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Author Topic: Annoying probability question of the day
Badvok
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quote:
Originally posted by TomDavidson:
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. [Smile]
The Monty Hall problem is totally different because there is a direct causal link.
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JoshuaD
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Jordan: I was up all night thinking about this, and I'm now nearly certain that you're partially wrong. :-)

It's not the asking of the question that matters. It's whether the question filtered the input.

Imagine the following scenario: I have a room of 1,000,000 dads who have two children and at least one son. I ask each of them "Is your other child a boy?" How many will answer yes? Approximately 333,333.

Now, same scenario. A room full of 1,000,000 dads who each have two children and at least one is a son. Instead I first ask, because I'm trying to kill time, "What day of the week was your son born on?" and receive an answer. I then continue to ask "Is your other child a boy?" How many will answer yes? Approximately 333,333. Nothing has changed from the first scenario.

Now, the final scenario. I have a room of 1,000,000 dads who each have two children, one of which is a boy who was born on Tuesday. I ask each of these men "Is your other child a boy?" How many will answer yes? Approximately 481,481.

See below for some more interesting contrasted scenarios. I'm about 99% sure this is correct. What do you think Jordan?

  • Scenario 1: I have a room of 1,000,000 fathers with 2 children. I stand them in a line and ask them one at a time "Are your children the same sex?" Approximately how many of them will answer yes? Approximately 500,000.
    -
  • Scenario 2: I have a room of 1,000,000 fathers with 2 children. I ask them one at a time "Do you have at least one son?" If they answer no, they are asked to leave and their answer is not tallied. If they answer yes, I continue: "Are your children the same sex?" Approximately how many of them will answer yes? Approximately 750,000 will have been asked the 2nd question, and approximately 250,000 will have answered yes to the second question. Approximately 250,000 will have been asked to leave without being tallied. One third of those asked the second question will have answered yes. One quarter of the entire group will have answered yes.
    -
  • Scenario 3: I have a room of 1,000,000 fathers who have 2 children where one of the children is a boy. I stand them in a line and ask them each: "Are both your children boys?" How many of them will answer yes? Approximately 333,333.
    -
  • Scenario 4: I have a room of 1,000,000 fathers who have 2 children where one child is a boy. I stand them in a line and ask them each "What day was your son born on?" and receive an answer. I then continue to ask "Are both your children boys?" How many of of them will answer yes? Approximately 333,333.
    -
  • Scenario 5: I have a room of 1,000,000 fathers who have 2 children where one child is a boy. I stand them in a line and ask them each "What day was your son born on?" If answer any day other than Tuesday, I ask them to leave. If they answer Tuesday I then continue to ask "Are both your children boys?" How many will answer yes to this question? Approximately 68783. (1,000,000 * 1/7 * 13/27).


[ July 26, 2010, 10:50 AM: Message edited by: JoshuaD ]

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TomDavidson
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quote:
The Monty Hall problem is totally different because there is a direct causal link.
What is the causal link? There is either a goat behind the door or there is not. Opening a door does not "cause" a goat to be placed; it simply removes one possibility from the set of unopened doors.

In the same way, telling someone that one of your two children is a boy removes one possibility from the set of possible children (namely, that you have two girls).

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Badvok
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quote:
Originally posted by TomDavidson:
quote:
The Monty Hall problem is totally different because there is a direct causal link.
What is the causal link? There is either a goat behind the door or there is not. Opening a door does not "cause" a goat to be placed; it simply removes one possibility from the set of unopened doors.

Nope, it doesn't "cause" the goat to appear but the contestant's choice directly affects Monty's choice - which in turn affects the odds faced by the contestant (edit to clarify: because Monty will always pick a goat).
quote:
In the same way, telling someone that one of your two children is a boy removes one possibility from the set of possible children (namely, that you have two girls).

True, but that isn't the probability space. John has at least one son, so he can either have two sons or a son and a daughter - 50:50. The gender of one child doesn't alter the gender of the other child.

[ July 26, 2010, 11:16 AM: Message edited by: Badvok ]

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JoshuaD
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Badvok: It does alter the probability space. You are going to be disregarding 1/4 of all fathers of two children as you ask this question. If you walks up to him and ask him if he has at least one son and he says no, you're going to simply walk away. It's only in the other 3 scenarios [BG, GB and BB] that you stay and take a guess at the gender of his other child.
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Badvok
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quote:
Originally posted by JoshuaD:
Badvok: It does alter the probability space. You are going to be disregarding 1/4 of all fathers of two children as you ask this question.

