I don't like this puzzle but I think it is fair to share it with you because somewhere on your life journey you will meet or hear or read someone challenging you to this. And then - you will impress everyone with the right explanation.

A person said: "Of my two children at least one is a boy."

What is the probability that both are boys?

Image by plushoff, distributed under CCL.

A person said: "Of my two children at least one is a boy."

What is the probability that both are boys?

Image by plushoff, distributed under CCL.

Answer ideas accepted any time until midnight on Saturday November 19th (EST), on our Family Puzzle Marathon. They will be hidden till then and everyone who submitted something reasonable will get a puzzle point.

## 36 comments:

Answer: 1/3

With two children, there are four possible combinations of Boy and Girl:

BB

BG

GB

GG

Since at least one is a boy, GG is not a possibility. Three equally likely combinations remain, one of these both boys. So the chances of both being boys is 1/3.

Let's define the event A: having a girl and the event B having a boy with same equal chance of happening and independent events.

We could have

Girl. Girl

Girl. Boy

Boy. Girl

Boy. Boy

The probability of G/G, G/ B, B/B, B/ G is 1/4

Now,

Prob of at least a boy. Prob both boys

girl. Girl

Girl. Boy

Boy. Girl

Boy. Boy

I couldn't complete my answer on the same window so this is the rest....

The inside of my table is

0. 0

1/2. 1/8

1/2. 1/8

1. 1/4

The probability that they are both boys is 1/8 +1/8+1/4 which is 1/2

Sorry for that.

I was so hoping to do some statistics this week end ( irony). Thanks for the puzzle.

This the following of my answer. I could not continue to work with the first window!

The inside table is

0. 0

1/2. 1/8

1/2. 1/8

1. 1/4

The probability to have both boys is 1/2 ( adding the probabilities of the last column)

I was far of dreaming to do some statistics during this week enk but thanks for the puzzle, it is refreshing.

There are several answers, I think.

Ignoring the statement of the parent person since it may not be true, the probability of having two boys would be about 1 in 4.

Accepting the statement of the parent as true, then the probability of the other child being male is about 1 in 2.

However it could depend, also, what country they're living in, since there are variations in m/f birthrates, and sadly infant mortality and infanticide.

I think if I'm going to get one wrong, it is likely this one. I'll post again if I change my mind.

The tempting answer is 1/3. But that only uses B G, G B, and B B The last one is the successful one out of three. It's the one I'd choose if I was talking about my own family.

However the binomial theorem (probability) offers us another far fetched solution. I don't like it much because it does not appeal to my logic.

The trick here is interpretting the meaning of "at least" mathematically. Your "success" can be one boy or two. Using the binomial distribution, you get two calculations.

Calculation one "at least one"

Trials = 2

success = 1

(2 C 1) * P^1* (~ p)^1

p = 1/2 meaning a boy or girl is equally likely.

so ~p = 1 - p = 1 - 1/2 = 1/2

2 * 1/2 * 1/2 = 1/2

Calculation 2

Trials 2

sucesses 2

(2 C 2) * (1/2)^2 * (1/2)^0

1 * 1/4 * 1 = 1/4

The total success rate = 1/2 + 1/4 = 3/4

If you ever got an anti intuitive result, this is it. As Mark Twain said

"There are lies, damn lies, and statistics."

There are 3 possibilities for the two kids

- 2 girls

- 2 boys

- 1 boy, 1 girl x 2

When a person says that at least one of their kids is a boy, that gives you information that eliminates one of the possibilities (ie. 2 girls).

Therefore, there are 3 possibilities left and of those, only 1 matches.

1 / 3 chance that the person has 2 boys.

(This is Dennis, of Dennis and Katrina)

We know there are two children. The possibilities for this are:

B-G

G-B

B-B

G-G

We know that at least one is a boy, so that eliminates the last possibility (G-G), leaving us with:

B-G

G-B

B-B

With three possbilities and one that meets the criteria, the probability is 1/3.

