G this is typically false, since account must be taken of how parents choose to stop having children. As an extreme example, suppose every family goes on procreating until they have two boys (and any number of girls). Then knowing that a couple has exactly two children, I can infer with certainty that they have two boys.
Philip Dawid
Jul. 1, 2010 at 3:11am
• I think the problem lies with our understanding of determinism. The problem is not with probability but with our understanding of what probability is telling us. We talk of luck, chance, and randomness when these things don't exist. Probability, until Bart Kosko came along, was a bastard step child in math. Sure it was used a ton and the practical applications of it do not need to be listed here, but it only existed because of logical hand waving. It fit(s) our view of the world but there was no foundation for this branch of mathematics. It had no derivation, and there fore was not really a branch of math in my mind. When Kosko stepped in and derived the basic probability formulas using Fuzzy Logic/Set theory, it not only provided a logical and theoretical base to justify its existence, but also fundamentally changed how it has to be viewed in relation to the real world. In this case we can see how this plays out. Sure probability fit to an extant our view of the world, but how much of that is just a psychological trick invented to survive a harsh world. We often see in shades of gray the world around us, AND we have this notion of luck and randomness. When the two collide we can have problems.
james ehrhart
Jun. 30, 2010 at 4:54pm
• Peres's answer is quite correct and Foshee et al are not. Consider the problem,

"There are four urns, one with two black balls, one with two white balls, and two with one black and one white ball. You randomly select an urn and randomly select a ball from that urn. You see it's black. What's the probability the other ball in that urn is black?"

The answer is obviously 1/2. Notice this problem is equivalent to the boy/girl puzzle if we assume boys and girls are equally likely and independent and that the sex of one of the children is randomly revealed to us.

The puzzle has no unambiguous answer if the someone who tells us they have a boy is doing something other than randomly revealing a child's sex. Consider the urn problem,

"There are four urns as in the preceding problem. Fred randomly selects one, looks in, and selects a ball. He shows you the ball is black. What's the probability that the other ball in that urn is black?"

If Fred chooses randomly, we still get 1/2. We get 1/3 if Fred follows the rule: `I will show a white ball if this is the urn with two white balls, otherwise I will show a black ball.' We could get anything between zero and one with depending on what we assume about Fred's behavior.

None of this reasoning changes with the "born on Tuesday" modifier. If we're just observing the sex and day of birth of a randomly selected child, the probability the other kid is a boy is still 1/2. Depending on the behavior of someone giving us this information, the probability the other kid is a boy could be anything between zero and one. We only get 13/27 under some rather odd assumptions about the behavior of someone revealing information to us.

I think people tend to arrive at the "clever" answer because they start by defining the sample space up in such a way that each outcome is equally likely (in an obvious notation, P(BB)=P(BG)=P(GB)=P(GG)=0.25). They then note that at least one boy means P(GG)=0. Then there are three possibilities remaining, and in only one is the other kid a boy, so 1/3. But that's wrong: the conditional probabilities P(BB|X), P(BG|X), and P(GB|X) are not necessarily equal. If we're seeing a randomly selected kid, for example, P(BB|X) = 2P(BG|X).

Chris A
Jun. 30, 2010 at 12:24pm
• I should probably add that the opposite argument makes sense to me as well: The chance of two boys in a two kid family is 25%, but if we learn that this is not a two-girl family this just seems to imply that the chances must be better than 25%, and applying the calculations I certainly agree it ends up with 1/3. I also understand from this perspective the "intuitive" conclusion described in the article that if we know one of them is a boy, the odds for the other one being a boy seems by itself to determine the odds for both of them being boys making the chance for two boys go up from the "unconditional" 1/4 to the 1/2. At that point the offered 1/3 appears almost comforting.

And then the brain jumps around again and I start seeing everything from the perspective I described in my previous comment. I suppose only one of them can be correct.

I think my basic assumption which causes me to reset into my first perspective all the time is that if there is no real correlation between A and B, conditional probability shouldn't be able to provide an improved answer (it would have to exploit a relationship that isn't there) and so if the answer is different it must in fact be the degraded. I wonder what is wrong in (one or both) of the ways I'm thinking about this...
mhelander
Jun. 30, 2010 at 2:21am
• Let me try to tackle it a slightly different way.

