# Two-Child Problem (when one is a girl named Florida born on a Tuesday)

Estimated read time (minus contemplative pauses): 33 min.

A classic probability riddle goes:

A couple has two children, one of whom is a girl. What is the probability both children are girls?

It’s usually credited to Martin Gardner who, in a 1959 issue of Scientific American, posed essentially this question but involving two boys. It’s commonly solved by working through a sample space in which gender outcome is conveniently assumed to be 50-50: GG, GB, BG, BB

We know the couple has a daughter, so I crossed out BB. Each sibling-pair outcome has the same probability—i.e., 1/4—of occurring. One out of the three remaining outcomes satisfies the GG condition. We thus conclude: Given that one sibling is a girl, there’s a 1/3 chance both siblings are girls.

This result contrasts with the intuitive answer of 1/2, which many give on the faulty grounds that the unmentioned child’s gender is equally likely for boy or girl. It turns out, though, that there’s an alternative interpretation that leads to a valid 1/2 result, more about which shortly. My goal in writing this post is to help myself and others develop an intuitive understanding of this problem and its variations.

[NOTE: I’ve posted a shorter explanation of the Two-Child problem here.]

In his 2008 book, The Drunkard’s Walk: How Randomness Rules Our Lives, Leonard Mlodinow offers a variation: What if you learn one of the children is a girl named Florida? He claims this information changes the probability of both children being girls from 1/3 to practically 1/2.

At first glance, this result struck me as so counterintuitive that I rejected it outright. But then it occurred to me that the information could make a difference when interpreting the problem in the same way that gets 1/3 in the standard Two-Child problem—which is indeed the interpretation Mlodinow applies. I still couldn’t accept, however, that it would make any difference under the aforementioned valid alternative interpretation; more precisely: when randomly observing that one daughter in a two-child family is named Florida, the chance of two girls should remain exactly 1/2. I’ll demonstrate this mathematically in a moment.

What strikes me as especially interesting here is that learning about the name Florida updates the 1/3 answer to practically 1/2, which is closer to both the intuitive (though ill-founded) 1/2 answer and the valid 1/2 answer.

Before getting into the nuts and bolts of what distinguishes the 1/3 and 1/2 interpretations, I’m going to introduce a problem that’s similar to, but conceptually easier than, the “Florida” example. At the 2010 Gathering 4 Gardner convention, puzzle maker Gary Foshee presented this problem:

I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?The answer given is 13/27 (or about 48%). This answer is arrived at under the same interpretation that leads to 1/3 for the standard Two-Child problem. Notice that introducing the “Tuesday” information raises that 1/3 to nearly 1/2, just as with the “Florida” variation (though the “Florida” variation is closer to 1/2). And, once again, the aforementioned alternative interpretation leads to a valid answer of exactly 1/2, with or without learning about Tuesday.

These differing results depend on differing interpretations about how information is obtained. In the common interpretation, which leads to 1/3 in the standard Two-Child problem, information is gathered deterministically so that the sample space includes all two-child families that contain at least one girl (born on a Tuesday or what have you). In the alternative interpretation, which yields 1/2, the sample space forms randomly so that you’re getting all two-child families that have two daughters, but only half of all two-child families with mixed-gender sibling pairs.

Gardner, in his 2001 The Colossal Book of Mathematics (pages 277–282), notes the importance of “specifying the randomizing procedures” in probability problems, citing the Two-Child problem as a prime culprit: “Many readers correctly pointed out that the answer [to his original presentation of the problem in 1959] depends on the procedure by which the information ‘at least one is a boy’ is obtained.” A few pages later, he writes:

The problem of the two boys, as I said, must be very carefully stated to avoid ambiguity that prevents a precise answer. In my Aha, Gotcha I avoided ambiguity by imagining a lady who owned two parrots—one white, one black. A visitor asks the owner, “Is one bird a male?” The owner answers yes. The probability both parrots are male is 1/3. Had the visitor asked, “Is the dark bird a male?”, a yes answer would have raised the probability that both birds are male to 1/2.

In other words, there’s no obvious reason to favor one interpretation over the other in statements of the problem that don’t specify how information is obtained.

I’ll now go through several variations of the problem with different approaches for solving. I belabor these solutions in an effort to evade the lingering confusion that seems to trail most discussions I’ve seen of the problem. I’ll begin with a quick demonstration of the above “parrot” examples.

First, the visitor gets a “yes” to the explicit question of whether one bird is male. Where p = parrot; b = black; w = white; F = female; M = male, we rule out the FF case, leaving MM as one of three remaining possibilities, so the answer is 1/3:

pbF & pwF
pbF pwM
pbM pwF
pbM pwM

Next, the visitor gets a “yes” to the question of whether one is male and black, so we can eliminate yet another possible outcome, which updates the 1/3 to a 1/2 chance both are male. Notice that this information is particularly useful due to already knowing that one is black and one is white. This would be like learning the that the older of two siblings is a boy in the Two-Child problem (I revisit this point in the coin examples below).

pbF & pwF
pbF pwM
pbM pwF
pbM pwM

A Game: Here I’ll use theoretical frequencies involving 120 families. I’ve seen the claim made that the probability of the “other child” being a girl is either 1 or 0, due to that child’s already having been born. We can avoid this confusion by formulating the question as: “If you guess both children are girls, what is the probability you’ll guess correctly?” The following silly game bears out this formulation.

One hundred and twenty families with two children have been chosen at random. A mother will be randomly chosen to meet with you, and will tell you the gender of one of her children. You win by correctly guessing, in one guess, whether the mother has two children of the same gender.

