[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]

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

I came up with (1/3)(6/7) which is 6/21 or 2/7.

[Guest]

If you consider the set of all men who have 2 children, then narrow it down to the ones who satisfy the conditions in each question, then find out how many of these men have 2 sons, you'll find the solutions mentioned before: 1/2, 1/3 and 13/27.

However, I found in a website that there's another way to look at it: if you meet a man and his son and he tells you that he has 2 children, it doesn't matter if he tells you which one is the oldest or that one of them was born on a Sunday, the probability is always 1/2. So it's an ambiguous question.

[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.

It's very subtle, and I am having a hard time understanding it, but I'm pretty sure 13/27 is actually correct.

Try making a 14x14 grid, where horizontally is child 1's state:

(boy, sunday), (boy,monday) ... (boy,saturday), (girl, sunday), ...

and vertically is child 2's state with the same labels.

Question 1 corresponds to dividing the top-left quadrant by the top half of the grid. This is 1/2 of the area.

Question 2 corresponds to dividing the top-left quadrant by the area of the top-right, top-left, and bottom-left quadrants. This gives 1/3

Question 3 corresponds to dividing the left and top perimeter (1 square thick) of the top-left quadrant by the left and top perimeter of the entire grid. This fraction of area is 13/27.

The difference comes from the overlap created when the allowable areas (determined by separately assuming each child has the condition of being born on Sunday) are superimposed.

When there is no overlap, 1/2 is the answer.

The more overlap there is, the more deviation from 1/2.

You cannot count the same area twice in the grid. If you were able to, the answer would always be 1/2. But since you cannot count a given spot more than once, the answer is different.

This only occurs for questions that do not distinguish which child has the given condition (unlike Q1 which states that a particular child (the older one) has the particular condition).

I still don't understand it intuitively as I would like, but I believe it.

Maybe I can make a picture tomorrow or something.

Edited September 7, 2010 by mmiguel1

[Guest]

Does the mum eat cereal for breakfast every day?

[Guest]

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

Ans. 1/2

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

Ans. 1/3

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?

Ans. 1/3

[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?

There are two different ways to look at this. First what is the probability he has a second son given he had a first son: this will always be 1/2 since the probability of a second child does not depend on the first child. This is the same answer for all three cases.

The other way to look at this is given 4 sets of children (Boy, Boy) (Boy, Girl) (Girl, Boy) (Girl, Girl), you pick the correct set based on the given information:

1. 1/2, either (Boy, Boy) or (Boy, Girl)

2. 1/2, either (Girl, Boy) or (Boy, Boy)

3. 1/3, either (Boy, Girl), (Girl, Boy) or (Boy, Boy) since you don't know which was born first on a Sunday and it doesn't specify that the day the other child was born.

Of course these probabilities also don't look at the fact that working with radar or genetic issue can change the probability of the man having more X or Y Chromosone Sperm. This hasn't been studied enough for us to make an informed decision about the change in probabilities.

[Guest]

I agree with the previous respondant; in each case, if there is one son and two children, there is a 50/50 chance that the second child will be either male or female.

[Guest]

Let's look at the groups of men who have 2 children

Group A: 2 boys

Group B: 1 boy then 1 girl

Group C: 1 girl then 1 boy

Group D: 2 girls

If we consider that the ratio between boys and girls is 1:1 (it's actually about 1.06:1 but anyway), each group has the same amount of men

Now, if I tell you that a man has 2 kids, he could be anywhere in these groups, so the probability that he's in group A is 1/4

If I tell you that the man has at least one son, he could be anywhere in groups A,B or C, so the probability is nA/nA+nB+nC or nA/3nA or 1/3.

That's how Marilyn vos Savant interpreted the question, and according to her, it's the only right answer.

However, some claim that there's another interpretation which leads to a different answer.

