posted
Aris isn't getting confused; he is legitimately and correctly observing that the question, as phrased, is ambiguous, and thus subject to two interpretations.

Most of us are interpreting the question as, essentially, this:

quote:There exists a family with exactly two children, at least one of whom is a boy. What is the probability that both of the children are boys?

However, as it was actually expressed, you're left wondering if the guy decided to tell us about a particular child before asking us, instead of intending to tell us if he does (or doesn't) have at least one boy. In that case, the probability is 1/2.

That's sort of what I thought was up last night, thus my remark about assuming point children. We're taking a real scenario expressed in conversational English, and reformulating it into a strictly mathematical interpretation. Aris, quite rightly, observes that the question may easily be understood as asking:

quote:I have two children. The particular child I'm thinking of right now is a boy. What is the probability that the other one is a boy?

As I said last night, I suspect more people would arrive at the "correct" answer if the question were phrased right.
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posted
Again; why is it the case that of the 27 families with two children who've had a boy on a Tuesday, 13 have two boys? That's simply not true.
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posted
You are correct, Vegimo. As you further and further fix the known boy from the sample combinations with additional information, the probability of the other child being a boy gets closer to a half.
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posted
Of the 27 families with one boy born on a Tuesday, 9 have a second boy. Why do you think otherwise?
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quote:Originally posted by TomDavidson: Again; why is it the case that of the 27 families with two children who've had a boy on a Tuesday, 13 have two boys? That's simply not true.

Here is the complete list of possibilities:

BMon BTue

BTue BMon

BTue BTue

BTue BWed

BTue BThu

BTue BFri

BTue BSat

BTue BSun

BWed BTue

BThu BTue

BFri BTue

BSat BTue

BSun BTue

Thirteen in total. Seven combinations in which the first boy is born on Tuesday, seven combinations in which the second boy is born on Tuesday, with one shared combination in which both boys were born on Tuesday.

I promise that I'm not trying to confuse or trick you in any way, shape or form, Tom.

Edit: if there are only nine, then you should be able to identify the four combinations in my list of thirteen in which one of the children is a girl, or neither of them is born on a Tuesday.

Further edit:you are just too smart not to get this, Tom!!

[ July 24, 2010, 04:16 PM: Message edited by: Jordan ]
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posted
Maybe a visualization is in order (and start with Jordan's plea regarding setting aside assumptions): imagine a world where the likelihood of any birth being a boy or a girl is exactly 50%. Imagine a world where the chance of any particular child being born on any given day is exactly 1 in 7.

Now imagine each of the two above statistics are completely independent such that boys and girls are equally likely to be born on any given day.

We then get to Mariner's 196 possibilities (on average). Imagine that one sample of each equally likely combination (Sunday Girl/Sunday Girl, Sunday Girl/Monday Girl... Saturday Boy/Friday Boy, Saturday Boy/Saturday Boy) is used.

Now here is the visualization: imagine a gatekeeper that only allows certain fathers into your area (the filter alluded to by Jordan): Two girls? Nope. Fathers of a boy or boys, but not or neither born on a Tuesday? Nope.

The gatekeeper allows exactly 27 fathers through in orde to speak to you. Of these 27, 7 are firstborn Girl, younger Boy (Girl born on any of 7 days, Boy on Tuesday) 7 are firstborn Boy, younger Girl, 6 are firstborn Boy (not Tuesday) and younger Boy (Tuesday), 6 are firstborn Boy (Tuesday) younger Boy (not Tuesday) and 1 is firstborn and younger Boy (both Tuesday). 13 of the 27 are boy-boy combinations.

If you change the filter to be 'boy born Sunday through Friday) you end up with 48/132 (approx 36%)
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posted
Tom, list the matrix to show the possibilities, then restrict the population and count the results. Probability theory is a predictor for the results, but actual populations are what prove the theory.

Jordan, my guess was about the numbers (167/335 and 4031/8063). I arrived at those numbers through:

denominator = one less than the product of all the possibilities of each restrictor numerator = one less than half of the original denominator

for no restrictor: ((1/2 * 2 * 2) - 1) / ((2 * 2) - 1) = 1/3

quote:And the most powerful piece of information yet, that the eldest child is a boy, eliminates a huge selection of combinations from consideration and increases the probability to 1/2

Just checking - you realize this was not part of the original question, right?
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posted
OK Tom, how about this. I've uploaded a spreadsheet with some lists so you can see how I generated the combinations, and what those combinations are. I'm not asking you to go that far, just try listing what your list of 27 possibilities is, and the nine possibilities in which there are two boys. I'm not sure where we're missing each other.
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quote:And the most powerful piece of information yet, that the eldest child is a boy, eliminates a huge selection of combinations from consideration and increases the probability to 1/2

Just checking - you realize this was not part of the original question, right?