1. It is given that there are 2 children, each born on a particular day of the week. That means there are 14 possibilities for each child. 14 * 14 = 196 different cases, and each case is equally likely to occur.

2. Also given is that one child is a son born on a Tuesday. So out of the world of 196 different possibilities, how many meet that requirement?

2a. Tuesday-Son is one of the 14 possibilities of the 1st child. If the 1st child is the one that was "given" in #2, and there are 14 possibilities for the 2nd child, then this gives us 14 cases (out of the 196) that would meet the Tuesday-Son requirement.

2b. Tuesday-Son is also one of the 14 possibilities of the 2nd child. If the 2nd child is the one that was "given", and there are 14 possibilities for the 1st child, then this gives us 14 cases (out of the 196) that would meet the Tuesday-Son requirement.

2c. However, the one case where BOTH children are Tuesday-Sons was counted in both 2a and 2b above. That means we've identified 27 distinct cases, not 28, that meet the requirements given.

3. Now that we have 27 cases that meet the requirements, lets look at how many of those cases have two sons.

3a. Of those 27 cases, consider the 14 that came from 2a, where the 1st child was a Tuesday-Son. Of those 14, 7 of them had a son as the 2nd child as well. Doesn't matter which day of the week the 2nd child was born on.

3b. Next consider the 14 that came from 2b, where the 2nd child was a Tuesday-Son. Of those 14, 7 of them had a son as the 1st child as well.

3c. Just like before, there is one case included in both 3a and 3b. It's the case where both children are Tuesday-Sons.

3d. Therefore, there are 13 cases (7 + 7 - 1) where there are two sons.

4. Out of 27 cases where there are two children and at least one is a Tuesday-Son, 13 of the cases have 2 sons. The probability is 13 / 27. The
Dave B
Jun. 29, 2010 at 4:40pm
• And, we are of course assuming the children being mentioned are the currently living children. It would be common to not mention those who have died. The probabilities for deaths of children are probably not evenly distributed for each sex, and probably vary according to age.

You then need to apply the distribution of these death probabilities by sex and age to the distribution of possible child ages for the set of families with non identical twins.

And, we haven't even mentioned transgendered children.
Bob spence
Jun. 29, 2010 at 3:34pm
• Here's the part I don't get with this Tues puzzle. When order was considered in the analysis (ie allowing boy/girl, girl/boy), the puzzle was changed and there is a missing option.

There are really 5 possibilities when considering order:
1. Boy(Tues), Boy
2. Boy, Boy(Tues)
3. Girl,Boy
4. Boy,Girl
5. Girl,Girl

Eliminate #5, and you have 2/4 or 50%.

If you don't consider order, there are only 3 options:
1. Boy, Girl
2. Boy, Boy
3. Girl, Girl

When you eliminate #3, it is still 50%.
Andrew Selkirk
Jun. 29, 2010 at 9:23am
• I happen to agree with Torbjørn AmundsenJun.

1) Boy, Girl
2) Boy, Boy
3) Girl, Boy

X= Boy
Y= Boy or Girl

So X is Always Boy, so that means number 3 is the same as number 1 since you know there is one boy. There is only a 50% change of getting a boy or a girl. When you are thinking logically you have to be careful not to allow your logic to become flawed as then it will skew and be far off from reality.

I will give you something to think about...

My dad had 2 boys, my brother had 2 girls, I had 3 boys, my wife's brother had 3 girls. What are the odds any of us would have another it would not be the one we already have.

Sorry the answer is still 50%.
Nathan Stanford
Jun. 29, 2010 at 8:00am
• This is a gambler's fallacy [wikipedia.org] problem. The more tangents you throw at it, the closer you get to .5 (50%), while never reaching it. This is the limit, why? Because there's only two potential outcomes for the other child: boy or girl.