What question? John states that he has one son! So the probability space only concerns the other child.
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Pete at Home
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quote:
Originally posted by TomDavidson:
quote:
If he randomly selected a kid whose gender he wanted to reveal, then the fact he mentioned it was a boy, leaves the other one's chance at 50%, not 33%.
Um....As far as I can tell, you're confusing this with an actual Monty Hall problem.

In the original Monty Hall problem, it matters whether Monty knows which doors have goats behind them because he is required to open a door that contains a goat. If he were not required to do so, and in fact selected doors at random, then it's true that his opening a door has no effect on the probability that the original door selected contains the car.

And how exactly does the goat transform into a car? Higher math? [Wink]
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JoshuaD
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quote:
Originally posted by Badvok:
quote:
Originally posted by JoshuaD:
Badvok: It does alter the probability space. You are going to be disregarding 1/4 of all fathers of two children as you ask this question.

What question? John states that he has one son! So the probability space only concerns the other child.
Read my long post above.

The unspoken assumption of this probability question is that you wouldn't be having this conversation with John at all if he had two daughters.

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DonaldD
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quote:
What question? John states that he has one son! So the probability space only concerns the other child
Badvok, you are ignoring that birth 'order' matters. You don't know whether 'John' is talking about his elder child or his younger child. There are 3 possible sequences of birth in this case: first born is Girl, second is Boy; first born is Boy, second is Girl; first born is Boy, second is also Boy.

That's 2 ways for one girl to be born and only one way for both children to be boys.

Or think about it this way: in a two-child family, the chance of having one boy and one girl is 50%, two boys is 25% and 2 girls is 25%.

By removing the 2-girl option, you are left with the other 3 options, where having a mix is still a higher proportion.

Or, imagine 4 doors: one with 2 girls, one with 2 boys, one with a boy and a girl, one with a girl and a boy.

Monty opens the door with 2 girls. How many doors are left with 2 boys, and how many are left with only one?

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Badvok
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Apologies to all - I seem to have gone down the wrong track here, I thought this long meandering thread was talking about:
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?
The answer to that question is simply 50:50. There is no other answer, we are simply asking for the probability a child is a boy or a girl.
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DonaldD
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Err, no,it's not [Smile]
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Badvok
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quote:
Originally posted by DonaldD:
Badvok, you are ignoring that birth 'order' matters. You don't know whether 'John' is talking about his elder child or his younger child. There are 3 possible sequences of birth in this case: first born is Girl, second is Boy; first born is Boy, second is Girl; first born is Boy, second is also Boy.[/QB]

Birth order cancels itself out. You are only counting boy-boy once - it should be twice Boy1-Boy2 and Boy2-Boy1.
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DonaldD
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No, you really shouldn't.

Unless you are suggesting that your first born can be younger than your second born child.

But even so, in such a time space you first born daughter could also be younger than your second born son, so that also cancels out.

Think of it in real-world terms to clarify the situation. You have two children: an elder child and a younger child. In how many ways can they be born?

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JoshuaD
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Badvok: No. If birth order cancels out, then the odds of me having two children of the same sex is 2/3: [B G], [B B], [G G]. This is incorrect. The odds of me having two children of the same sex is 1/2: [B G], [G B], [B B], [G G].

Please try to refrain from knee-jerk responses. Some of this mathematics is unintuitive. Take some time to think about the responses you receive before plowing forward with your insistence; this will save un-necessary repetition in the thread and it will also give you more time to formulate your objections in a coherent and complete way rather than shooting them from the hip as they come to mind.

I was suspicious of the 13/27th reasoning for nearly 4 days before I finally posted my reasoning on why I think it's not exactly correct. The thread will be here when you're done thinking, I promise. :-)

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Badvok
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OK, to clarify how birth order is irrelevant:

Child1:Child2
Eldest+Boy:Youngest+Boy
Youngest+Boy:Eldest+Boy
Eldest+Girl:Youngest+Boy
Youngest+Girl:Eldest+Boy

= 4 combinations

[ July 26, 2010, 12:05 PM: Message edited by: Badvok ]

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DonaldD
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You forgot about:

Youngest+Boy:Eldest+Girl
Eldest+Boy:Youngest+Girl

= 6 combinations

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Badvok
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quote:
Originally posted by JoshuaD:
The odds of me having two children of the same sex is 1/2: [B G], [G B], [B B], [G G].