Sorry, I should have quoted or linked. I was referring to Brian's modified question in the second post:

quote:Brian: Would it change the odds if he said 'my oldest child is a boy'?

posted
Ah- finally spotted what's been twitching me here looking at that spreadsheet. We have to fully account for each child being independent of the other- so {BTue, BTue} isn't complete:

Both boys being born on Tuesday needs to be counted twice, because you don't know whether you're talking about the first child or the second child. So this really brings it back to 14 / 28 or 50%

But we've also accidentally eliminated a bunch of other cases by not considering the 1st child and 2nd child separately. I'll bet that if we step back a stage and account for that with all boy/girl combinations, we'll find another 14 possibilities to bring us back to 1/3 (while adding back in all of the girl/girl possibilities brings us to a 25% chance of a boy/boy set)
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posted
No, each possibility is distinct, and none should be counted twice. There are only 27 possibilities where either child is a boy born on Tuesday. 13 of those have 2 boys.
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quote:Originally posted by Pyrtolin: Both boys being born on Tuesday needs to be counted twice, because you don't know whether you're talking about the first child or the second child. So this really brings it back to 14 / 28 or 50%.

First, I'm really glad you're looking at the data carefully.

Second, you're right (and quite perceptive) to pick up on the fact that {BTue BTue} is a special case that isn't being counted twice; your only mistake is in thinking that it should. Look back at the original problem which doesn't involve days of the week: {B B} only appears once. {B B} and {BTue BTue} being special cases that account for two children meeting our precondition, yet are counted only once, is important to understanding the problem fully.

More directly, in the list of the possibilities we have, it doesn't appear twice because there is only one case in which both children are boys born on Tuesday, not two. {BTue BTue} only happens if the first and second children are both boys who are born on Tuesday, and there's only one case in which this happens.

Aside from anything else, the fact that this stood out to you means you're well on the way to getting the answer.
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quote:there's only one case in which this happens.

Only one absolute case? Sure. But that case is twice as likely to occur because there are two ways that it could be arrived at- one for each boy that the original asker might be referring to.

The probability has to account for the fact that he could be referring to either child, so it means that you have to count each possibility once for each child, rather than for the absolute distribution of cases.
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posted
Further- The actual equivalent from the original problem is the way that Boy/Girl is "counted twice".

It gets double weight becuase B/G and G/B are actually distinct, independent sets, in the same way That B1T/B2T and B2T/B1T are completely independent sets here.

If you can't count BT/BT twice here, you can't count B/G twice in the original set and are limited to just BB, BG, GG as your possibilities, which isn't accurate. The same principle that applied there applies here.
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posted
Actually, I'm wrong there, but leaving it for thought process. The problem is actually that we're using the wrong initial set because it accidentally happens to reduce properly.

The right basic set is: B1 -> 1 G2, 1 B2 B2 -> 1 G1, 1 B1 G1 -> 1 G2, 1 B2 G2 -> 1 G1, 1 B1

Because we don't know which child he's referring to. The distribution ends up the same 2/8 BB, 4/8 GB, 2/8 GG, but the actual set is secretly larger.

Once he says that one child is a boy, then we have

lose the 2 GG scenarios, and are left with 2/6 BB and 4/6 GB. Again, those reduce properly so it's easy to accidentally undercount he full set of possibilities.

As Aris has noted- what he's talking about matters, not just the basic distribution.
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posted
The initial mistake enters because we're incidentally using 2 things from a 2 item set of possibilities. (If there were 3 sexes and he had two kids, the mistake wouldn't slip by so easily) Once you have 2 and 7, they don't reduce so easily, so the same shortcut doesn't work, and trying to use it accidentally factors out cases that need to be accounted for.

posted
No, the reason the B-Tue,B-Tue should only get counted once is because it is only one family. The father in question is not twice as likely to show up as any other father. There are only 196 distinct families, thus only 196 distinct fathers. Each one has the same possibility of being the father you happen to meet. Once he limits the population by specifying the restricting factors, he is still only one father with only one family, and is not counted twice. None of them are. There are only 27 families with a B-Tue included, and 13 of those have two boys.
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posted
Hm. In working this out for myself, it turns out to be quite amusing to substitute "born on Tuesday" with "prefers red to blue." And, indeed, I misunderstood. *laugh* Once you have any distinguishing characteristic beyond sex, this breaks down the same way. As Jordan observed on the first page (which I completely failed to understand), the addition of another distinguishing characteristic produces more situations in which both parties might share the same characteristic, thus skewing the totals further away from an expected percentage.

quote:Pyrtolin: Only one absolute case? Sure. But that case is twice as likely to occur because there are two ways that it could be arrived at- one for each boy that the original asker might be referring to.

