[Guest]

Alright, we've all seen the first couple of these, but the third one is kind of interesting.

1. A man has two children, the older of which is a son. What is the probability that he has two sons?

2. A man has two children, at least one of which is a son. What is the probability that he has two sons?

3. A man has two children, one of which is a son who was born on a Sunday. What is the probability that he has two sons?

[Guest]

0.5 for all

[Guest]

Alright, we've all seen the first couple of these, but the third one is kind of interesting.

1. A man has two children, the older of which is a son. What is the probability that he has two sons?

2. A man has two children, at least one of which is a son. What is the probability that he has two sons?

3. A man has two children, one of which is a son who was born on a Sunday. What is the probability that he has two sons?

(1) 1/2

M = Male, F = Female

Select MM from MM, MF

(2) 1/3

Select MM from MM, MF, FM

(3) 6/13

"one of which" implies that the other child is not a son who was born on Sunday.

Either this other child is a daughter born on any day of the week or a son born not on sunday.

There are 6 male possibilities and 7 female possibilities for this other child corresponding to days born on.

6+7=13

So this other child is a male with probabality 6/13

[Guest]

nice interpretation of the 3rd , tricky but nice

didn't consider the order in 2nd.

[Guest]

(3) 6/13

"one of which" implies that the other child is not a son who was born on Sunday.

Either this other child is a daughter born on any day of the week or a son born not on sunday.

There are 6 male possibilities and 7 female possibilities for this other child corresponding to days born on.

6+7=13

So this other child is a male with probabality 6/13

He said "Once of which" not necessarily meaning "Only one of which"...

Edited September 5, 2010 by Anza Power

[Guest]

The third question didn't ask anything about the days the children were born. It just mentioned it in the explanation.

So I think it's .5 for all of them.

[Guest]

Consider as the possible variables the 14 sex/day possibilities for a particular child (B-Mon. G-Mon, B-Tues., etc.). Of the 14*14=196 possibilities, there are 27 permutations that include a boy born on Sunday:

In exactly one of the possibilities, both children are boys born on Sunday. In this case, obviously, the man has two boys.

If there is only one B-Sun, and he is the oldest, there are 13 sex/day possibilities for the younger child (since the chance of a second B-Sun is excluded). In 6 of the 13, the younger child is also a boy.

If there is only one B-Sun, and he is the youngest, there are 13 sex/day possibilities for the older child (since the chance of a second B-Sun is excluded). In 6 of the 13, the older child is also a boy.

So, the chance of two boys given that one child is a boy born on Sunday is (1+6+6)/(1+13+13)=13/27

[Guest]

He said "Once of which" not necessarily meaning "Only one of which"...

That's an ambiguity in the English language, it can be interpreted either way.

In my interpretation, "one of which" means that there is one child that satisfies the following.

If he meant more, he could have said "at least one of which".

On the other hand, it can be interpreted as, considering a particular child, this child satisfies the following.

Both interpretations are equally valid I would say. I can't find anything in the sentence to clearly allow the reader to choose between one of them.

Let me know if you find something that does that.

Edited September 5, 2010 by mmiguel1

[Guest]

The semantics of the puzzle are not that difficult. A man has two children: there are only three possible combinations, two boys, two girls, or a boy and a girl. The second condition in each statement eliminates "two girls" as a possible combination: in the first case age is a red herring but the fact that one of the two is a boy is enough; in the second statement we again eliminate two girls as a possibility leaving us with just two outcomes; and again in the third statement, it's enough to know that one is a boy to reduce the outcomes to one of two possible combinations. In all three cases there's a fifty-fifty chance of two boys.

[Guest]

To clarify, "one of which" says nothing about the other child. That is, the other could also be a son born on a Tuesday.

The correct answers have all been given now.

1. 1/2

2. 1/3 (find explanations for these all over the place...suffice to say that P(BG) is not the same as P(BB), so 50% is not correct. Simulate it with a computer and you'll see.)

3. 13/27 (Mtheadedd's explanation copied below)

Consider as the possible variables the 14 sex/day possibilities for a particular child (B-Mon. G-Mon, B-Tues., etc.). Of the 14*14=196 possibilities, there are 27 permutations that include a boy born on Sunday:

In exactly one of the possibilities, both children are boys born on Sunday. In this case, obviously, the man has two boys.

If there is only one B-Sun, and he is the oldest, there are 13 sex/day possibilities for the younger child (since the chance of a second B-Sun is excluded). In 6 of the 13, the younger child is also a boy.

If there is only one B-Sun, and he is the youngest, there are 13 sex/day possibilities for the older child (since the chance of a second B-Sun is excluded). In 6 of the 13, the older child is also a boy.

So, the chance of two boys given that one child is a boy born on Sunday is (1+6+6)/(1+13+13)=13/27

So while it looks like the day of the week shouldn't have anything to do with it, it really affects the problem greatly. Actually, if add some event to one of the births, the probability of having two sons approaches 1/2 as the probability of that event decreases. For example, if, instead of saying one of the children is a son born on a Sunday, we say that it's a left-handed son (we'll assume that's a 1/10 chance), the probability of having two sons increases to 19/39.