If you meet a man and his son and he tells you he has 2 children, you know that he has to be in group A, B or C, but he's more likely to be in group A than in the previous scenario, because men from group A will always be with a son, while men from groups B and C will sometimes(50% of the times, for the sake of logic) be with a daughter. So the probability is nA/(nA+nB/2+nC/2). Since nA = nB = nC, that is nA/2nA or 1/2

So basically the problem is how you obtained the information that the man has 1 son. I don't know why you would assume that you've met the man and his son, since this is never stated in the question, but that's a valid solution anyway. I guess people just want to prove that Marilyn is wrong.

So the moral of the story is: if you think you're smart, never take an IQ test and just tell everyone you're dumb like a door. Otherwise people will keep trying to outsmart you, and that's really annoying. Sorry for typing such a big text. I was bored.

[Guest]

The question never asks about older/younger or what days of the week anyone was born on. It only asks about the probiblity of a son. The changes of a son is 50% (either a son or a daughter). Way too much reading between the lines into each one on the questions.

[Guest]

No, the question doesn't ask what is the probability of the other child being a boy. The questions ask what is the probability of both children being boys. It makes a big difference.

The answers above have demonstrated the math, but let me try to get at the issue more conceptually for the second question. Imagine an infinite line of men. The moderator states: if you have children step forward. Then, if you have exactly two children step forward. Then, if you have at least one son, step forward. Now take all the fathers the front line and ask: what percentage of them have two sons. The answer is not 1/2.

Of the fathers with two children, an equal percentage of them (25%) would have had (in order of birth), BB, BG, GB, and GG. When the moderator asked the fathers with at least one son to step forward, all but the GG fathers did so. So, the front line is composed equally of BG, GB, and BB fathers. What percentage of them are BB? 1/3.

[Guest]

I've been thinking about this on and off again as my schedule allows me.

I've picked up a little more insight I think on this problem.

I think the trick with Q3, is that it significantly changes the question in a subtle way.

Consider a similar question, in which the change in the question is much easier to see.

A man has two children. At least one of the children is a boy OR is born on Sunday.

What is the probability that the man has two sons?

Using an OR instead of an AND makes it much more obvious to me that the probability to change by introducing days of the week.

After all, the guy doesn't even need to have any sons, as having at least one daughter born on Sunday would make make him a qualified possibility.

I can quickly see that adding these extra cases will change the probability that the man has two sons.

This new probability will be 49/(49+49+49+6+6+1) = 49/160 which is down from 1/3 when the days of the week were not introduced (Q2).

Q3 uses an AND instead of an OR.

This AND doesn't add new possibilities like the OR did, but rather eliminates possibilities that were available in Q2. Possibilities where the man has 2 sons, but neither is born on Sunday are eliminated. This change goes unnoticed to the problem solver at first glance, but after some thought, most would agree that this might lead to a change in the probability. This is because the same possibilities are subtracted from both the target cases (2 sons, the numerator), and the possible cases (having at least one son born on sunday, the denominator).

It is not easy to maintain a constant ratio when the numerator and the denominator of a fraction are decreased by the same amount.

(x-z)/(y-z) = x/y

y(x-z) = x*(y-z)

xy - yz = xy - xz

if z != 0

then

x=y

It is only possible to maintain the same probability when adding or eliminating possible cases if the probability was already 1!

Therefore adding cases with an OR (z<0 in my example), or removing cases with an AND (z>0 in my example) HAS to result with a change in the probability.

Extra input or more insight anyone?

[bonanova]

Monkey wrench, anyone?

This question was

with yours truly and others insisting the answer is 1/3.

However, consider the following argument that we don't have enough information!

That is, there is a built-in, unspecified probability linking the information, given to us

by our informant, to the gender distribution, that is known by the informant. That is,

Our informant followed some [unspecified] algorithm when s/he gave us the information.

Possibilities include, but are not limited to the following:

1. I will pick a child at random. If it's a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter.
   Probability of two sons is 1/2.
   
   
2. If either child [in the inclusive sense] is a boy, I will say "at least one child is a son."
   Otherwise I will [have to] say "at least one child is a daughter."
   Probability of two sons is 1/3.
   