I'm afraid not. What we're counting are families, not how many ways he could be asking the question.

quote:As Aris has noted- what he's talking about matters, not just the basic distribution.

If you go by Aris' interpretation of the question, the conditional (i.e. the whole thing that makes the question mathematically interesting) disappears; every question we've examined so far reduces to, "What's the probability that a child is a boy?" and the answer in every case is ½.

The alternative interpretation, and the one you're "supposed" to take (in order to get the more interesting answers of 1/3 for the first question and 13/27 for the second) is easier to understand if you rephrase the question more precisely:

quote:A family has two children. Given that at least one of them is a boy (or, given that one of them is a boy who was born on a Tuesday), what is the probability that both of the children are boys?

The subtlety in Aris' interpretation is that we aren't told how the man in question came to be telling us that he had one son, so the question may be "fixed" before you even start; the stricter formulation isn't ambiguous and is the more interesting interpretation that I (and others) are going by to get our figures.

[ July 24, 2010, 07:43 PM: Message edited by: Jordan ]
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posted
You have to explicitly account for all combinations of "Is he talking about the first child" and "Is he talking about the second child" or you're not getting all the possible sets. The fact that you don't know which one he's referring doubles the problem space. When the problem space is a multiple of two, that factors out, but when it's not, you can't factor it out.

What you're saying would be true if he said "My oldest child is a boy" , because that resolves a specific, relevant element the problem that was otherwise unknown, in the same way that if you were just looking at the B/G matrix, saying that the first child is a boy would resolve the probability to 50/50.
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quote:Tom: Hm. In working this out for myself, it turns out to be quite amusing to substitute "born on Tuesday" with "prefers red to blue." And, indeed, I misunderstood. *laugh* Once you have any distinguishing characteristic beyond sex, this breaks down the same way.

Oh, sweet! I knew you'd get it soon.

It's quite delightful, actually. Even something like, "is named after a famous comedian" affects the probabilities, bringing them yet closer to a half; the more specific the conditional, the more families are eliminated and the closer we come to just asking, "what's the chance that this specific child is a boy".

Or, the more specific you are about one of the children, the more likely it is that your precondition only applies to one of the two children. Questions like "the eldest child" can only apply to one of them off the bat, which is why the probability is exactly ½.
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quote:Pyrtolin: What you're saying would be true if he said "My oldest child is a boy" , because that resolves a specific, relevant element the problem that was otherwise unknown, in the same way that if you were just looking at the B/G matrix, saying that the first child is a boy would resolve the probability to 50/50.

You're quite close here to grasping it intuitively. In the same way that saying that one of the children is both male and the eldest fixes the first column (by definition of "eldest"), saying that one of the children is both male and born on a Tuesday fixes most of the first column, except for the one case where both children were boys born on Tuesday.

Consider the question using the "at least one child" phrasing instead of asking how he "selected" one particular child might help.
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posted
Jordan - I was talking about this thread with a friend, and I couldn't get my head wrapped around how it's NOT 1/2, but when I thought about it this way it made sense: We are not asking 'what are the chances that this second child will be born a boy' we are asking the question "what are the chances that the man fathered two boys (since we know that he already fathered one)." Which is exactly the same as asking "what are the odds of flipping a coin twice and getting heads each time?" and not "I flipped a coin and got heads, what are the chances I'll get heads again." because the child has already been born.

When I saw it that way, I saw the math: 1/2 x 1/2 which is 1/4. There are 4 out comes to flipping a coin twice, and this has already been stated in this thread: HH, TT, HT, TH. since we can eliminate TT (since we know the first flip was heads) there are only three other possibilities, making it 1/3.

It is definitely not 1/2 though. That's just math, it really can't be argued as a matter of opinion.

I haven't quite got the "Tuesday" part of it yet, but I think it works from the same principle. Something about subtracting the time where BOTH are boys and BOTH are born on Tuesday figures in, but I'm not that strong on this sort of thing.
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posted
Or maybe re-phrase to "what are the chances of someone giving birth to a boy and then another boy?" Which is 1/4, but then we say "ok we had two kids, one was a boy, what are the chances they are both boys?"

Eh I'm having trouble articulating this, yet I can see it in my head without all these damn words getting in the way. I guess that's the issue with the whole thread.
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posted
Jordan: I suspect the following two cases are distinct. Do you agree?