(Assuming, of course, that the probabilty a baby is a boy is the same as the probabilty of a baby being a girl)

If you have two children, you have four possibilities:

Two boys

Boy first, then girl

Girl first, then boy

Two girls

These possibilities we can denote as

BB, BG, GB, GG

As you can see, if you have at least one boy, you have only three possibilities:

BB, BG, GB

and only one of those three cases is two boys, so

given that you have at least one boy, the probability that both children are boys is 1/3.

The answer to the puzzle is intended to be 1/3.

We would have to assume that of all 2 child families, that there are exactly 1/4 of them with 2 boys, 1/4 with two girls, 1/4 with a boy older than a girl, and 1/4 with a girl older than a boy.

Of these 4 equally likely possibilities, only 3 of them have "at least one boy." Of those 3 equally likely possibilities, only 1 has both boys.

Although I don't have a source to quote, I believe that in reality families with 2 children have the same gender more than half the time, as the combination of many variables involving each specific father and mother systematically favor one gender over the other.

the chance of getting two boys is 50%.

NAW

Any possible answer to this problem involves a number of assumptions, so I think the problem has no definite solution.

If you have two children and one in a boy there are four possibilities (and we assume that generally for you as a parent of two children, each of these possibilities is equally likely)

(a) Ist child: boy, second child: boy

(b) 1st child: boy, second child: girl

(c) 1st child: girl, second child: boy

(d) 1st child: girl, second child: girl

(1) If it is assumed that in any of the cases where you have a boy (cases a, b, and c), you are equally likely to say, "I have at least one boy," then the probability that you have two boys if you say you have at least one boy is 1/3.

2) An alternative assumption that could be made is that you no matter what your distribution of children (a, b, c, or d above), you always say you have "At least one boy/girl". Assuming no further preference towards boys or girls when you make this statement, if you have one boy and one girl, you say you have "At least one boy" half of the time (probability 1/2) and you say you have "At least one girl" half of the time (probability 1/2). If you have two boys, you say you have "At least one boy" all the time (probability 1). Then the probability that you have two boys if you say you have at least one boy is 1/2.

3) If you do have a preference towards girls over boys, then you could be likely to say that you have "At least one girl" in cases b, c, and d above, and only say that you have "At least one boy" when you have two boys (case a). With this assumption, if you say you have "At least one boy" then there is 100% probability that you have 2 boys.

4) If you do have a preference towards boys over girls, then you could be likely to say that you have "At least one boy" in only in cases b or c, and if you have two boys, you would probably say so, in which case, the probability that you have two boys when you say you have "At least one boy" is 0.

I vaguely remember seeing a similar type of problem before, so if I happen to be correct I don't know if I deserve full credit :-). Anyway, I think the way to think about probability problems of this sort is to list all possible specific scenarios. With two kids, we have four scenarios (and the order matters because each kid has their own probability of being of one or the other gender). Here they are: boy+boy, girl+girl, boy+girl, girl+boy. Since the person said at least one is the boy, we remove girl+girl case, leaving the total of 3, out of which only one corresponds to both kids being boys. therefore the probability of both boys is 1/3. I believe that if we replace kids gender question with simple coin tosses, the problem is still equivalent and the answer is the same.

I warned you that it is an obnoxious puzzle.

First trouble is that the intuitive answer of 1/2 is not correct. One is already a boy; and there is a 50/50 chance of another child being girl or boy. So, it should be 1/2. Nope, cross this out.

Look at the bigger picture. Most of you offered four two-kid scenarios:

GG

BB

GB

BG

If at least one is a boy GG is not an option, we are left with 3 and only one - BB - gives the desired result. 1/3 is the correct answer and most of you have mastered this out.

Some of you decided that the puzzle is too simple and added a twist of complexity. Tom questioned parent's honesty and raised question of infant mortality (probably after watching some local news). SteveGoodman18 questioned the 50/50 probability of each gender. And I agree that there is something in it as some families tend to produce mostly boys or girls, but let's keep it simple.

I think everyone participating deserves a puzzle point.

But wait, this puzzle can get even worse.

TracyZ emailed me that she found online a related but more complicated problem, known as the Tuesday's child problem

A man has two children. One is a boy born on Tuesday. What is the probability that he has two boys?