Edited September 5, 2010 by Chuck

[Guest]

The fact he has one son may mean he is physically disposed to having sons only. This possibility would have to be factored in.

[Guest]

To clarify, "one of which" says nothing about the other child. That is, the other could also be a son born on a Tuesday.

The correct answers have all been given now.

1. 1/2

2. 1/3 (find explanations for these all over the place...suffice to say that P(BG) is not the same as P(BB), so 50% is not correct. Simulate it with a computer and you'll see.)

3. 13/27 (Mtheadedd's explanation copied below)

So while it looks like the day of the week shouldn't have anything to do with it, it really affects the problem greatly. Actually, if add some event to one of the births, the probability of having two sons approaches 1/2 as the probability of that event decreases. For example, if, instead of saying one of the children is a son born on a Sunday, we say that it's a left-handed son (we'll assume that's a 1/10 chance), the probability of having two sons increases to 19/39.

So in the third one, why does it matter who's older or younger? I'm confused about that.

[Guest]

0.5 for all

I say you are correct

[Guest]

The fact he has one son may mean he is physically disposed to having sons only. This possibility would have to be factored in.

How would that even be possible? If that were true, then we wouldn't really have any difficulty determining the probability of him having two sons now would we?

[Guest]

So in the third one, why does it matter who's older or younger? I'm confused about that.

It matters for pretty much the same reason that it matters in the second problem.

In problem 2, most people catch onto the fact that having at least one son means that the man could not have two daughters. Assuming order doesn't matter, he could have one boy and one girl, or two boys. What many people fail to see (and as a result, assume that there is a 50% chance that the other child is a boy), is that, the probability of having one boy and one girl is TWICE as much as that of having two boys. Let's change the problem a bit and say that we have 4 doors: one red, one blue, and two green. One door has a prize behind it, and you have to guess its color in order to win. You have a 50% chance of winning if you guess right now. This is the same as the probability that, given that a man has two kids, exactly one is a boy and one is a girl. However, if I say that he has at least one boy, then you know he can't have two girls. This is like me telling you that the prize is not behind the red door. Now, if you guess "green," you have a 2/3 chance of winning. Similarly, there is now a 2/3 chance that he has one boy and one girl, given that he does not have two girls.

In short, the order matters in the third problem because when the man has one boy and one girl, because he could have had the two kids in either order, both possibilities must be accounted for.

Those of you who still say 1/2, do this:

Flip a coin twice. Let's say that heads means that the man had a son, and tails means he had a daughter. Flip the coin twice, and...

A) If it comes up with two heads, then the man had two sons. Record this as a successful trial.

B) If it comes up with one head and one tail (in either order), then the man had one son and one daughter. Record this trial as a failure.

C) If it comes up with two tails, then the man had two daughters. Do not count this trial, since we know the man had at least one son.

Run this experiment as many times as you can (preferably at least 25 times), or write a program to do it for you. See how many successes and failures you get. You will probably get about twice as many failures as successes.

[Guest]

Easy one:

1. 1/3

2. 1/3

3. 1/3

[Guest]

So in that case the solution for #1 could be the same for #2. Still doesn't explain the added probability for difference in age.

[Guest]

So in that case the solution for #1 could be the same for #2. Still doesn't explain the added probability for difference in age.

For question 3, you didn't factor the initial chance of having a boy or girl, however in question two you did.

Similarities: Question 3: We do not know whether the boy is older or younger.

Question 2: We do not know whether the boy is older or younger.

If I read the explanation correctly for question 3, you assumed that the chance of having a boy versus a girl was 50/50. Now it may be that i'm just too dumb to realize this, but could someone explain why you don't put the 1/3 2/3 chance of having a boy vs. girl in?

[Guest]

I believe that the answer to question 3 is 2/7. 13/27 is making the same mistake as picking 50/50 for question 2.

[Guest]

This result is counter-intuitive.

I've been trying to reconcile it with my intuition.

What if we thought of all three questions including day of the week information.

1. A man has two children, the older of which is a son. What is the probability that he has two sons?

2. A man has two children, at least one of which is a son. What is the probability that he has two sons?

3. A man has two children, one of which is a son who was born on a Sunday. What is the probability that he has two sons?

Q1)

2 children, older is a son born on one day of the week. What is the chance that the younger is a son born on any day of the week?

B = Boy

G = Girl

N = Sunday

M = Monday

T = Tuesday

W = Wednesday

R = Thursday

F = Friday

S = Saturday

1 = Older child

2 = Younger child

- = Any possible value

Possibilities:

For 1:

BN, BM, BT, BW, BR, BF, BS

For 2:

BN, BM, BT, BW, BR, BF, BS

GN, GM, GT, GW, GR, GF, GS

An outcome is taking any possibility from 1 and any possibility from 2.

Person,Gender,Day of Week

Select:

(1B-,2B-) from (1B-,2--)

Assuming equal probabilities for each outcome, the answer is 50%.