   
3. If both children are boys, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1.
   
   
4. If the older child is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2.
   
   
5. If the taller child is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2. Maybe. See note following case 7.
   
   
6. If the child whose name comes before the other's in the alphabet is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2.
   
   
7. I will ask the children to raise their right hand. If the hand that is raised first belongs to a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is unknown.

Why is case 7 unknown? The speed with which a boy raises his hand might differ from a girl's speed.

We can't assume that in the mixed cases of BG and GB, that a boy's hand will be raised first 50% of the time.

This is a case where the elements of the sample space don't necessarily have equal a priori probability.

This might apply to case 5 as well: if all boys are taller than all girls, then case 5 becomes 1/3, equivalent to case 2.

If some girls are taller than some boys, but boys tend to be taller than girls, then case 5 lies somewhere between 1/2 and 1/3.

[Guest]

Monkey wrench, anyone?

This question was

with yours truly and others insisting the answer is 1/3.

However, consider the following argument that we don't have enough information!

That is, there is a built-in, unspecified probability linking the information, given to us

by our informant, to the gender distribution, that is known by the informant. That is,

Our informant followed some [unspecified] algorithm when s/he gave us the information.

Possibilities include, but are not limited to the following:

1. I will pick a child at random. If it's a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter.
   Probability of two sons is 1/2.
   
   
2. If either child [in the inclusive sense] is a boy, I will say "at least one child is a son."
   Otherwise I will [have to] say "at least one child is a daughter."
   Probability of two sons is 1/3.
   
   
3. If both children are boys, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1.
   
   
4. If the older child is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2.
   
   
5. If the taller child is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2. Maybe. See note following case 7.
   
   
6. If the child whose name comes before the other's in the alphabet is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2.
   
   
7. I will ask the children to raise their right hand. If the hand that is raised first belongs to a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is unknown.

Why is case 7 unknown? The speed with which a boy raises his hand might differ from a girl's speed.

We can't assume that in the mixed cases of BG and GB, that a boy's hand will be raised first 50% of the time.

This is a case where the elements of the sample space don't necessarily have equal a priori probability.

This might apply to case 5 as well: if all boys are taller than all girls, then case 5 becomes 1/3, equivalent to case 2.

If some girls are taller than some boys, but boys tend to be taller than girls, then case 5 lies somewhere between 1/2 and 1/3.

Personally, I think it is safe to assume that the informant's influence on transmitting the information to the reader should be as transparent as possible. The informant was never introduced as a character, and the informant may well just be the writer of the riddle itself. If the writer says something, then for the sake of the riddle, the statement is real no matter what, and any thought of separating "reality" as far as the riddle is concerned and what the writer is saying should be extinguished.

So, the writer says at least one child is a boy.

This next part is where the reader makes a logical jump.

It is common convention to assume that if the probabilities in a sample space are unspecified, then the reader should by default assume that each physical outcome in the sample space has equal probability.

I believe that your claim bonanova, says that this convention may be unjustified because there are many possible ways in which the statement could be made truthfully which yield different probabilities for certain events.

Who is to say equal probabilities for all outcomes is the right thing to assume if it unspecified?

You therefore say that there is not enough information to answer the question (while meanwhile agreeing that 1/3 is the best answer).

I think this is a language issue, not a logical issue.

If the writer wanted us to use equal probabilities for each outcome, he can say nothing, knowing that by convention we will assume what he wants. Either way, one can come to a logical answer.

Edited September 22, 2010 by mmiguel1

[Guest]

#3; If one male was born on Sunday then the only poss. left for Sunday is a female. Given there are six days left in a week, we COULD say he had a poss. of having either a boy/or girl(50/50) for Monday through Saturday. IF we calculate it this way, would we not have 6/7.

[Guest]

ok, days of week =7

possible male births left in week=6

possible female births left in week=7

13 opportunities. 6 of them could possibly be male=6/13

#3; If one male was born on Sunday then the only poss. left for Sunday is a female(1). Given there are six days left in a week, we COULD say he had a poss. of having either a boy/or girl(50/50)or (12 chances) for Monday through Saturday. IF we calculate it this way, would we have 6/13 or the probability of .4615384615?