A number of mathematicians and math-related blogs have looked at the Tuesday boy problem and suggested that the correct answer to this problem is 13/27. But this leads us to the darker corners of math kingdom. Enter on your own risk, following the link that TracyZ sent below:

Peter Cameron

Sean Carmody

Tanya Khovanova

My answer of 3/4 is correct if the statement reads

"What is the probability of getting at least 1 boy from 2 children?" The difference is quite subtle (and does not really need the binomial formula)

There are 4 ways to get 4 children; with this restriction only 1 is not possible, so the answer is 3/4.

But you would have to doubt a father knowing what he has.

The answer is 1/2. And no, it is not because "the other child has a 1/2 chance to be a boy." That is a wrong *solution*, but it doesn’t mean the *answer* 1/2 must be wrong.

To see that it must be so, try this variation. A person says "I have two children. What is the probability they both have the same gender?" That seems easy: 1/2. But what if they add "Of my two children, at least one has the gender that I wrote in this sealed envelope. Now what is the probability they both have that gender?" By the logic presented above, it would seem that you should answer 1/3 regardless of what is written in the envelope. But that would mean the answer to the first question is also 1/3.

This apparent paradox is resolved by recognizing that the person had no choice what to write (or tell you in Maria's original question) if the children share a gender, but had to pick between two genders if they do not. So, while the probability a random person has a BG family is 1/4, the chances they would have that family AND write (or tell you) "boy" is 1/8. The answer is (1/4)/(1/4+1/8+1/8)=1/2.

The answer would be 1/3 if the conversation had gone like this:

A: "I have two children."

B: "Oh really? Any boys?"

A: "Yes."

This way, the parent of a boy is required to tell you about a boy. Similarly, the answer to the "Born on Tuesday" variation is 1/2 if the parent told you without being asked, and 13/27 if you asked the parent "Oh really? Any boys born on a Tuesday?" (Of 196 family combinations, 27=14+14-1 include a Tuesday Boy, and 13=7+7-1 have two boys and include a Tuesday Boy. These formulas add the families where the first child qualifies, to those where the second qualifies, then subtracts one for double counting the one where both do.) It changes not because being born on a Tuesday affects the sibling's gender, but because it is nearly twice as likely you will find a Tuesday boy in a two-boy family.

JeffJo

The answer is 1/2. And no, it is not because "the other child has a 1/2 chance to be a boy." That is a wrong *solution*, but it doesn't mean the *answer* 1/2 must be wrong.

First, try this variation. A person says "I have two children. What is the probability they both have the same gender?" That seems easy: 1/2. But what if they add "Of my two children, at least one has the gender that I wrote in this sealed envelope. Now what is the probability they both have that gender?" By the logic presented above, it would seem that you should answer 1/3 regardless of what is written in the envelope. But that would mean the answer to the first question is also 1/3.

This apparent paradox is resolved by recognizing that the person had no choice what to write (or tell you in Maria's original question) if the children share a gender, but had to pick between two genders if they do not. So, while the probability a random person has a BG family is 1/4, the chances they would have that family AND write (or tell you) "boy" is 1/8. The answer is (1/4)/(1/4+1/8+1/8)=1/2.

The answer would be 1/3 if the conversation had gone like this:

A: "I have two children."

B: "Oh really? Any boys?"

A: "Yes."

This way, the parent of a boy is required to tell you about a boy. Similarly, the answer to the "Born on Tuesday" variation is 1/2 if the parent told you without being asked, and 13/27 if you asked the parent "Oh really? Any boys born on a Tuesday?" (Of 196 family combinations, 27=14+14-1 include a Tuesday Boy, and 13=7+7-1 have two boys and include a Tuesday Boy. These formulas add the families where the first child qualifies to those where the second qualifies, then subtracts one for double counting the one where both do.) It changes not because being born on a Tuesday affects the sibling's gender, but because it is nearly twice as likely you will find a Tuesday boy in a two-boy family.

I have been reading what JeffJo wrote about ten times in the last two days and I finally get it.