The selected mother tells you, “one of my children is a girl.”

You run the numbers. When 120 families have their first child, 60 are boys and 60 are girls. When they have their second child, they get the same result. So, 30 of the first-born boys get sisters and the other 30 get brothers, while 30 of the first-born girls get sisters and the other 30 get brothers. You know this mother has a daughter, so you’ll cross off all two-boy sibling pairs. This leaves 90 pairs (in bold above). You divide that into the 30 girl-girl pairs (underlined above). That’s 30/90, or 1/3.

You thus conclude that a brother is twice as likely as a sister, and are about to make your guess. But then it occurs to you that you can evaluate this probability as “the chance the mother has two daughters given that she mentions a girl.” Assuming she truthfully follows the rules of the game: she’ll definitely mention a girl when she has two daughters; when it’s GB or BG, she might just as well mention a boy; when it’s BB, she won’t mention a girl. This means that, once she’s mentioned a girl, she’s twice as likely to be in the GG condition—a 1/4 chance to begin with—than either the GB (1/8, given that she could have mentioned a boy) or the BG (also 1/8) conditions. You divide the winning condition by the candidate conditions: (GG)/(GG + GB + BG) = (1/4)/(1/4 + 1/8 + 1/8) = 1/2. Not 1/3. Yikes.

To confirm this, you run through the expected frequencies again. Of the 120 families present, there will be 30 in each condition: GG, GB, BG, BB. You rule out BB. In the GG condition, you’ll be told 30 times of a girl. In GB and BG, you’ll hear “girl” 15 times each, for 30 total. That means you’ll hear a girl mentioned 60 times. You must be in one of those 60 conditions. Thirty of those are the GG condition. So guessing GG will be correct 30/60 times. That’s 1/2.

But then it occurs to you that if instead of the mother offering you information, you had asked her, “Is at least one of your children a girl?”, and the mother had said “yes,” you’d be back at 1/3. This is because by asking you are guaranteeing that a particular gender will be revealed, so this way you don’t lose half the GB and BG groups to the mother mentioning a boy, thereby further restricting the sample space.

You again confirm with expected frequencies. Of the 120 families, there are 30 each in GG, GB, BG, and BB. You rule out BB. In GG condition, you’ll hear “yes” 30 times. In GB and BG, you’ll hear “yes” 30 times each, or 60 times. That means you’ll hear a girl mentioned 90 times. You must be in one of those 90 conditions. Thirty of those are the GG condition. So guessing GG will be correct 30/90 times. That’s 1/3.

Here’s a diagram demonstrating the two approaches:

An instance of 1/3: For a study, you collect all the world’s two-child sibling pairs containing at least one girl. If you now randomly select a pair from that collection, what’s the chance the pair contains two sisters?

The theoretical expectation is that 1/3 of all pairs in the study will be GG. So the answer is 1/3.

When you explicitly ask a mother of two if at least one of her children is a girl, this is the sort of theoretical project you engage in. When you wait for such a mother to mention having a girl, you end up with a smaller proportion of BG and GB sibling pairs in your “study” (half as many in each group, in fact), because she could have mentioned a boy.

An Instance of 1/2: You know Mr. Smith has two children. You encounter Mr. Smith walking in the park with his son. What is the chance both his children are boys?

This common variation works the same as when the mother happens to mention her daughter in the game above:

GG: You will not see Mr. Smith walking with a boy.
GB: You could see Mr. Smith walking with a boy or a girl, presumably with even odds; so the probability of being in this scenario is half of 1/4, or 1/8.
BG: Same as BG: 1/8.
BB: You will only see Mr. Smith walking with a boy, so the chance you’re in this scenario is 1/4.

Now simply divide the satisfying scenario (underlined) by the sum of all the candidate scenarios (in bold):

(BB)/(BB+BG+GB) = (1/4)/(1/4+1/8+1/8) = (1/4)/(1/2) = 1/2

Here’s a probability tree showing the same thing:

Demonstrations with Coins: I’ve seen commenters argue that birth order isn’t important. If you’re in that group, you might find the importance more intuitive by considering coins, though this point isn’t obvious to everyone when it comes to coins either. As George Johnson points out in this 2008 New York Times review—”Playing the Odds“—of Mlodinow’s book: “The great 18th-century mathematician Jean Le Rond d’Alembert thought the answer [to how many outcomes are possible when flipping two fair coins] was obvious: there are three possibilities, zero, one or two heads. So the odds for any one of those happening must be one in three.”1 Johnson, by the way, expresses skepticism about Mlodinow’s “Friday” example—a healthy response, I think. Hopefully by the time we get back to that example, its nuances will be clearer.

Now consider the sample space when flipping two fair coins:

HH
HT
TH
TT

Each of these distinct results has a chance of HH = HT = TH = TT = 1/4.

Suppose I flip one coin and you see it land Heads. I then flip a second coin but hide the result. What is the chance the second coin also shows Heads? In other terms, what is the chance a guess of HH will be correct? (Notice that I get to assign a chance of 1 to what’s showing, as I can see the result.) You know the first throw was Heads, so you adjust the sample space as follows:

HH
HT
TH
TT

It can be either HH or HT, so its 1/2 that both landed Heads—in other words, it’s just the probability that the second coin came up Heads. This demonstrates the importance of order in the TH and HT cases (and, isomorphically, the BG and GB cases).