What you (or the website you copied and pasted the ratio from) fail to take into account (and why it's a Gambler's fallacy problem) is that when involving chance, anything that happened in the past is completely irrelevant to future probables. I could roll a die 99 times, and get 6, the probability of getting 100 6's when I've already got 99 6's is still 1 out of 6, not 6^100.

The reason the chi square doesn't come into play here is because it doesn't MATTER the order. Has she said "What is the probability my SECOND-BORN was a boy?" it would be perfectly logical to write the square because the boy who was born on Tuesday could be either the first born or the second born, she never stipulated.

We can say that the boy, who was born on a tuesday, was also a Gemini. Does this change the ratio? No, the probability of having two boys is still 50-50%, because the unknown only has two possible outcomes: boy or girl.
Heathen Unbeliever
Jun. 29, 2010 at 7:51am
• X = A boy born on Tuesday
P(X|boyboy) = 1/7*1/7 + 1/7*6/7 + 1/7*6/7 = 13/49
P(X|boygirl) = 1/7
P(X|girlboy) = 1/7
P(X|girlgirl) = 0
P(boyboy) = P(boygirl) = P(girlboy) = P(girlgirl) = 1/4
P(X) = (13/49 + 1/7 + 1/7 + 0) * 1/4 = 27/196
Reverend Bayes:
P(boyboy|X) = P(X|boyboy)*P(boyboy)/P(X) = 13/49 * 1/4 * 196/27 = 13/27
=48.1%

From 33.3% to 48.1% just because of the Tuesday criterion? The key intuition here is that that Tuesday is a constraint on boys. A boy-boy reality is more likely to survive a boy constraint than a girl-boy or a boy-girl reality.
Jun. 29, 2010 at 7:47am
• I must be tired too. There seems to be a clear assumption in the articles math that may be wrong. Just because they discovered three different situations, what demonstrates they have equal probability of occurrence? Torbjorn's analysis appears to shows the boy,boy situation is 2X as likely to occur then the other 2 situations.
Dace Hummel
Jun. 29, 2010 at 6:59am
• This seems a bit flawed, but my first impression might not be trusted. The whole point is the order of birth, which suggest the following reasoning:

Boy mentioned, girl
Girl, boy mentioned
Boy, (boy mentioned)
(boy mentioned), boy.

It still equals 1/2.

Or am I too tired? :)
Torbjørn Amundsen
Jun. 29, 2010 at 5:48am
• Even with identical twins, one is still born first, and hence it fits into the above analysis.
Mystify
Jun. 29, 2010 at 5:21am
• "I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"

Do identical twins not exist in this maths universe? Considering what was ignored otherwise (statistical differences between rates of birth on weekdays and weekends), we can ignore it when twins are born either side of midnight, and say that if your son was born on a Tuesday, then the odds of his identical twin brother being born on a Tuesday is 100%.

Non-identical twins can also both be boys.

So your homework is to go onto wikipedia, find out the rates of birth of identical and non-identical twins, and allow for that in the answer.

BTW, I've only got an AS-level in maths, and that included no stats modules, so don't flame too hard ;)
e5rgyhbjnkmo e5rgyhbjnkmo
Jun. 29, 2010 at 5:02am
• The "(and only one)" qualification suggested by Ralph Dratman is _not_ required. Indeed, in the first case of the analysis, "older child is a boy born on Tuesday", the possibility that the younger child is also a boy born on Tuesday is explicitly included and counted. The hypothesis for the second case does exclude the possibility of both being boys born on Tuesdays. The two cases are mutually exclusive and exhaustive.

Note that if the puzzle had included the "(and only one)" qualification, then the possibility count would have been 13 (6 for boy and 7 for girl) in both cases, and the probability drops to 12/26.
David Vanderschel
Jun. 29, 2010 at 4:05am
• There is a mistake (a flaw) in the puzzle as it is stated here. In order to fit the proposed analysis, the problem should have read,

"... one (and only one) of whom is a boy born on Tuesday."

But the problem as stated ("one of whom is a boy born on a Tuesday") does not exclude the possibility that both children are boys and that each boy was born on a Tuesday.