JeffJo - thank you for making this a thorough step-by-step explanation (3 puzzle points even so you were late). It seems to make sense, however I prefer to not claim 1/2 as the final correct answer yet and leave the space below open for any additional arguments, clarifications and thoughts. It is apparent that in this two-line puzzle every word is important and a subtle change in the dialog may lead to a very different solution.

There is a good chance that there will be both boys. I have done research, and found out that exactly 94,232 more boys than girls were born in the U.S. during 2004. That doesn't mean you will always have a boy though, there is still the chance of it being a boy and girl or both girls. There is still the chance of having a girl, take a look below:

Boy and Boy

Boy and Girl

Girl and Boy

I would also like to put my hypothesis in fraction form: 3/4 chances you will have a boy. Many people say 1/2 chances ( equal ), but I'm thinking different.

The order of the children makes no difference since both are mutually exclusive cases.

We know that one is a boy, and then there is a 50-50 chance for the other to be a boy. Thus, it follows that, because of knowing one IS a boy, the probability of having 2 boys is 50% for the unknown-sex child.

@Maria:

First, I should point out that I didn't originate that argument. It's a variation of Bertrand's Box Paradox, published in 1898 by French mathematician Joseph Bertrand as a cautionary tale about the dangers of assuming that the information you have represents a requirement on the selection. The logic is the same, but the proportions differ a little.

The Box Paradox is more often compared to the Monty Hall Problem, or to the Three Prisoners Problem, where both logic and proportion are the same. Mathematicians universally agree that the answers to those problems are not determined by the proportions, yet most of the same ones will say the Two Child Problem is. Yet it is easy to show that the answer of "1/3" for the TCP is making the same assumption as the answer of "1/2" in the other two problems.

@First Anonymous:

There are all sorts of corrections you could make, if you knew the data. More boys are born, but more boys die young (a morbid thing to address in a puzzle). Identical twins are possible. Some parents are naturally more likely to produce one-gender families than others. And these factors aren’t constant - they vary racially and geographically. The point is, this is just a thought experiment, not a scientific survey that would have to take all these things, and more, into account. So we simply assume a 50-50 split of the genders, and independence in siblings.

@Second Anonymous:

The order won’t make a difference to the answer, but it does make a difference to the counting. Say you have 100 families. Count the families with an older boy by assigning numbers to them - you can expect to assign the numbers 1 thru 50 this way. Then do the same with younger boys, assigning the numbers 51 thru 100. You will have counted to 100, but there will be 25 families that got two numbers, and 25 that didn’t get a number at all.

You are saying that the answer is 1/2 because you assigned 100 numbers, and 50 of them were assigned to two-boy families. But adjusting for that double counting, you only assigned numbers to 75 families, and 25 of those families have two boys. The *proportion* of two-boy families within all families with a boy is 25/75=1/3, but that isn’t the necessarily the same thing as the probability.

I have one comment (to start with) that has to do with the assumptions made by people presenting information in such a way that it will "colour" the way one interprets what the second child is. TracyZ made the point that

3) If you do have a preference towards girls over boys, then you could be likely to say that you have "At least one girl" in cases b, c, and d above, and only say that you have "At least one boy" when you have two boys (case a). With this assumption, if you say you have "At least one boy" then there is 100% probability that you have 2 boys.

The point she is making is that if you have a bias to report facts one way then it compromises the results. But that is not math's fault: it is embedded in people's psychology. A math problem is going to ignore psychology unless it is also a statistics problem. By the way Tracy, please forgive me if this sounds too personal. I don't intend it that way. I only want to say that information should not be conveyed in such a way that it is not changed by the habits of the person conveying the facts.

By the way, the only way I can see this problem being 1/2 is if you say "my oldest child is a boy." What is the chance that the other one is a boy? That would be 1/2 wouldn't it?

I'm a bit leery of the 0.25/(0.25 + 1/8 + 1/8) answer because I feel that that choosing one condition and then asking what are the chances of this coming up neglects the other 2 possibilities.

I can however be talked out of this because I'm not sure.

A two-parter @ Jerome:

Changing the details can sometimes help you avoid preconceived ideas about the problem. The following is the same problem, really, but each member of the pair can have more than two values.