1) A man walks up to you and tells you he has a son and another child. You then ask him what day of the week his son was born and he tells you it was Tuesday. Finally, he asks you the probability for the sex of his other child.

2) A man walks up to you during a meeting of "men who have sons who were born on Tuesday". He informs you that he has two children. Finally, he asks you the probability for the sex of his other child.

Do you believe this situations have identical answers or different answers?

quote:edgmatt: We are not asking 'what are the chances that this second child will be born a boy' we are asking the question "what are the chances that the man fathered two boys (since we know that he already fathered one)."

Yes. The crucial thing is, he's not asking, "What's the chance that this particular child is a boy?" but "What's the chance that both my children are boys?"

quote:I haven't quite got the "Tuesday" part of it yet, but I think it works from the same principle. Something about subtracting the time where BOTH are boys and BOTH are born on Tuesday figures in, but I'm not that strong on this sort of thing.

You're dangerously close. If the numbers game is getting in the way, you can get quite close to understanding the why of it intuitively instead of mathematically.

Let's try an easier one. We're talking about coins again. You get a friend to toss two coins one after the other (without watching), and you need to work out what the chances are that he got two heads. You get to ask him one question about his toss, and that's it.

The key is that the more specific you are about which coin you're asking about, the closer the question becomes to, "what's the chances that one particular coin came up heads?" which is always ½.

To start with, you don't know anything about the coins. So the chance that both came up heads is 1/4.

You ask: "Did at least one of the coins come up heads?" He says yes. So now you know that at least one is heads, but not if it was the first one or the last one. So the possibilities now are:

{Heads Tails}, and he was talking about the first one;

{Tails Heads}, and he was talking about the last one; or

{Heads Heads}, and he could have been talking about either of them.

So the chances are 1/3 that both his coins came up tails.

You ask: "Did the first coin come up heads?" He says yes. Now, that's pretty darn specific! It basically eliminates one of the coins completely. So now you're looking at:

{Heads Tails}; or

{Heads Heads}.

In other words, there's a ½ probability that the toss came up {Heads Heads}.

You ask: "Did a silver coin come up heads?" He says yes.

Let's assume that coins are either silver or copper, and that both are equally common. Your question is a bit more specific than just asking if one coin came up heads, but not as specific as narrowing it down to the first coin. The possibilities here are:

{SilverHeads CopperTails};

{SilverHeads SilverTails};

{SilverHeads CopperHeads};

{CopperTails SilverHeads};

{SilverTails SilverHeads};

{CopperHeads SilverHeads}; and

{SilverHeads SilverHeads}.

Count them: there are seven combination of heads and tails, and each is equally likely; however, there are only three cases where both coins came up heads. So the probability is 3/7.

The reason that it's not ½ is that there's a small chance that both coins came up heads and both were silver, and in that case he could have been talking about either of the two coins. The coin being silver gives us more information, but not enough to distinguish between them when both of the coins are silver.

Just a final one to amuse you. Let's say that you know that in a tiny number of coins, Roosevelt is clearly wearing a feather boa. You ask: "Did a coin with Roosevelt wearing a feather boa come up heads?" He says yes. Now, you know that it's pretty darn unlikely that he has two of those coins, so you can almost be sure that he was talking about a specific coin and there's only one coin left to guess at. But, in the improbable event that two of his coins had the beboaed Roosevelt, he could have been talking about either of the two.

The idea is that the more likely it is that you've narrowed it down to just one of the coins, the closer you get to only having to guess at what one of the coins is.

[ July 25, 2010, 10:26 AM: Message edited by: Jordan ]
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quote:JoshuaD: I suspect the following two cases are distinct. Do you agree?

No, don't pollute my beautiful maths with tricksy semantics!

The first one is very ambiguous. My first reading of it is that the man has to actually select a son to talk about in order to answer your direct question; in this case, it reduces to Aris' interpretation and the probability is 50%.

The second one is very clear, and it comes to 13/27 as per the dictates of conditional probability.

So my first reading is that they're distinct; but if you had asked, "tell me the day that at least one of your sons was born on" in the first question, then I'd say the scenarios were identical. The key thing is that he's narrowed down which of his children he might be talking about a bit, but not all the way.
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posted
Jordan: My point with those two scenarios is that in the first case the day the person was born on didn't matter. The man and I would be having that conversation regardless. In the second scenario, we'd only be having the conversation iff he had at least one son born on Tuesday.

The first case doesn't filter, the second case does. Do you think this matters?
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posted
I see. No, it wouldn't matter. If you replied, "Tell me a day of the week on which at least one of your sons was born," the number of eventualities comes out the same regardless of what day of the week he replied with (and there would be less ambiguity that he was selecting the son to tell you about beforehand).