Edited September 22, 2010 by peanut

[Guest]

ok, days of week =7

possible male births left in week=6

possible female births left in week=7

13 opportunities. 6 of them could possibly be male=6/13

I guess that we were supposed to interpret "one of which" in #3 as meaning "at least one of which".

So in fact, possible male births left in week = 7

Edited September 22, 2010 by mmiguel1

[Guest]

The trick is to realize that all outcomes where the father does not have a son born on Sunday are eliminated.

All outcomes: male or female, any day of the week:

# of outcomes for any gender, any day of the week:

(child 1 male or female) * (child 2 male or female) * (child 1 sunday through saturday) * (child 2 sunday through saturday):

2*2*7*7 = 196

# of outcomes where at least one child is a male born on sunday:

(both children are male born on sunday) + (first child is male born on sunday)*(second child not male born on sunday) + (second child is male born on sunday) * (first child is not male born on sunday)

(1) + (1)*(13) + (1)*(13) = 27

# of outcomes where zero children are both male and born on sunday:

(first child is not male born on Sunday)*(second child is not male born on Sunday)

(13)*(13) = 169

Note that 169 + 27 = 196

So of the original 196, because we know that the father must have at least one son born on sunday, we reduce the number of possible outcomes from 196 down to 27. That is, we eliminate 169 other outcomes.

Of the 27 remaining outcomes,

how many of them result in the father having two sons?

(both children are male born on sunday) + (first child is male born on sunday)*(second child is male but not born on sunday) + (second child is male born on sunday) * (first child is male but not born on sunday)

(1) + (1)*(6) + (1)*(6) = 13

Thus 13 out of 27 possible combinations result with the man having two sons.

If we assume each of these outcomes is equally likely, then the answer is

13/27

Edited September 22, 2010 by mmiguel1

[Guest]

The trick is to realize that all outcomes where the father does not have a son born on Sunday are eliminated.

All outcomes: male or female, any day of the week:

# of outcomes for any gender, any day of the week:

(child 1 male or female) * (child 2 male or female) * (child 1 sunday through saturday) * (child 2 sunday through saturday):

2*2*7*7 = 196

# of outcomes where at least one child is a male born on sunday:

(both children are male born on sunday) + (first child is male born on sunday)*(second child not male born on sunday) + (second child is male born on sunday) * (first child is not male born on sunday)

(1) + (1)*(13) + (1)*(13) = 27

# of outcomes where zero children are both male and born on sunday:

(first child is not male born on Sunday)*(second child is not male born on Sunday)

(13)*(13) = 169

Note that 169 + 27 = 196

So of the original 196, because we know that the father must have at least one son born on sunday, we reduce the number of possible outcomes from 196 down to 27. That is, we eliminate 169 other outcomes.

Of the 27 remaining outcomes,

how many of them result in the father having two sons?

(both children are male born on sunday) + (first child is male born on sunday)*(second child is male but not born on sunday) + (second child is male born on sunday) * (first child is male but not born on sunday)

(1) + (1)*(6) + (1)*(6) = 13

Thus 13 out of 27 possible combinations result with the man having two sons.

If we assume each of these outcomes is equally likely, then the answer is

13/27

Yes..I see. I am not sure how much I like it, but I see. Thanks

[Guest]

Each of the three statements begin "A man has two children". We assume that the man has only two

children, but, in fact, two children might be a subset of three or more children. Of course without having a bound on a number the answer would be undefined, therefore, in order to provide a definitive answer we need assume the number of children is exactly two.

In all three statements we are given that one is a son. One might make an assumption that the possibilities are limited to either son or daughter, but then they are excluding the possibility of a child being intersexed - neither son nor daughter. Of course, current laws in most, if not all, nations require a child to be defined as male or female. Therefore, we may accept the assumption by legal terms that the other child is a son or daughter.