Say I take two normal, six-sided dice. One is white, one is red. I roll them behind a screen, and tell you that one landed on a five. What is the probability that I rolled doubles? There are many approaches you could take:

1) The information is unimportant; the answer is 1/6 because the probability of rolling doubles is 1/6.

2) There are 36 possible combinations WR, where W means the white die and R the red one, and both can be one thru six. Of those 36, 11 include a five. Of those 11, one is a double five. So the answer this way is 1/11. This corresponds to the answer of 1/3 to Maria's question. (In general, the answer is 1/(2N-1), where there are N values.)

3) You could assume that, before rolling, I decided to tell you whatever the red die landed on. Then the probability the white die is also a five is 1/6. This corresponds to "oldest child is a boy."

4) You could assume I decided, again ahead of time, that I favor some numbers over others. Like if I tell you about the highest value of each pair. Then 1/11 would have been correct if I had said "one is a six," but not when I said "one is a five." There are still 11 combinations with a five, but with two of them I wouldn't have said "one is a five;" The answer for "five" is 1/9. And it goes up as the number I tell you about goes down; 1/7 for "one is a four," 1/5 for "one is a three," all the way up to 1/1 for "one is a one."

5) You could assume that I have no preferences whatsoever. If I roll doubles, I have to tell you about that number. But if I roll non-doubles, there are two possibilities. I must use some random method that picks a different dies each time (an example might be whichever stops rolling first). In this case, while there are still 11 combinations with a five, 10 include a different number as well and I will choose the five half of the time. In this case, the answer is 1/(1+10/2)=1/6. This is equivalent to the answer that uses the 1/8 you talked about - actually, it is 1/(1+2/2).

Since I will roll about one doubles about every 6 rolls, a "correct" method must produce answers that average out to 1/6 over many repetitions. By any method that gives a fixed value - which means all but #4 - the average will be that value.

#4 does not have a constant answer. But I won’t say each value the same number of times, either, which affects the average. I will say "six" in 11 of 36 cases, "five" in 9 of 36, etc. The result is that the variable probability averages out to 1/6, which is what we need to happen.

#2 cannot be a correct solution because it does not average out to 1/6. Another way to see that, is to make it a betting game. If I bet you $1 that I rolled doubles, the fair odds are that you should pay me 1/P dollars back if I win. Would you be willing to pay me $11, based on me telling you a number, if you know I will roll doubles once every 6 tries?

The reason #2 is incorrect, is because it makes the same sort of assumption as #4, but it changes the "most favored" value to what I tell you in each roll. But that is impossible. I made a point, in each method, of saying that the method had to be picked ahead of time. That's because you can't use a method to pick a value from the pair, and then turn around and use that value to determine the method.

This is the point that is difficult to see when there are only two values - boy and girl - and only the one example in the problem statement. But you are implicitly making the assumption that the person in Maria's problem favors telling us about boys, because that person told us about boys.

Finally, three of these methods give the answer 1/6, but they are not all the same. #1 says it because the probability of doubles is 1/6. #3 says it because the probability of the same value on "the other" die is 1/6. But #5 makes no unwarranted assumptions - the only method that can claim that - and also gets 1/6.

I'm with Anonymous #2. I don't see how order makes a difference in the way the problem was worded, and of the four possibilities that have been referred to (i.e. GG, BG, GB, BB) BG and GB are the same for the purpose of this particular problem. And as JeffJo stated in his response to Anon. #2, the counting which results in a 1/3 answer and the probability is not the same. While all the other factors are interesting and food for much thought, I'm trying to keep it simple, and hopefully, not too simple!!

Oops! Order doesn't matter but having two possibilities for different genders does! I'll switch to 1/3!

"BG and GB are the same for the purpose of this particular problem." No, they are not. Families don't change just because you are counting them in different ways.

Example: Say you find 60 two-child familes. If you want to count them based on order, you should get 15 in each group BB, BG, GB, and GG.

Now take same 60 families, and count them ignoring order, in groups with 2 boys, 1 boy, and 0 boys. Do you think you should get 20 in each group? That is different than in the previous example.