Without that exclusion, we are back to the simpler problem, with 1/3 chance that both children are boys.
Ralph Dratman
Jun. 29, 2010 at 1:03am
• This article is shameful and full of errors. You can use past events to establish an a priori probability (like in average 1/2 of people are boy and 1/2 are girls) and use it determine the probability of future event (the probability that a new born in the future is a boy is 1/2 and a girl is 1/2) but you CANNOT use it to compute the probability of events that are part in the past and part in the future (the probability of having two boys is not 1/2*1/2=1/4 because one of the boys is already born so he is a boy with certainty). This is the first pitfall students are taught (or should be taught) in probability and statistics 101.
Massimo Di Pierro
Jun. 29, 2010 at 9:30am
• Okay, I need some help.

Can someone please point out what Tuesday has to do with the equation? There is NO remark about the probability of a second son born on a given day, just the question about the probability of a second son. Where in those two little sentences does it ask for the probability of a second son born on ANY of the other seven days? It seems like people are extending the equation without the logic to justify the act of doing so. Seriously, help me out.

~Q
Mr. Questnfool
Jun. 29, 2010 at 9:32am
• An inappropriate comment was removed from this discussion.
Eva Emerson
Jun. 29, 2010 at 11:59am
• How would things change if the problem said "Tuesday at 11:19am" or "Tuesday at 11:19am and 34.593499940000044523232 seconds"? Would the answer get asymptotically closer to 0.5000000?
Jerry Vandesic
Jun. 29, 2010 at 2:32pm
• Ok, that didn't work. the site cuts off the end of the comment if it's too long. I've posted the code I wrote at the following link:
g42 DOT org /xwiki/bin/view/Main/TwoChildProblem
Bob Smith
Jun. 29, 2010 at 2:39pm
• I didn't believe the results of the story, so I wrote a short Java application that duplicates what they are talking about. Believe it or not the result turns out to be 33% for the simple version of the two child problem!

The site won't let me post all the code (it's too long for a message apparently), so I've posted it at:
g42 DOT org /xwiki/bin/view/Main/TwoChildProblem
Bob Smith
Jun. 29, 2010 at 2:31pm
• Jerry: Exactly.
Jun. 29, 2010 at 3:03pm
• Much more simply, if I throw a dice, the probability of getting a 6 is 1/6, but if I throw a dice a 99 times and get a 6, the probability of getting a 100th 6 is not 1/6

So past occurrences and information about those does alter the probability of a future event. Its about the way you look at the problem. In my example, the probability of getting a 6 in the 100th turn, without considering the past results would still be 1/6, but it would be different from the "probability of getting a 100th sixes if I already have 99 sixes".

I am no mathematician, but here are my 2 pence of simple logic. I also did get a C+ in my math exam in grad school. So please be kind. :)
Prashant Singh
Jun. 29, 2010 at 3:00pm
• Maybe this is because I'm not a native speaker in English, but to me the chance of him having two boys is just 25% (g/g; b/g; g/b; b/b) I suppose I should read the question as: 'What is the probability that my other kid is a boy too?' But that's not what is asked. So 25% it is :-)
Jules Stoop
Jun. 29, 2010 at 2:04pm
• Heathen Unbeliever and Massimo Di Pierro are, in my estimation, correct. The probability is 50%, obviously and plainly. The "logic" used by Devlin is statistical sophistry, suitable for confusing freshman Liberal Arts students. Tuesday has nothing to do with it. The other child's gender has nothing to do with it. 50%. End of story.
Cletis
Jun. 29, 2010 at 2:04pm

If someone says, "I have two children. One of them is a boy." The odds of the other one being a boy is 1/3. If that confuses you, look up the Monty Hall Problem on Wikipedia.

If someone says, "This is little John. I have another child." The odds of the other one being a boy is 1/2.

Saying, "I have two children. One of them is a boy born on Tuesday," is a lot like saying, "This is little John." "Boy born on Tuesday" almost, but not completely, identifies the child, because there's a possibility that the other child was a boy born on Tuesday. That small possibility skews the odds slightly away from 1/2.