Now imagine I flip both coins hidden from your view. I randomly pick one of the coins and announce it to you: Heads. But I don’t tell you whether it landed first or second. (To be clear, they could land at the exact same time and this would still work, but I’m differentiating with time because it’s more intuitive and more like the BG, GB cases. To see this, imagine I’m using a quarter and a nickel. This is also like designating a white and black parrot above.) What is the probability both coins show Heads? Adjust the sample space:

HH
HT
TH
TT

As with Mr. Smith walking in the park, the probability here is 1/2: for HH, I could only mention Heads; for each HT and TH, I could have equally told you at least one landed Tails. So, there is a 1/4 chance we’re in the HH condition and a 1/8 chance each for the HT and TH conditions. As usual, we set the ratio as (HH)/(HH+HT+TH), and we get (1/4)/(1/2) = 1/2.

And again, had you asked, “is one of them Heads,” and I answered “yes,” the probability of your guessing “HH” correctly would be 1/3. In other words, were we to run the game over and over again, and each time you ask “is one of them Heads?”, 1/3 of those cases in which I answer “yes” will be HH cases.

As already noted, most presentations of this sort of problem assume 1/3 conditions, despite not specifying how information is learned. Here’s an example from Probability Demystified:2

Two coins are tossed. Find the probability of getting two heads if it is known that one of the coins is a head.

Solution: There are three ways to get at least one head. They are HT, TH, and HH. (Note: HH is included since the problem does not say exactly one head.) There is one way to get 2 heads: P = P(2 heads)/P(at least 1 coin is a heads) = 1/3.

The problem can also be done as follows:

P = P(2 heads)/P(at least 1 coin is a heads) = P(2 heads and 1 coin is a head)/P(at least 1 coin is a heads) = P(2 heads)/P(at least 1 coin is a heads) = (1/4)/(3/4) = (1/4)×(4/3) = 1/3.3

## A Girl Born on a Tuesday

The usual answer given to the “Tuesday” problem is 13/27. This, however, assumes a sample space restricted in the same way as in the generic 1/3 Two-Child cases. One way this can be demonstrated is to list out every candidate outcome, then take the ratio of the satisfying candidates to the overall candidates. You can see this done at the Blank Slate blog, where the author, Vishal, considers the same question but for a girl born on a Friday.

Instead, I’ll work it out with straightforward math that amounts to essential the same thing, and then with a tree diagram examining 196 families, accompanied by a step-by-step verbal account.  Then I’ll similarly demonstrate the 1/2 result. I’ll then end this section with an even more straightforward mathematical approach.

The 13/27 result using straightforward math:

Let:
g = younger sister
G = older sister
b = younger brother
B = older brother
t = born on a Tuesday
~t = born on some day other than a Tuesday
P(of something) = Probability of something

For example: P(gt and G~t) means “the probability of the younger sister being born on a Tuesday and the older sister being born on not-Tuesday.”

My goal here is to set a ratio of {the sibling pairs that satisfy the “two girls with at least one born on a Tuesday” condition} to {all those sibling pairs that contain at least one girl born on a Tuesday}.

For the denominator, I’ll add up the relevant probabilities—i.e., those cases in which at least one girl is born on a Tuesday:

P(gt and Gt) = 1/144×1/14 = 1/196
P(gt and G~t) = 1/14×6/14 = 6/196

P(g~t and Gt) = 1/14×6/14 = 6/196
P(Gt and b) = 1/14×1/2 = 1/28 = 7/196
P(B and gt) = 1/14×1/2 = 1/28 = 7/196

Add all these together for the denominator: 27/196

The numerator will be the sum of those probabilities that satisfy the girl-girl condition (in bold): 13/196

Now just set the ratio and simplify: (13/196)/(27/196) = 13/27

The 13/27 result using a tree diagram and considering theoretical frequencies for 196 two-child families:

In the above diagram, candidates are in bold, satisfying candidates are underlined. Theoretically, 196 couples’ first-borns will consist of 98 boys and 98 girls. Of those 98 boys, 49 will get sisters, 7 of whom are born on a Tuesday. Of the 98 first-born girls, 14 will be born on a Tuesday and 84 will be born on some other day. The 14 Tuesday-girls will get 7 brothers and 7 sisters. Of those 7 sisters, 1 will be born on a Tuesday, while 6 will be born on some other day. The 84 non-Tuesday-girls will get 42 brothers and 42 sisters, 6 of whom will be born on a Tuesday.

This comes out to 27 candidate pairs of which 13 satisfy the condition “two girls, at least one of whom was born on a Tuesday.” In other words: 13/27.

Remember, 13/27 is technically what you’ll get when you ask the question yourself (or apply some similar method), as with the 1/3 result before we applied the “Tuesday” evidence. Things change, however, when we are in a scenario in which you learn of the Tuesday-born girl by some random means. Then the answer is 1/2. To show this, I’ll follow the same steps as above.

First, find the denominator by adding up all the candidate conditions:

P(gt and Gt) = 1/14×1/14 = 1/196; in the condition, mother always mentions Tuesday-girl, so probability is 1/196.
P(gt and G~t) = 1/14×6/14 = 6/196; in this condition, mother mentions Tuesday-girl half the time, so probability is half of what we have here: 3/196
P(g~t and Gt) = 1/14×6/14 = 6/196; as directly above, keep half: 3/196
P(gt and b) = 1/14×1/2 = 1/28 = 7/196; again, keep half: 3.5/196
P(b and Gt) = 1/14×1/2 = 1/28 = 7/196; again, keep half: 3.5/196

Add all these together for the denominator: 14/196

The numerator will be the sum of those probabilities that satisfy the girl-girl conditions, which are in bold: 7/196

Now just set the ratio and simplify: (7/196)/(14/196) = 7/14 = 1/2

This time we did exactly as when starting with 196 families in the 13/27 condition, but we further restricted the sample space to reflect the probability of mentioning a boy or a non-Tuesday daughter:

We see that in the {Gt, gt} cases, only a Tuesday daughter will be mentioned. So we keep all such cases, of which there is 1.