Yes, and?
quote:
Originally posted by JoshuaD:
Please try to refrain from knee-jerk responses.

I'll throw the same back at you!

I SAID birth order cancels out and is irrelevant NOT that the probability of same sex offspring is 2/3 - I don't know where you got that from!

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Badvok
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quote:
Originally posted by DonaldD:
You forgot about:

Youngest+Boy:Eldest+Girl
Eldest+Boy:Youngest+Girl

= 6 combinations

Eh? I think I listed those.
Edit: Ooops sorry, no I didn't - need to go home now!

[ July 26, 2010, 12:19 PM: Message edited by: Badvok ]

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DonaldD
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If birth order is irrelevant, then the only possibilities of child combinations are, as JD mentions, [B B], [G G] and [B G]. If order is truly irrelevant, then each of these options would be accorded the same probability, no? In which case, each situation would occur 1/3 of the time. [B B] + [G G] = 1/3 + 1/3 = 2/3.

I believe JD was attempting reductio ad absurdum.

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DonaldD
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quote:
Eh? I think I listed those
Nope, you listed their inverse cases only.
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DonaldD
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I think you may be getting caught up in Aris' option from the previous pages - specifically, that you are thinking of a particular child. Yes, if you rephrase the question as "John Doe comes up to you and says: "I have two kids. My son John is a boy..." then you end up with a 50% answer (unless George mentions his child George, in which case all bets are off.)
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DonaldD
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quote:
Birth order cancels itself out. You are only counting boy-boy once - it should be twice Boy1-Boy2 and Boy2-Boy1.
To focus on this just a little further - if Boy1 and Boy2 are not temporal distinctions, but rather actual names, think of it like this:
Boy1 = John
Boy2 = Sam
But then, you would need to name the girls as well:
Girl1 = May
Girl2 = Jan

In which case, you are more obviously missing some combinations. If you can have [B1 B2] (John/Sam) as well as [B2 B1] (Sam/John) then you should also accept both [B1 G1] (John/May) as well as [G1 B1] (May/John), [B1 G2] (John/Jan) and [G2 B1] (Jan/John)

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vegimo
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Badvok:
The reason birth order does not cancel out and remains relevant is the same reason you can't count the boys twice. The probability that any particular child will be a boy is 1/2. In a large 2-child-family population though, there will be 1/4 of the families with 2 boys, 1/4 of the families with 2 girls, and 1/2 the families with a boy and a girl. If we exclude the 2-girl families (because of the condition of the problem), there are now 1/3 of the families with 2 boys and 2/3 of the families with one of each. Just because a father has 2 boys does not mean he is twice as likely to be the particular father who approaches you. Any of the fathers could be the proud papa, and when he shticks his boast on you, there is a 1/3 chance that he is one of the fathers with 2 sons and a 2/3 chance that he is the father of a son and a daughter.

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Brian
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Tom:
quote:
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.
That was kinda my point. If the specific boy mentioned is older, then the odds are 50:50. If the specific boy mentioned is younger, then the odds are 50:50. Do I have that right so far?

If so, then those are the only two possibilities. That specific boy is either younger or older. Unless his last name is Schroedinger. [Smile]
If the only two choices lead to a 50:50 chance, why would the intermediate odds be different just because we don't know which one it is yet?

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scifibum
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My thought after reading this thread: it's a lot harder to figure out how probability rules apply to informally described questions than it is to do the math.
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DonaldD
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Brian, the range of truthful statements possible is not limited to 'my eldest is a boy' or 'my youngest is a boy'.

In the population in question, there is also the possibility of saying 'my eldest is a girl' and 'my youngest is a girl'. This occurs in 1/3 of the cases in the population in question.

So, 50% (boy or girl after identifying a boy) * 2/3 (likelihood of father chosing to identify a boy in this population) still gives 1/3.

Whereas the chance of the 'other' child being a boy when the father identifies a girl (in this population) is 100%. 100% * 1/3 is also 1/3.

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vegimo
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Brian,

(Another wording of DonaldD's response)
Your issue with the problem leads back to what the question actually asks. We are not trying to figure out whether either child 1 is a boy or child 2 is a boy. We are trying to figure out whether the father belongs to the portion of the population with 2 boys or the portion of the population with a boy and a girl.

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PSRT
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Well, DonaldD, we don't really know what the population in question is. How was the man chosen? We're told he approaches us. But that doesn't give us enough information about the population he is approaching us from. If he is drawn from the population of all fathers with two children, the answer is 1/2. If he is drawn from the population of fathers with two children, one of whom is male, the answer is 1/3. If he is drawn from the population of fathers who have a son born on Tuesday, the answer is 13/27.