It doesn't matter how you come by the additional information, it's the fact that you have that additional information that counts. As soon as he tells you what day of the week one of his sons was born on, it increases the amount of information you have about his children.

Edited to add: looking at the "friend flipped two coins" question might help to clarify.

[ July 25, 2010, 01:38 PM: Message edited by: Jordan ]
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posted
I didn't really write this in the flow of the conversation, so I apologize if I'm reiterating anthing that's been said already... I just needed to get it out, though it looks like it fits in well with Jordan and JoshuaD's discussion.

There's something to this problem that I think is being missed. It's extremely subtle, and it's bugging the heck out of me.

First, while I would agree that the math is correct for a particular interpretation of the statement made by the father, there remains some significant ambiguity in how the statement should, in fact, be interpreted. Consider the following generic form of the father's statement: "I have two kids. One of them is a boy. He was born on [day of the week]." This is trivially true for any boy, as all boys are born on some day of the week. So let's consider the original problem in which the day of the week is excluded: "I have two kids. One of them is a boy." The probability that the other is a boy is 1/3. Now let's say that after the fact, before revealing the identities of his children the father states: "he was born on [day of the week]." Does the probability suddenly change? He has to have been born on some day of the week, so it's not as if any new information is being offered. So where's the difference coming from?

Here's the deal: when considering two boys with different birth days of the week, the father must choose which boy's birthday to use in statement. What's to stop the father from using Monday in the statement (assuming it's not a lie), when the other boy's birth day of the week is Tuesday? If the father is the one choosing the day, we're left with no new information that would lead to a new probability (unless we know something more about how the father would choose). To illustrate this, imagine that the father uses the following selection criteria: if the father has two sons he will only use Tuesday in the statement if both boys were born on Tuesday (he'll use the other son's birthday in every other case). The father has not lied in any way here, or in my estimation been unfaithful to his statement, and the resulting probability would have to be, if my math is correct, 1/15 (subtract all the cases with a second boy not born on a Tuesday). Of course, if the father were to make the same statement with a different day of the week which would have to have different selection criteria the probability would be different. Just to give another example, if the father always stated the birth day of the week of the older brother we'd be back to 1/3 regardless of the day of the week stated, and if he always states Tuesday if any of the boys are born on Tuesday we get 13/27. I suspect, but haven't worked it out, that we would get 1/3 if he chose at random which boy's birthday to use. Furthermore, there doesn't appear to be any way to constrain the father's original statement in a practical way (without it taking on the flavor of a legal disclaimer) that would give the appropriate interpretation for the original mathematically derived result. On the other hand, if Disney World opened its doors one day only to families with two children, one of whom was a boy born on a Tuesday, we'd be in business. To be more explicit about it, if the father were somehow constrained to acknowledge Tuesday birthdays above all others, or the population has already been filtered then the math is right. Otherwise, I don't think it is.
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posted
If I understand Jordan correctly, the more information we have, the more likely it is to be 50%. That we start out with no extra info, so we have to go with pure probability about a theoretical family, but that as we get more info about this particular family, we can narrow it down. Okay. That makes sense.

But no-one has answered my last question yet. (Or if you did, it went completely over my head)

If you have exactly two kids, and one is a boy, then either the elder is a boy, or the younger is a boy. And everyone agrees that in both of those cases, it becomes 50:50 as to the sex of the other.

So why are the actual odds 1:3 with that level of information? Does deduction count as information?
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posted
OMG! What on earth is this thread on about? I can't believe no one has simply pointed out that if the outcomes of two events are not linked then you can not mathematically combine the probabilities! The probability that John has two boys is 50:50 = the probability the unknown child is a boy. The gender of the unknown child is not dependent on the gender of the known child (excepting any biological factors).
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quote:The gender of the unknown child is not dependent on the gender of the known child (excepting any biological factors).

Hee. Badvok, Google the problem we discuss at the beginning of this scenario, the Monty Hall problem.
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quote:If you have exactly two kids, and one is a boy, then either the elder is a boy, or the younger is a boy. And everyone agrees that in both of those cases, it becomes 50:50 as to the sex of the other.

So why are the actual odds 1:3 with that level of information?

Elder Boy Younger Girl

Elder Girl Younger Boy

Elder Boy Younger Boy

As you can see, there is one chance in three, because the case of "Elder Boy/Younger Boy" actually collapses the two potential cases "Boy 1 older/Boy 2 younger" and "Boy 2 older/Boy 1 younger." When you know which one is older, you can specifically say "Boy 1 is older" and thus eliminate the special case.
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