In each of the six {Gt, g~t} cases, the mother could have mentioned the g~t daughter. So we keep half: 3.

Similarly, we keep half of the {G~t, gt} groups: 3.

Those counted so far constitute our satisfying candidates. Their sum is our numerator: 7.

We now count half of the remaining candidate groups: 7.

The denominator is the total of all candidate groups: 7+7=14.

We end up with: 7/14 = 1/2.

And there we have it.

Now I’ll demonstrate a quicker way to get to the same results in the 1/3-scenario. I’ll move quickly, as I think anything I say here can be intuitively verified against what we did above. We want to find the fraction of cases in which there are two girls, at least one of whom is born on a Tuesday, and then work out what proportion that fraction is of all cases in which at least one of two siblings is a girl born on a Tuesday. In other words, this is just a straightforward conditional probability calculation:

$\displaystyle \frac{\textrm{P(two girls AND at least one of those two girls is born on a Tuesday)}}{\textrm{P(at least one of two siblings is a girl born on Tuesday)}} =$

$\displaystyle \frac{\textrm{P(two girls)} \times \textrm{P(at least one of those two girls is born on a Tuesday)}}{\textrm{P(at least one of two siblings is a girl born on Tuesday)}}$

I could further refine this notation (e.g., to relate it to Bayes’ theorem, which I’ll briefly mention again below), but what we have now is plenty for arriving at an intuitive solution. For the numerator, we need two probabilities. The first is easy. The probability of two girls is 1/4.

Then we need the probability that at least one of those girls is born on a Tuesday. This is just the complement of the probability that neither of them is born on a Tuesday: (probability the first child not born on a Tuesday) × (probability the second child is a girl and is not born on a Tuesday) = (6/7) × (6/7) = 36/49. This is the fraction of scenarios that fail to meet our desired condition of there being at least one of two sisters born on a Tuesday. So we subtract this from 1 to get our desired condition: 1 – 36/49 = 13/49.

So our numerator is (1/4) × (13/49). (I’ll leave it unsimplified.)

Usually, the quickest way to find the probability that “at least one” of something to happens is to find the probability that it happens zero times and subtract that from 1. We’ll do this to find the denominator as well, where we need the probability that at least one of two siblings is a girl born on Tuesday, including when it’s either two boys or a boy-girl mix. That’s just: (probability that the first child is a boy OR is a non-Tuesday girl) × (probability that the second child is a boy OR is a non-Tuesday girl). Both of these will be the same probabilities, so we can just find the first one and square it.

This works out to: (probability the child is a boy) + (probability the child is a non-Tuesday girl) = 1/2 + (1/2)(6/7) = 1/2 + 6/14 = 13/14. That’s the number we need to square: (13/14)2 = 169/196. Recall that this is the probability of failing to get what we’re looking for, so we must take the complement: 1 – 169/196 = 27/196. And that’s our denominator.

We now just plug those in. I’ll go through all those steps again so it’s clearer:

$\displaystyle \frac{\textrm{P(two girls)} \times \textrm{P(at least one of those two girls is born on a Tuesday)}}{\textrm{P(at least one of two siblings is a girl born on Tuesday)}} =$

$\displaystyle \frac{(1/4) \times (1-(6/7)^{2}}{1-(13/14)^{2}} = \frac{(1/4) \times (1-(36/49)}{1-(169/196)} =\frac{(1/4) \times (13/49)}{(27/196)} =$

$\displaystyle \Big(\frac{1}{4}\Big) \times \Big(\frac{13}{49}\Big) \times \Big(\frac{196}{27}\Big) = \frac{13}{27}$

## A Girl Named Florida

According to Mlodinow, 1/3 is the answer to the generic Two-Girl problem. After further restricting the sample space on account of the name Florida, he updates that 1/3 to (practically) 1/2.

Had he applied the alternative—i.e., the “happens to mention” or random—interpretation, those probabilities would have been 1/2 and 1/2, respectively. Mlodinow does not discuss the fact that different ways of obtaining information may yield different sample spaces.

I won’t demonstrate that math. You can find an analysis in a 2010 paper by Stephen Marks & Gary Smith: “The Two-Child Paradox Reborn?” They conclude, “Thus, if the mother mentions a girl, the probability of two girls is one-half, no matter what her name is. The answer is the same as in our original Bayesian analysis: there is no paradox” (page 7). They also indirectly make the point that obtaining information in different ways may yield different sample spaces, noting that “Mlodinow does not answer the question that he poses, about the probability that a given family will have two girls, if we learn that it has a girl named Florida. Instead, he gives an approximate answer to a different, hypothetical question: among all BG, GB, and GG families that have a daughter named Florida, what proportion are GG?” (page 7).

I recommend reading Mlodinow’s chapter (and indeed entire book) to get the most out of the discussion. As Marks & Smith point out: “despite these issues, his analysis is instructive” (page 7).