The statement preceding the first question adds that the given child is older. The age of the child has no bearing on the question of probability unless one is using statistics or unless order is important. There is no large sampling group and order is not a criterion in the question, therefore the answer is one-half. One might infer that order is a criterion by the mention of age, yet the answer would remain one-half even if it were.

The second question's preceding statement gives that "at least one of which is a son". If one is to incorrectly assume that order is a criterion, one might answer one-third, but there is no reference in the question that order is one of the criteria, and there is not inference in its preceding statement. The only implication that order might be a criterion is in the preceding, but separate statement and question. Yet, as it is indicated to be a separate question, order is not a criterion, and the answer is one-half.

The third question's preceding statement gives that "one of which is a son who was born on a Sunday". Again, the qualifier has no bearing on the question of probability. Thus, the answer may be one-half.

The difference here is one might infer from the phrasing of the preceding statement that ONLY one child was born on a Sunday, in which case the other child would be born on a day other than Sunday. We then have to make the reasonable assumption that Sunday is exactly one day of an integral sequence of seven days. Thus, with reasonable assumptions, and as order is not a criterion, the answer would be six-thirteenths.

Edited September 23, 2010 by Dej Mar

[Guest]

Each of the three statements begin "A man has two children". We assume that the man has only two

children, but, in fact, two children might be a subset of three or more children. Of course without having a bound on a number the answer would be undefined, therefore, in order to provide a definitive answer we need assume the number of children is exactly two.

In all three statements we are given that one is a son. One might make an assumption that the possibilities are limited to either son or daughter, but then they are excluding the possibility of a child being intersexed - neither son nor daughter. Of course, current laws in most, if not all, nations require a child to be defined as male or female. Therefore, we may accept the assumption by legal terms that the other child is a son or daughter.

The statement preceding the first question adds that the given child is older. The age of the child has no bearing on the question of probability unless one is using statistics or unless order is important. There is no large sampling group and order is not a criterion in the question, therefore the answer is one-half. One might infer that order is a criterion by the mention of age, yet the answer would remain one-half even if it were.

The second question's preceding statement gives that "at least one of which is a son". If one is to incorrectly assume that order is a criterion, one might answer one-third, but there is no reference in the question that order is one of the criteria, and there is not inference in its preceding statement. The only implication that order might be a criterion is in the preceding, but separate statement and question. Yet, as it is indicated to be a separate question, order is not a criterion, and the answer is one-half.

The third question's preceding statement gives that "one of which is a son who was born on a Sunday". Again, the qualifier has no bearing on the question of probability. Thus, the answer may be one-half.

The difference here is one might infer from the phrasing of the preceding statement that ONLY one child was born on a Sunday, in which case the other child would be born on a day other than Sunday. We then have to make the reasonable assumption that Sunday is exactly one day of an integral sequence of seven days. Thus, with reasonable assumptions, and as order is not a criterion, the answer would be six-thirteenths.

Order is a criterion.

This is because the children are unique from one another.

We may pick one child at random and call him/her child 1, and pick the other and call him/her child 2.

This decision can be made completely independently of the age of the children (or not).

Writing child 1's gender first and child 2's gender second:

MF is NOT equivalent to FM.

They would be equivalent only if the two children were indistinguishable in the final result of the probabilistic experiment.

An example of that sort of case would be if we were counting the number of males.

In that case, the two DISTINCT physical outcomes MF and FM would both produce a value of 1 for the random variable (number of males).

As far as (number of males) is concerned, MF and FM are indistinguishable states.

In our case, the two children are in fact unique, and MF and FM are distinct physical outcomes.

For convenience, I would let child 1 be the older and child 2 be the younger, as the attributes older and younger are characteristics that any two children must have (unless they both emerged in the hospital at the exact same time and blah blah blah blah --- completely departing from practical reasoning).