Q2)

For 1:

BN, BM, BT, BW, BR, BF, BS

GN, GM, GT, GW, GR, GF, GS

For 2:

BN, BM, BT, BW, BR, BF, BS

GN, GM, GT, GW, GR, GF, GS

Select (1B-,2B-) from (1B-,2-- OR 1--,2B-)

OR represents a union of possibilities

(1B-,2-- OR 1--,2B-) =

{

}

[Guest]

This result is counter-intuitive.

I've been trying to reconcile it with my intuition.

What if we thought of all three questions including day of the week information.

1. A man has two children, the older of which is a son. What is the probability that he has two sons?

2. A man has two children, at least one of which is a son. What is the probability that he has two sons?

3. A man has two children, one of which is a son who was born on a Sunday. What is the probability that he has two sons?

Q1)

2 children, older is a son born on one day of the week. What is the chance that the younger is a son born on any day of the week?

B = Boy

G = Girl

N = Sunday

M = Monday

T = Tuesday

W = Wednesday

R = Thursday

F = Friday

S = Saturday

1 = Older child

2 = Younger child

- = Any possible value

! = Not

Possibilities:

For 1:

BN, BM, BT, BW, BR, BF, BS

For 2:

BN, BM, BT, BW, BR, BF, BS

GN, GM, GT, GW, GR, GF, GS

An outcome is taking any possibility from 1 and any possibility from 2.

Person,Gender,Day of Week

Select:

(1B-,2B-) from (1B-,2--)

Assuming equal probabilities for each outcome, the answer is 50%.

Q2)

For 1:

BN, BM, BT, BW, BR, BF, BS

GN, GM, GT, GW, GR, GF, GS

For 2:

BN, BM, BT, BW, BR, BF, BS

GN, GM, GT, GW, GR, GF, GS

Select (1B-,2B-) from (1B-,2--) OR (1--,2B-)

OR = Union = Sum - Intersection

Sum = (1B-,2B-) + (1B-,2G-) + (1G-,2B-) + (1B-,2B-)

Intersection is (1B-,2B-)

Union = (1B-,2B-) + (1B-,2G-) + (1G-,2B-)

Probability = 49/(49 + 49 + 49) = 1/3

Q3)

Select (1B-,2B-) from (1BN,2--) OR (1--,2BN)

Union = Sum - Intersection

Sum =

(1BN,2G-) + (1BN,2BN) + (1BN,2B!N) +

(1G-,2BN) + (1BN,2BN) + (1B!N,2BN)

Intersection = (1BN,2BN)

Union = (1BN,2G-) + (1BN,2B!N) + (1G-,2BN) + (1B!N,2BN) + (1BN,2BN)

Select

(1BN,2B!N) + (1B!N,2BN) + (1BN,2BN) from the union (1BN,2G-) + (1BN,2B!N) + (1G-,2BN) + (1B!N,2BN) + (1BN,2BN)

Select

(6 + 6 + 1) from (7 + 6 + 7 + 6 + 1) = 13/27

The deviation from 1/2 comes from subtracting the intersection.

If we did not subtract the intersection we would have

(6 + 6 + 1 + 1) from (7 + 6 + 7 + 6 + 1 + 1) = 14/28 = 1/2

The smaller the intersection, the closer to 1/2.

If no intersection is subtracted in (Q2), then

we would have (49+49)/(49+49+49+49) = 1/2

Interesting...

[Guest]

That makes no sense. They would all be 50% umm one is a son the other can be a girl, or boy, in which 50, 50! they are all asking the probability of the exact same thing! You are all making a mountain out of a mole hill! Nothing matters but their gender!

Edited September 6, 2010 by brrr

[Guest]

That makes no sense. They would all be 50% umm one is a son the other can be a girl, or boy, in which 50, 50! they are all asking the probability of the exact same thing! You are all making a mountain out of a mole hill! Nothing matters but their gender!

That's what people said to Marilyn vos Savant (the most intelligent woman in the world). Then they made computer simulations and found out she was right.

[Guest]

I think Q3 is 2/7, I don't see how it's 13/27. If you minus the day, you get 1/2. However, it is not stated which order the children are in. I believe your solution to Q3 is wrong.

I too have been trying to find some sense in #3. The question is almost identical, but answers are different.

Edited September 7, 2010 by firefight

[Guest]

3. A man has two children, one of which is a son who was born on a Sunday. What is the probability that he has two sons?

Can somebody explain the significance of the son being born on a sunday. Otherwise its the same Q as 2)

List of possibilties. 2*2=4

These are MM, MF, FF, FM

1)MM, FM are the only applicable choices. So p(both m | second m) = 1/2

2)MM, MF, FM, are the only applicable choices so p(both m | at least 1 m) = 1/3

[Guest]

3. A man has two children, one of which is a son who was born on a Sunday. What is the probability that he has two sons?

Can somebody explain the significance of the son being born on a sunday. Otherwise its the same Q as 2)

List of possibilties. 2*2=4

These are MM, MF, FF, FM

1)MM, FM are the only applicable choices. So p(both m | second m) = 1/2

2)MM, MF, FM, are the only applicable choices so p(both m | at least 1 m) = 1/3

The significance is that another boy cannot be born on a sunday. ( I know, the question is flawed.)

Edited September 7, 2010 by firefight