To be clear, the “issues” they refer to are less about Mlodinow answering “a different question” and more about him not specifying how information is obtained in his initial statement of the problem. In other words, he starts with a generic statement of the problem, which, as I and others (e.g., Gardner) have repeatedly noted, poses ambiguity about whether it is a 1/3 or a 1/2 circumstance. Mlodinow assumes a 1/3 interpretation, while Marks & Smith assume 1/2.

Interestingly, Mlodinow goes on to give a more detailed account in which several families are asked to stay or leave depending on whether they satisfy a given criterion (e.g., having a daughter named Florida). This is indeed a scenario demanding the 1/3 approach. Marks & Smith, however, begin their critique of Mlodinow with the story of a mother who “mentions” the gender of one of her two children. In other words, they concoct a valid 1/2-type circumstance in order to critique Mlodinow’s interpretation, which, at least in its more detailed form, amounts to a valid 1/3-type account.

I think both writings would be improved by explicit discussion of the two interpretations.

## Conclusion Part I: Is there a paradox in going from 1/3 to 1/2?

Suppose you meet a new neighbor. She mentions her two kids. You ask, “is one of them a girl?” If she says “yes,” is it really true that your probability for correctly guessing that both children are girls will be different than had you gotten a “yes” to “is one of them a girl born on a Tuesday?”—i.e., 1/3 versus 13/27, respectively? It seems it is.

The vexing thing here, however, is that the mother is going to tell you some day of the week, so might you just as well always assume the 13/27 from the start?

That’s a terribly thought-provoking question! The answer is no.

One way to see this is to imagine having a list of all the two-child sibling pairs in the world. Getting confirmation of a girl would allow you to cross off all the world’s two-boy pairs. A third of those left on the list will be two-girl pairs. If you then get confirmation that one is a girl born on Tuesday, you will be able to additionally cross off all pairs not containing a girl born on Tuesday.

You don’t have such a list, but you can calculate theoretical frequencies based on relevant probabilities, which do indeed yield the probabilities in question.

Perhaps this isn’t enough to develop an intuition about the problem.

A start for doing so is to notice that you’re only guessing at whether there are two daughters when the mother confirms that she has a daughter born on a Tuesday. Were you to ask many such mothers, most would say “no.” Notice that a “no” means you can now cross off all pairs that have a girl born on a Tuesday, but the children might still be BB. A “no” to “is one of them a girl?” tells you it is a BB pair, which is to say it gives more information than a “no” to the “Tuesday” question (the opposite of getting a “yes” to those questions).

In other words, these probabilities are conditional on correctly guessing that the mother has at least one girl versus correctly guessing that she has at least one girl born on Tuesday. Were you to start asking these questions of all the two-child mothers you meet, you’ll get a “yes” to the first question 75% of the time, but a “yes” to the second only about 14% of the time. One in three of the former will be GG cases, while 13 in 27 of the latter will be.

Remember, if the parent just randomly mentions, “I have two children, one of whom is a girl born on a Tuesday,” there is precisely a 1/2 chance both will be girls. The potential paradox under consideration applies only to scenarios in which you’ve explicitly asked that question and gotten a “yes” (or have similarly deterministically restricted the sample space).

So, we see that there is no paradox here. You simply have to take into account how many times you’d find yourself in the relevant situation, and then calculate the expected frequency of two-girl pairs in that situation.

Finally, this might be made easier to understand by noticing that, as you get more information, the probability that both are girls gets closer to the intuitive answer of 1/2, especially when the chance for a “yes” is small. This aligns with the intuitive observation that “one is a girl” gives less information than “one is a girl born on a Tuesday.”

Let’s test this with the low-chance scenario of at least one of two children being a girl born on a Tuesday at 8:15 and 22 seconds p.m. (One out of 1,209,600 newborns will be girls in that condition, though, admittedly, this presumes some hypothetical standard for determining an event, down to the second, that counts as being born.) I did some quick calculations. As expected, for the “You Ask” circumstance, I got practically 1/2, and in the “They Happen to Mention It” circumstance, I got exactly 1/2. I popped the numbers into Desmos with a convenient denominator (you’ll derive these fractions if you start with 9,676,800 families):

It’s tempting to generalize this to show that “They Happen to Mention It” cases tend to 1/2 as you get more information, and the “You Explicitly Ask” cases always come out precisely 1/2. You can see the former done in the Wikipedia entry on this topic: “Boy or Girl Paradox,” where the “Tuesday” example is plugged it into Bayes’ theorem (thus streamlining much of the sort of work I’ve been doing here) to get 13/27:

“If ε [i.e., the probability of being born on a Tuesday] is now set to 1/7, the probability becomes 13/27, or about 0.48. In fact, as ε approaches 0, the total probability goes to 1/2, which is the answer expected when one child is sampled (e.g. the oldest child is a boy) and is thus removed from the pool of possible children. In other words, as more and more details about the boy child are given (for instance: born on January 1), the chance that the other child is a girl approaches one half.”

After simplifying, the resulting formula (which also shows up in the aforementioned Marks & Smith paper) is (2–ε)/(4–ε). It clearly goes to 1/2 as ε approaches 0, and to 1/3 as ε approaches 1.

This is a lovely abstraction that must of course be applied with care, so that ε represents some class of properties that seem to provide no significant evidence for the gender of the mentioned child’s sibling. You would not, for example, plug in the probability of having a sibling named Elizabeth, or who’s in the girl scouts, or who’s pregnant. Learning such things about a mentioned child’s sibling must yield a greater than 1/2 probability for that person being a daughter. Proper application of Bayes’ theorem, not to mention common sense (though common sense is what we’re attempting to align with formal probability!), will take care of these. Obvious, perhaps, but I figured it bears mentioning.