Older and younger are thus convenient markers for distinguishing the two children (which whether we like it or not, are already physically distinct, if by nothing else than by location, I assume they do not occupy the same space in the universe). Use whatever characteristics you like to distinguish the two children, including randomly assigning them different numbers.

Now, that I have explained myself, I think it's fine to just call child 1 = the older child, and child 2 = the younger child.

Question 1: given: child 1 is a male. what is prob that both children are male. Equivalent to what is prob that child 2 is a male? Assume male and female are equally likely. Prob is 1/2. Agreed.

The second question's preceding statement gives that "at least one of which is a son". If one is to incorrectly assume that order is a criterion, one might answer one-third, but there is no reference in the question that order is one of the criteria, and there is not inference in its preceding statement. The only implication that order might be a criterion is in the preceding, but separate statement and question. Yet, as it is indicated to be a separate question, order is not a criterion, and the answer is one-half.

I disagree here because the distinction is inherent in the physical problem. It does not need to be explicitly specified. The only cases where two physical outcomes can be treated as indistinguishable is when we are not dealing with actual physical outcomes, but with some measurement in which the distinct physical outcomes produce identical results (e.g. summing stuff, multiplying stuff, commutative operations, ...).

Here we are calculating the probability for physical outcomes and therefore MF is distinct from FM.

We also continue to use the classic assumption that all physical outcomes have equal probabilities.

This gives 1/3, not 1/2.

You would only get 1/2 if MF is indistinguishable from FM.

I am a male, and I have a sister. I can definitely tell you that in an alternate universe where I was a girl, and my sister was a boy, the outcomes would be distinguishable from our universe.

For question 3, the trick is to realize that all physical outcomes in which the man does not have a son born on Sunday are eliminated. (e.g. sons born on Monday and Tuesday are not possible).

This changes the number of physical outcomes in which the man has two sons, and the number of physical outcomes that the reader believes are possible.

This gives a prob of 13/27 assuming all physical outcomes (before and after eliminating a bunch of them) are equally likely.

Edited September 23, 2010 by mmiguel1

[Guest]

Order is a criterion.

This is because the children are unique from one another.

We may pick one child at random and call him/her child 1, and pick the other and call him/her child 2.

This decision can be made completely independently of the age of the children (or not).

Writing child 1's gender first and child 2's gender second:

MF is NOT equivalent to FM.

They would be equivalent only if the two children were indistinguishable in the final result of the probabilistic experiment.

An example of that sort of case would be if we were counting the number of males.

In that case, the two DISTINCT physical outcomes MF and FM would both produce a value of 1 for the random variable (number of males).

As far as (number of males) is concerned, MF and FM are indistinguishable states.

In our case, the two children are in fact unique, and MF and FM are distinct physical outcomes.

For convenience, I would let child 1 be the older and child 2 be the younger, as the attributes older and younger are characteristics that any two children must have (unless they both emerged in the hospital at the exact same time and blah blah blah blah --- completely departing from practical reasoning).

Older and younger are thus convenient markers for distinguishing the two children (which whether we like it or not, are already physically distinct, if by nothing else than by location, I assume they do not occupy the same space in the universe). Use whatever characteristics you like to distinguish the two children, including randomly assigning them different numbers.

Now, that I have explained myself, I think it's fine to just call child 1 = the older child, and child 2 = the younger child.

Question 1: given: child 1 is a male. what is prob that both children are male. Equivalent to what is prob that child 2 is a male? Assume male and female are equally likely. Prob is 1/2. Agreed.

I disagree here because the distinction is inherent in the physical problem. It does not need to be explicitly specified. The only cases where two physical outcomes can be treated as indistinguishable is when we are not dealing with actual physical outcomes, but with some measurement in which the distinct physical outcomes produce identical results (e.g. summing stuff, multiplying stuff, commutative operations, ...).

Here we are calculating the probability for physical outcomes and therefore MF is distinct from FM.

We also continue to use the classic assumption that all physical outcomes have equal probabilities.

This gives 1/3, not 1/2.

You would only get 1/2 if MF is indistinguishable from FM.