Our assumptions about the English language matter in solving this problem.

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Jordan
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quote:
JoshuaD:
Jordan: I was up all night thinking about this, and I'm now nearly certain that you're partially wrong. :-)

You are very possibly correct, this could become a situation where the teacher becomes the student. I'll get back to you tomorrow after having a proper think. [Smile]
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DonaldD
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quote:
Well, DonaldD, we don't really know what the population in question is
PSRT: Actually, we do. Right now, we are working with the simplified population of father with at least one boy.

If we were talking about the population of Father with a Tuesday boy, the math gets more complicated but the methodology would be similar (and yes, we would end up with the 13/27 probability)

The fact that we have used the father's statement to identify the population does not preclude the possibility of different truthful statements being made about the children in question.

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Mariner
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Cool, I dropped a bomb, ran away, and created lots of interesting discussion.

Thanks Jordan, for articulating clearly the issue that I was gradually understanding as I read the thread. We have a range of probabilities being argued here: 1/3, almost 1/2 (13/27), and 1/2. The reason the second one (Boy born on Tuesday) is almost 1/2 is because, as Jordan said, by adding more information it becomes more and more likely that you're describing a specific child rather than making a general observation.

So with that said, let's pick some scenarios in which we can (mostly) all agree that the probability Joe has 2 boys is 1/2:
- Joe comes up to you and says: "I have two kids. The oldest is a boy."
- Joe comes up to you and says: "I have two kids. The youngest is a boy."
- Joe comes up to you and says: "I have two kids. One of them is named Jake." (barring the highly improbable occurence that he named a girl Jake or that he named both sons Jake)
- Joe comes up to you and says: "I have two kids." A boy then walks up and says "Hi Dad" to Joe.

In each of these cases, Joe is referring to a specific child when the sex of one of them is revealed. When that happens, the question of whether or not Joe has two boys really boils down to the question of whether or not Joe's OTHER child is a boy. And there, we know its 50-50. That's why the "born on Tuesday" bit is relevant under Jordan's interpretation of the question. It increases the odds that the "at least one is a boy..." is referring to a specific boy. That wasn't clear to me at first, but now I understand.

But I don't think Jordan's entirely right that the issue is only due to the ambiguity of the English language. It does come down to when and where the selection is made. Aris is right in that the intention of Joe is important. Or more exactly, I think the key here is less the ambiguity of the English language, and more that you don't know all the rules of the game. I'm going to switch to flipping two pennies here because it's easier to comprehend:

New scenario: Joe flips two pennies. He says "At least one is heads" Jordan responds "I bet you have one heads and one tails" Jordan then provides his reasoning. There are four initial possibilities:
HH - 25% chance
HT - 25% chance
TH - 25% chance
TT - 25% chance
100% total

Joe then specifically eliminated the final possibility. In other words, Joe truthfully saying "at least one is heads" can only happen 75% of the time. Therefore, you eliminate the final possibility, and divide each of the other chances by 75%
HH - 33% chance
HT - 33% chance
TH - 33% chance

Jordan notes that the combination of one head and one tail is twice as likely as the head-head combination, and therefore concludes that its a safe bet.

But is that what actually happens? Let's change the rules of the game a bit. Once again, Joe flips two coins. He then feels absolutely compelled to say "At least one is [heads/tails]" (and he must speak truthfully, obviously). Now, there are four possibilities of what can happen:

HH - Joe is forced to say "At least one is heads" - 25% chance
HT - Joe can choose whether he wants to say "At least one is heads" or "At least one is tails" - 25% chance
TH - Joe can choose whether he wants to say "At least one is heads" or "At least one is tails" - 25% chance
TT - Joe is forced to say "At least one is tails" - 25% chance

Assuming Joe is not biased in terms of picking heads or tails, there are now a grand total of 6 possibilities:
1) HH - "At least one is heads" - 25% chance
2) HT - "At least one is heads" - 12.5% chance (the original 25% multiplied by 50% of choosing to say heads)
3) HT - "At least one is tails" - 12.5% chance
4) TH - "At least one is heads" - 12.5% chance
5) TH - "At least one is tails" - 12.5% chance
6) TT - "At least one is tails" - 25% chance

Joe says "At least one is heads" Looking at the above, you eliminate options 3, 5, and 6. In other words, Joe truthfully saying "At least one is heads" only happens 50% of the time (note that this is the key difference between this and Jordan's logic: Jordan says that he CAN say it 75% of the time, but here I say that he WILL say it 50% of the time). So, once again, we take the available options and divide them by the probability that this happens, and we get:
1) HH - "At least one is heads" - 50% chance
2) HT - "At least one is heads" - 25% chance
4) TH - "At least one is heads" - 25% chance
Now, if Jordan says that he bets that 1 is heads and 1 is tails, he only has a 50-50 shot.