That in mind, I’ll close this section with the the following observations:

A neighbor gets on the elevator and mentions having…

…two kids. Chance both are girls is 1/4.

…two kids and one is a girl. Chance both are girls is 1/2.

…two kids and one is a girl born on a Tuesday. Chance both are girls is 1/2.

…two kids and one is a girl born on a Tuesday at 8:15:22 pm. Chance both are girls is 1/2.

…two kids and one is a girl named Florida. Chance both are girls is 1/2.

A neighbor gets on the elevator and you ask if she has…

…two kids. Chance both are girls is 1/4.

…two kids and one is a girl. Chance both are girls is 1/3.

…two kids and one is a girl born on a Tuesday. Chance both are girls is 13/37, which is getting quite close to 1/2.

…two kids and one is a girl born on a Tuesday at 8:15:22 pm. Chance both are girls is 2419199/4838399, which is .49999989666, or practically 1/2.

…two kids and one is a girl named Florida. Chance both are girls is practically 1/2 (per Mlodinow’s analysis).

## Conclusion Part II: Is there a paradox in ever getting 1/3 rather than 1/2?

At the start of this post, I noted that 1/3 result “contrasts with the intuitive answer of 1/2, which many give on the faulty grounds that the unmentioned child’s gender must be equally likely for boy or girl.” But is it really faulty, or does the intuitive answer reveal problems with formal interpretations? More precisely, is it problematic here to relate single instances of a trial to expected frequencies across several such trials?

Consider a variation on one of the coin examples. I flip two fair coins, keeping both concealed from you. You ask, “Is one Heads?” I say “yes,” and I show you the coin that landed Heads. What is the probability the still-concealed coin is also Heads? It’s 1/3. One way to bear this out is to imagine running this experiment 120 times. You’ll expect to hear “yes” and see a Heads 90 times, 30 of which will be in an HH condition. In other words, I’ll be concealing a Heads 30 times and concealing a Tails 60 times.

So, the concealed quarter will be Heads 1/3 of the time over several runs of the game. That’s clear. But what also seems clear to many people is that, in any single instance of the game, I’m concealing a single coin, and the chance it landed Heads should be 1/2, not 1/3.

Someone might try to support the latter point by noting that the sample space becomes restricted to {“the seen coin is Heads, the concealed coin is Heads”, “the seen coin is Heads, the concealed coin is Tails”}. Landing Heads or Tails is 50-50, so those two outcomes have equal chances, making it 1/2 that both are Heads. That seems valid. But it’s not, because the concealed coin doesn’t have even odds for being Heads or Tails! Remember, when you flip two fair coins, a Heads-Tails mix is twice as likely as getting two Heads. So, assigning even odds would be like assigning 1/2 to the chance of rolling a 4 on a fair die, on the grounds that you’ll either get a 4 or you won’t. People become confused in the coin instance, however, due to an independent throw of a quarter being 50-50 for Heads-Tails.

To see this more clearly, consider three observations:

First, the coin results are already settled. It is indeed 1/2 for each coin as it’s being flipped. But, because a Heads-Tails mix is more likely, once both coins have been flipped, you’re more likely to have a concealed Tails than a concealed Heads. This experiment is about the probability of your guess being correct about multiple events occurring, not about the probability that a given coin landed Heads in the first place.

Second, suppose you’ve already flipped 100 quarters once each, and you have set aside all the instances of Tails. Let’s say there are 50 Tails (there’s only about an 8% chance of that happening in real life, but we’re using theoretically expected frequencies). Now, suppose you grab two random coins from that set of 50 Tails. What is the probability both are Tails? It’s obviously 1. Something similar occurs above, but it’s less obvious. Above, you’re pulling two quarters out of a set of settled outcomes where there are twice as many Heads-Tails mixed pairs than there are Heads-Heads pairs.

Third, suppose we do the experiment as previously described, where I flip two coins concealed from view, and you ask if one is Heads. When I say “yes” and show you a Heads, the still-concealed coin has a 1/3 probability of also being Heads. But suppose I then re-flip the concealed coin. The probability of its being Heads is now clearly 1/2, as this restricts the sample space to {HH, HT}. (The same is also true, in fact, if I re-flip the visible coin.) But, without re-flipping, the sample space remains {HH, HT, TH}, as you’ve been given no reason to eliminate the TH outcome.

In other words, and in simplest terms, in a single run of the experiment, you’re more likely to get one Tails than no Tails. There is no paradox here.

## Post Script: A further ambiguity?

The following strike me as 1/3 cases, but not always obviously so. Perhaps some are 1/2 cases, or are ambiguous. If I become fully convinced they’re all 1/3, I’ll delete this section.

Mr. Smith Takes His Son to a Father & Son Picnic: You know Mr. Smith has two children. You see him with his son at a Father & Son picnic. What is the probability he has two sons?

Throughout this writing, I’ve relied on theoretical frequencies. But this approach only feels obviously right to me when it makes sense to talk about many distinct, though significantly similar, events—e.g., when surveying many random two-child families.

In the case of Mr. Smith walking in the park with his son, theoretical frequencies—represented in a tree diagram, contingency table, simulation, etc.—might be said to involve encountering some random two-child parent walking with a child hundreds of times. Not always Mr. Smith, but many different parents of two. More generally, we might consider the problem to be about learning that some randomly encountered two-sibling pair includes at least one son, regardless of context.