I am a male, and I have a sister. I can definitely tell you that in an alternate universe where I was a girl, and my sister was a boy, the outcomes would be distinguishable from our universe.

For question 3, the trick is to realize that all physical outcomes in which the man does not have a son born on Sunday are eliminated. (e.g. sons born on Monday and Tuesday are not possible).

This changes the number of physical outcomes in which the man has two sons, and the number of physical outcomes that the reader believes are possible.

This gives a prob of 13/27 assuming all physical outcomes (before and after eliminating a bunch of them) are equally likely.

I strongly disagree. Order is NOT given to be a criterion. Thus, just as in set theory, the set {M, F} is considered the same as the set {F, M} we do not care about the sequence of the birth of the children and it matters not that the children may be unique as the question being asked is blind in that regard. But for argument’s sake, let us consider the first two cases as if it did matter.

For two children we then have 6 possible arrangements:

{F1, F2}, {F2, F1}, {M1, M2}, {M2, M1}, {M1, F1}, {F1, M1}

Note: If M,F and F,M are not considered the same, then we need consider M1,M2 and M2,M1 being not the same.

(1) Given the older child is a male, we are left with three possible arrangements:

{M1, M2}, {M2, M1}, {M1, F1}, thus there is a 2/3 chance the second child is a male.

(2) Given only one child is a male we have four possible arrangements:

{M1, M2}, {M2, M1}, {M1, F1}, {F1, M1}, and counting, we have a 1/2 chance the second child is a male.

These answers are in reverse order to those you incorrectly claimed as the correct answers. Therefore,

either you are wrong OR you are wrong... I will let you determine which as order may be important.

[Guest]

What is known in all of these questions.... 1 child = male.

If he has one son already, what are the outcomes of the next child no matter when or where it is born.... male or female. You don't need a matrix to see that he can either have a boy or a girl. 50% on all because you know he has a boy already and there can only be 2 outcomes to having a child (minus the whole hermaphrodite option)

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?

This question never specifies, leads on or says anything that he can't have another son on a sunday. You know he has 1 son so the other has to be either female or male. 50% prob both are male.

[Guest]

I strongly disagree. Order is NOT given to be a criterion. Thus, just as in set theory, the set {M, F} is considered the same as the set {F, M} we do not care about the sequence of the birth of the children and it matters not that the children may be unique as the question being asked is blind in that regard.

Let's make some common notation,

() parenthesis means an ordered tuple

{} curly brackets means an unordered collection or set

| pipe means given

! means not

P[A] means probability of event A

Question 2 asks P[{M,M} | {M,F}]

I fully acknowledge and have never denied that the question does not make a reference to order. As you see here, I used {M,F} instead of (M,F) or (F,M).

My point with this, is that at the fundamental level, we cannot create {M,F} in physical reality. We can only create (M,F) or (F,M).

Gender is only one characteristic of many characteristics that these children have.

In any real situation, we may safely assume that there exists some characteristics other than gender by which these two children differ. That is what I mean when I say these children are distinct.

Let's say one child is 10 pounds and the other is 8 pounds just to pick a characteristic for this example.

Now if we assign genders in all possible ways, how many physical outcomes do we have?

Outcome 1: 10 lb male and 8 lb male

Outcome 2: 10 lb male and 8 lb female

Outcome 3: 10 lb female and 8 lb male

Outcome 4: 10 lb female and 8 lb female

BTW, I don't care about order, you can put the 8 lb baby first in the list or the 10 lb one or whatever you want.

I care about distinction, not order, order is just a convenient representation for distinction.

Your event Dej Mar, {M,F} corresponds to the union of outcomes 2 and 3.

But both outcomes cannot exist simultaneously in reality.

I know the question never mentioned anything about weight.

It doesn't matter, you can do this same exercise with any characteristic that differs between the children.

You cannot argue that such a characteristic cannot exist, and at least one such characteristic must exist.

Therefore in physical reality, your unordered set {M,F} is not realizable.