By the way, in a third scenario, Joe flips two coins. Jordan asks "Is at least one heads?" Joe nods. Jordan can then go through his initial reasoning and correctly obtain the 2/3 probability of one head and one tail. This is functionally equivalent to the probability that Joe CAN truthfully say "at least one is heads".

So the question is, is Joe's statement that at least one of his kids is a boy born on Tuesday coupled with the original game (randomly selected two child families) or not? In other words, do we care whether or not Joe CAN say "One is a boy born on a Tuesday" or whether he WILL say "One is a boy born on a Tuesday" if he's going to mention the sex and the date?

To put it into precise terms for methematicians to use, we have two scenarios:

"A family has two children. Given that at least one of them is a boy who was born on a Tuesday, what is the probability that both of the children are boys?" Answer: 13/27

"A family has two children. Given that the father will tell you the sex and gender of one of them, what is the probability of both children being boys if he says that at least one is a boy born on a Tuesday?" Answer: 1/2

And from the original scenario, I don't know if you can with complete certainty choose which scenario is correct.

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scifibum
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Quibble:
quote:
- Joe comes up to you and says: "I have two kids." A boy then walks up and says "Hi Dad" to Joe.
Since you don't know whether this kid is the oldest or the youngest, this does not seem to me to narrow the possibilities beyond the generic "has two kids at least one of whom is a boy".

Edit: Unless knowing the kid's hair color, height, or approximate age has anything to do with it. Which it doesn't feel like they should. But my intuition has been wrong on this from the start, so...

(I'm comforted by the fact that it would be difficult to construct an accurate field of equally-likely variations on these particular attributes. Or would it? *head explodes*)

[ July 26, 2010, 04:53 PM: Message edited by: scifibum ]

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JoshuaD
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Mariner and I are in agreement on how this question is properly approached. Really cool lesson in probability. I spent years as a successful poker player, and this still sent me through a loop for a few days.
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DonaldD
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Mariner: in your second scenario, you have made a mistake: the fact that in the HT and TH cases the person chooses to state either heads of tails does not affect the probability of the combination actually occuring.

So even though there may only be a 12.5% likelihood of him saying 'H' in either case, it doesn't change the number of occurences of actual 'HT' or 'TH' combinations.

If, however, you are positing that the guesser knows what is going on inside the first guy's head... well in that case, we could make up just about anything. Essentially, the question comes back to the following: in the puzzle, what is the purpose of the father's initial statement? Is it to limit the population, or to trick you with language?

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Mariner
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Scifibum, the "oldest" vs "youngest" is just a useful tool to keep them separated, it doesn't actually matter. What matters is that sex determination is independent of one another. When we say that order matters, we don't necessarily mean chronological order. It could be in terms of height, or which one is furthest north at the moment, or which one managed to get through the most posts in this thread without their head exploding. It's just to insure that the two children are independent.

So by having one of the children appear, you are specifically defining one of the children. Because of that, you are specifically dealing with one unknown child when asked if there are two boys.

Imagine if a child was hiding under a cardboard box, and you had to guess whether it was a boy or girl. Imagine then if said child's brother walked in the room. Would that change your guess? No. You are still guessing about the sex of only one child. And if a boy comes up to Joe and says "Hi Dad", then there's only one kid left that you have to guess.

Donald, there is no mistake. Yes, the odds of HT appearing are 25%. HT appears in scenario 2 and 3 in my list above. If you add the probabilities of scenario 2 and 3 together, you get 25%. Exactly what you would expect.

What I did was no different then the mathematical way to solve the Monty Hall problem. Suppose I choose Door #1. There are 4 possibilities:
1) The prize is in #1, and door #2 is opened to show the goat
2) The prize is in #1, and door #3 is opened to show the goat
3) The prize is in #2, and door #3 is opened to show the goat
4) The prize is in #3, and door #2 is opened to show the goat

The odds of each occurring are 1/6, 1/6, 1/3, and 1/3 respectively. Thus, the odds that the prize are in door #1 is still 1/3, since you add the probability of scenarios 1 and 2 together. After a door opens (we'll say it's door #2), you collapse the options and redetermine the probability. In this case, it becomes 1/3 that the prize is in door #1 and 2/3 that it's in door #3. Same thing as I did above.