Context is significant in many cases, however, in terms of whether one would generally expect to see a son or daughter. Walking in a park or being observed at a window are apparently neutral contexts. Seeing Mr. Smith and son attending a Father & Son picnic is not. And so the probability that both his children are boys is 1/3 (though perhaps it’s not the best example: might we expect him to take both sons if he has two?). I feel fairly confident of this result, but less confident when context is less clear.

The Mother Who Always Mentions Her Daughter: Your new neighbor steps into the elevator and says, “One of my two children is a girl. This is what I always say when I speak of my children.” What is the chance she has two daughters?

This case provides a context that suggests a daughter will always be mentioned, but perhaps this is ambiguous. There’s nothing otherwise special about the context in terms of suggesting that a particular gender will be mentioned. Furthermore, the possibility for such biases is always present, even when they go unannounced. In standard formulations of the Two-Child problem it’s assumed this averages out: some parents may have a bias for mentioning a son, some for a daughter, and most probably have no bias. And so we indifferently assign even odds to mentioning a son or daughter.

But how to interpret a scenario in which the parent announces a bias? Would a simulation consist of 120 mothers who always mention their daughter? Or, rather, would it consist of 120 mothers, half of whom have a bias for mentioning their daughter, the other half for their son (provided they have one)? The former gives 1/3, the latter 1/2.

There are moments when it strikes me as obviously 1/3, and I feel I should delete this Post Script. But then doubt reemerges.

Perhaps it would help to think in terms of coins.

The Flipper Who Wants to Reveal a Heads:I flip two coins concealed from your view. I tell you I’ll reveal that one landed Heads, provided one lands that way. Otherwise, I’ll reveal a Tails. I then reveal a Heads. What is the chance both landed Heads?

Clearly, the probability here is 1/3. Were I to tell you that one landed Tails, you’d know both are Tails. Of the three remaining, equiprobable options, one is HH. This ratio of HH to HH+TH+HT would be borne out over many runs of the game.

Suppose, however, we play a version of this game in which I don’t tell you about my intention to reveal a Heads. From my perspective, I know you should assign a 1/3 probability to HH. But from your perspective, you assign 1/2, given that, as far as you know, I could reveal a Tails when one coin lands that way. Were we to play this game several times, you might start to notice my bias. But we play just once.

This variation exemplifies the point that, when determining probabilities, we work with the information we have. Indeed, even with something as basic as tossing a single quarter or die, we assume those objects fair until observed results indicate otherwise. Something similar goes on with the Two-Child problem when we assume Mr. Smith could have gone to the park with his daughter (a context crafted to be neutral, I admit), or that a mother could have equally mentioned her son or daughter given a neutral context. Still, this assumption underlies much debate about probability, and perhaps is in some large part responsible for the counterintuitiveness of probability and for my lingering doubt about whether some of the examples in this Post Script should be 1/3 or 1/2. Ultimately, it’s something I tend to belabor because I’m not as interested in how to solve probability riddles as I am in what those riddles tell us about how we should deal with chance and uncertainty in the real world.

That said, let’s return to the single instance in which I flip two coins and reveal a Heads, while telling you I intend to reveal a Heads. I put the solution at 1/3. But it’s not clear to me that this can be analogized to the Two-Child problem in which a mother intends to reveal a particular gender. It’d be absurd to run a simulation in which a mother has two children repeatedly while hoping each time to be able to reveal to random strangers that one of her children is a daughter. Rather, we imagine encountering many mothers who already have two children.

Given this, I wonder whether a coin-based simulation of that problem should involve 120 instances of a flipper (whether the same or distinct flippers) who wish to reveal a Heads (resulting in a 1/3 result), or should involve 120 flippers each of whom has some bias, split 50-50 for Heads and Tails (resulting in 1/2).

What creates this difficulty (for me) is that the mother in question already has two children and won’t exactly have an ongoing wish to reveal a particular gender so much as she’s developed a bias or habit (that may often go unspoken) to reveal a particular gender, for whatever reason. This may amount to a kind of long-term context, or it may reflect an attitude the mother adopted recently.

These points alone push me towards a 1/3 solution, but I hesitate because we know such unspoken biases may exist when concocting 1/2 scenarios, but we assume they average out. This seems particularly true when it comes down to a cognitive state that the parent is generally disposed to have, rather than an external context that promises a particular gender will be mentioned or observed.

The question boils down to: Do the biases still average out when they are spoken, or do we take the spoken bias to indicate a situation in which that specific bias is always spoken? If the latter is true, then we are applying a principle of indifference in the unspoken case that makes me wonder if the fact that, for example, Mr. Smith has gone to the park with his son might be appropriately counted as evidence that Mr. Smith has a tendency for taking his son, rather than daughter, to the park (due to some personal—e.g., cognitive or affective—bias, and/or circumstances that practically amount to a bias). This would increase the chance of seeing Mr. Smith with a boy in the BG and GB conditions. I won’t explore that here, but will note that one of its difficulties would involve estimating the proportion of parents who have such a (spoken or unspoken) bias, as well as how this affects the frequency of learning about a certain child (e.g., Mr. Smith might take his son to the park despite a bias towards taking his daughter). Speaking of which… on to the last item.

Another Game with 120 Mothers: You are in a real game with 120 mothers, each of whom will presumably go either way when mentioning a daughter or son (provided she has both). But one mother—perhaps the first, final, or someone in the middle—you meet expresses that she always intends to speak of a daughter. What is the probability she has two daughters?*

(*In this example, I’m wandering yet closer to the edges of my intuition about this problem and probability in general. Perhaps it’s not worth stating at all, but I’m not sure.)