You actually get 15 families with 2 boys, 30 with 1 boy, and 15 with 0 boys. The 1-boy group is bigger because it has to include 2 of the ordered groups, both BG and GB, which are still different even if you don't care about the order.

@Annie:

There's a couple of things you should realize about probability. First, you can include factors that don't matter, and the answer should still come out right. For example, you can calculate the probability a child will be born a boy by figuring out the probability the child will be born during the day and be a boy, and the probability the child will be born during the night and be a boy. So including order when it shouldn't matter can't make you get the wrong answer. But ignoring it when it does matter can.

Second, you have to use the same set of probabilities as the basis for questions that do require order, as for those that don't. Otherwise, you be basing some answers on the assumption that one in four two-child families have two boys, and others on the assumption that one in three such families have two boys. Can you see the difference?

In this problem, we need to use order to determine the relative number of (unordered) two-boy, one-boy, and zero-boy families. That ratio is 1:2:1 beacsue order is needed to count them.

Hey Maria as a footnote, has any other problem created so much comment??

I'm a little uncofortable with what Annie and Anonomous 2 are saying. The answer could be 1/2 but not for the reason both have stated.

What I am about to say is going to sound like I'm saying something out of both sides of my mouth at once.

Probability problems like this are an idealized model for something that could actually happen in the outside world. I'm no expert: I shy away from stats if I can just because it gets into discussions like this one. On this planet BG is much different from GB so that is how you have to count both. They are not the same and in a count of families, you will get 2/3 with one of each and 1/3 of the gender selected (in this case both boys). What you count from the universal set is the first job. Then you find the probability (I think).

Jeff I really have to tell you that I admire your grasp of the general problem and if I think of someway to answer some of it I will. I wouldn't hold my breath waiting though.

This discussion has been very instructive and thought provoking! I now see the error in my thinking and can comfortably convert my answer to 1/3. I'm open to Jeff's answer of 1/2 but am less comfortable with my understanding of his explanation. Thanks to all!!

@ Annie:

Imagine 300 proplrcome up to you and say either "Of my two children at least one is a boy," or "Of my two children at least one is a girl."

1) How many would you expect to make each statement? You'd need an explicit reason from my problem statement to justify answering differently for the two statements, and there is no succh reason. So I'll assume you answered 150 for each.

2) Of the 150 that said "at least one is a boy," how many do you expect have two boys? Of the 150 who said "at least one is a girl," how many do you expect have two girls? I'll assume you said 50 for each, since you seem comfortable with 1/3 as the answer to maria's question.

3) Of 300 random people who have two chikldren, how many do you expect to have two boys? Two girls? I'll assume you said 75, since you accepted my counting of such families.

4) How can the answers to #2 and #3 different? The fact is, they can't. But there is nothign wrong with #3; there is with #2. To get 1/3 for "two boys," you had to assume that every parent of a boy and a girl said "at least one is a boy." To get 1/3 for "two girls," you had to assume that every parent of a boy and a girl said "at least one is a girl." They can't say both. You get 1/2 if you assuem such parents flip a coin, and only tell you one thing.

JeffJo's last explanation has me now firmly convinced of that 1/2 is the right answer, and I think I understand this more than I did when I answered 1/2 as a possibility initially.

@Jerome, you are right about my comments involving potential biases in the parent answering -- they do not reflect straight probability, but psychology and statistics as you suggest (though the problem could have explicitly stated more of its inherent assumptions... ). In my work, I analyze data collected from participants in studies and surveys at my lab, and these data often involve biases. My comments in my initial answer reflect my own biases and thinking based on statistics and human psychology.

If I flip a coin and it comes out heads, what are the chances the next flip (the second child) will also be heads?

I am just looking at your comments. I am confident with my answer of 1/2 because of the expression "at least" in the problem which refers at the use of conditional probabilities.

I will continue to check the new comments..

Well there is only three possibilities

BB

GG

BG

But GG is obsolete as its already known that there is atleast one boy

so the probability is just 1/2

Alot of people were saying:

GB

BB

BG

But, BG is the same thing as GB just to clear that up.

-Hibah

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