And it is from physical reality that we make the assumptions that each of these outcomes has equal probability.

The way to look at {M,F} is to say that it is an event, not an outcome, in which outcome 2 or outcome 3 takes place.

In this way, if (M,F) and (F,M) each have a probability of 0.25, then the event {M,F} has a probability of 0.5 and is twice as likely as the event {M,M} (which is the same as (M,M) )

Your way, Dej Mar, when taken to the physical realm with any differing characteristic, such as weight, age, location of center of mass of baby with respect to some origin, ..., etc

comes to a road block:

With same example as before,

Outcome A: 10 lb male and 8 lb male

Outcome B: ? lb male and ? lb female

You then assign both of these equal probability, and say prob of A is 0.5 and prob of B is 0.5.

The takeaway is B cannot be physically realized as a single outcome and physical realization is the ultimate authority regarding how we assign probabilities. The list of things explicitly mentioned in the question is not the ultimate authority.

But for argument’s sake, let us consider the first two cases as if it did matter.

For two children we then have 6 possible arrangements:

{F1, F2}, {F2, F1}, {M1, M2}, {M2, M1}, {M1, F1}, {F1, M1}

Note: If M,F and F,M are not considered the same, then we need consider M1,M2 and M2,M1 being not the same.

(1) Given the older child is a male, we are left with three possible arrangements:

{M1, M2}, {M2, M1}, {M1, F1}, thus there is a 2/3 chance the second child is a male.

(2) Given only one child is a male we have four possible arrangements:

{M1, M2}, {M2, M1}, {M1, F1}, {F1, M1}, and counting, we have a 1/2 chance the second child is a male.

I have no idea how to interpret your notation here.

In all cases you have a 1 and a 2 (which I think would mean order of birth) except for the ones in which order of birth (as a valid distinction) would actually matter.

I'm guessing that this cannot be the case because {M1,F1} would not make sense (exact twins? to the picosecond? really?), no, it should be {M1,F2} or {M2,F1}

Also, according to me, we go by physical outcomes, and {M1, M2} and {M2,M1} are different ways of writing the same exact physical outcome, and this outcome should be counted once.

So to use your notation (at least I think so),

{M1,M2}, {M1,F2} are the possible outcomes for question 1

{M1,M2} {M1,F2} {M2,F1} are the three possible outcomes to question 2.

By order, I meant I was using order to serve the same purpose as your 1 and 2 --- that is each distinct child is uniquely identified by their position in the gender string (MM, MF, FM, FF) in my original case or by the 1 or 2 in your case.

These answers are in reverse order to those you incorrectly claimed as the correct answers. Therefore,

either you are wrong OR you are wrong... I will let you determine which as order may be important.

Very funny.

I am very confident that I am not wrong, but you may believe what you like.

[Guest]

A pair of coin flips results at least one head.

What is the probability that both flips were heads?

Flip 100 coins, throw away all cases where you have tails-tails, and compare the number of times you get one heads and one tails versus the number of times you get two heads. You will find that the number of times you get one heads is roughly twice the number of times you get two heads.

This answer comes from making assumptions not at the question specification level, but at the physical level.

Experimentally verify it if you want.

Bonanova listed some ways in which different assumptions on the part of receiving this information can change the probabilities and therefore the coin flip experiment as well, but I think the most conventional interpretation is captured in the above experiment.

[bonanova]

Two children gives a 196-case sample space

B[sun-sat] B[sun-sat] 49 cases

B[sun-sat] G[sun-sat] 49 cases

G[sun-sat] B[sun-sat] 49 cases

G[sun-sat G[sun-sat] 49 cases

A Sunday son reduces the sample space to 27 cases:

bb: B[sun] B[sun-sat] 7 + B[sun-sat] B[sun] 7 - B[sun] B[sun] 1 [double counted] = 13 cases

bg: B[sun] G[sun-sat] 7 cases

gb: G[sun-sat] B[sun] 7 cases

gg: 0 13 cases.

bb/(bb+bg+gb+gg) = 13/27