To evaluate, you might ask yourself: Do I run numbers for a simulation in which two-child mothers who have at least one daughter always intend to reveal a daughter? This would yield 1/3 for both being daughters. Or do I assume that each mother has such an intention, but what that intention is could equally go either way gender-wise, given mixed-gender sibling pairs? This would balance things back to 1/2. Or do I run numbers for a game in which 1/120 mothers have this sort of intention, given that this best reflects the data observed in the world before me? In which case maybe I stick with 1/2, as the 1/120 doesn’t seem to make much difference.

Were I to relate this to the coin example, would it be like having 120 flippers, and only one of them expresses a wish to always reveal a Heads (when one lands that way)? And so on.

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#### Footnotes:

1. D’Alembert also fell prey to the Gambler’s Fallacy, according to his Wikipedia entry.
2. I recommend it as a quick intro to probability. Succinct chapters, lots of focused practice problems, interesting sidenotes.
3. Bluman, Allan. Probability Demystified (2012, second edition), pp 66-67. McGraw-Hill Education
4. This is the P(being born a girl)×P(of being born on a Tuesday)=(1/2)×(1/7)=1/14.

## 6 Replies to “Two-Child Problem (when one is a girl named Florida born on a Tuesday)”

1. JeffJo says:

You can use Bertrand’s Box Paradox to address the Two Child Problem. After all, the only difference between the two Problems is the addition of a fourth box. But the paradox Bertrand referred to was not the problem itself, it was how to tell that an answer some people like can’t be accepted.

“A couple I know has two children, but I only know the gender of one. What is the probability both children have the same gender?”

The gender that I know can only be “boy” or “girl.” If it is “girl,” the question is identical to your first one so it must have the same answer. And if it is “boy,” the question is functionally equivalent, and again must have the same answer. So I don’t actually need to tell you the gender for you to give an answer.

The probability that a family of two children includes only one gender is 1/2. The fact that I know one can’t change that to 1/3 because I have given you no pertinent information about the genders. Changing would be a paradox, so it can’t change. The only acceptable answer is 1/2.

I didn’t say it was conclusively correct. As Martin Gardner pointed out, the problem statement is ambiguous if we are not told how we learned about one gender. But we can conclude, by reductio ad absurdum, that no other answer can be correct.

2. JeffJo says:

Great article. It addresses many of the same points I’ve raised for years, ever since reading Mlodinow’s book. But I have another complaint about his analysis, that you didn’t address. He implicitly assumes that there can be two girls named Florida in that same family. In some discussions later, he explains how the contribution of the FF family depends on f^2, where f is the fraction of girls named “Florida,” so its inclusion in his solution (which is (2-f)/(4-f)) is minor. The problem with this is that the other combinations he considered (BF, GF, etc.) all depend on f, so one factor of f divides out of his quotients. So while the two-Florida family’s effect is small, it isn’t as small as Mlodinow claims.

But there is another hidden effect of this assumption. He assumes that every name can be duplicated. To see how this affects Mlodinow’s solution, imagine a village where only three girls’ names are used: Mary, Mary Ann, and Mary Beth. “Mary Ann” and “Mary Beth” are equally likely to be chosen, but “Mary” is twice as likely as either. So for a first daughter, “Mary” has a 50% chance and the other two names each have a 25% chance.

But what about the second daughter? In the 50% of families where it is possible to name her “Mary,” there is a 2/3 chance that she will be. This makes a 5/6 chance that a two-girl family has a “Mary.” Similarly, there is a 7/12 chance that a two-girl family has a “Mary Ann,” and the same for “Mary Beth.”

Mlodinow’s solution for the uncommon name “Mary Ann” is (2-1/4)/(4-1/4)=7/15. But it should be (7/12)/(1/4+1/4+1/12)=7/13. This is not less that 1/2 as Mlodinow claims, it is greater. The reason is that two sisters are more likely to include a Mary Ann, than a pair of unrelated girls.

1. Dan Jacob Wallace says:

Thanks for this, JeffJo. The Mary A/B example is fascinating.

3. JeffJo says:

I appreciate the reply. If you are ever bored, it isn’t hard to derive a general formula based on the Mary’s. Say you have a set of names N, which map to a set of frequencies F.

Use all the normal assumptions of independence to get an expression for the probability of the second daughter’s name. Two terms that fall out of it are:

S(n):=F(n)/(1-F(n)). This can be thought of as measuring a name’s “power;” that is, how it, as a first daughter’s name, affects the probabilities for the younger.

S0:=sum(all n, S(n))-1. It’s kind like an average frequency for a name, so “common” names have S(n)>S0, and uncommon ones have S(n)<S0.

Then the probability of two girls, given that there is a girl with name n, is (2+S0-S(n))/(4+S0-S(n)). Compare this to Mlodinow's (2-f)/(4-f).

4. EdwinA says:

Great post. Can’t say I have digested all of it yet. But I was so struck by the girl Florida problem when I read Mwlodinov’s book at the time It seemed so counterintuitive. Now I understand why. Giving such importance to a rare name, is like giving importance to the date and time of birth. You were probably the only one born in a specific second in a particuluar country, so it makes you quite unique. But then everyone has a very rare property, but if everyone has it, it can’t mean anything!

5. Edwin, you just re-cast Bertrand’s Box Paradox (the actual paradox I described above, not the problem itself) to this problem. And summarized it quite succinctly.