And you're right that the question comes back to the purpose of the father's initial statement, but your options are misleading. It's actually: is it to describe a specific child (answer is 1/2), or is it to eliminate specific options (answer is 1/3).

Now to confuse myself further. What if Joe had said "I have two kids. But I don't have two girls." Now, it's clear that the purpose of his statement is to eliminate specific options, but it's functionally identical to "at least one is a boy". So now what?

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PSRT
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quote:
PSRT: Actually, we do.
Not from the question as posed to us. If we wish to specify which set of parents our hypothetical father has been drawn from, we can do that, and then get an exact answer. Without specifying, the question as worded is ambiguous enough that there are multiple correct answers, depending upon interpretation.
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scifibum
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Mariner, that's not satisfying. Back to this scenario:

A man has two kids, at least one of whom is a boy. The probability that he has two boys is 1/3.

The reason that it's 1/3 is that there are three distinct (and equally probable, for argument's sake) ways for the man to get two kids, at least one of whom is a boy. BB, BG, GB. The only thing that distinguishes the latter two scenarios is birth order.

By having a boy who forms one member of one of the three duos above walk into the room, you haven't eliminated ANY of the three scenarios. Whereas knowing the boy is the oldest DOES eliminate the last scenario.

[ July 26, 2010, 09:43 PM: Message edited by: scifibum ]

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Badvok
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OK, I got a little messed up yesterday in trying to put this straight.

I guess we need to go back to basics:

Example 1:
What is the probability of a child being a boy or a girl?
Answer: Two possibilities (in this simplified mathematical world anyway), therefore 1/2

Example 2:
If there are two children what is the probability of them both being boys?
Answer: Four possibilities (B+B, B+G, G+B, G+G), therefore 1/4

Example 3:
If there are two children what is the probability of one child being older than the other?
Answer: Almost 1, but for the sake of this question lets call it 1 (the chances of finding two children born at exactly the same moment in time is pretty slim and almost impossible if they are from the same mother - sorry girls no matter how horrible it sounds it is not totally impossible).

Example 4:
If there are two children what is the probability of them both being boys if one is a boy?
Answer: OK, this is the main question.
See example 1, we have 2 possibilities for gender of each child.
See example 2, we have 4 possibilities for combinations of two children: B+B, B+G, G+B, G+G.

Now, this is the contentious bit, for the sake of clarity we need to identify the children we are
talking about so we don't get confused, we'll use androgynous names and call them Child1 and Child2.

Child1 is either a boy or a girl, the probability of either is 1/2.
Child2 is either a boy or a girl, the probability of either is 1/2.

If we collapse the first statement to say that Child1 is a boy it has no effect on the second statement and we now have:

Child1 is a boy, probability 1.
Child2 is either a boy or a girl, probability 1/2.

Therefore the probability of both Child1 and Child2 being boys is 1 X 1/2 = 1/2.

Or if we collapse it the other way around:

Child1 is either a boy of a girl, probability 1/2.
Child2 is a boy, probability 1.

Therefore the probability of both Child1 and Child2 being boys is 1/2 X 1 = 1/2.

The classic mistake many are making is to take only part of the probability space from example 2 - only those that contain a boy - doing this is mathematically incorrect - you can't split probability spaces like that!

Example 5:
Does age or birth order affect the result of Example 4?

As we see from Example 3, the probability that one child is older than the other is 1 and therefore it will have no effect on the calculation.

Or to put it another way:

If Child1 is the elder, what is the probability that Child2 is the younger? Answer 1.

And Conversely:

If Child1 is the younger, what is the probability that Child2 is the elder? Answer 1.

If we leave aside real-life biology we can also say that birth order does not affect gender probabilities and therefore it has no relevance to a problem space concerning only gender.

But if we must examine this further to satisfy those who still have doubts then we see that we have the following permutations:

Child1 is eldest and a boy, Child2 is youngest and a boy
Child1 is eldest and a boy, Child2 is youngest and a girl
Child1 is eldest and a girl, Child2 is youngest and a boy
Child1 is eldest and a girl, Child2 is youngest and a girl
Child1 is youngest and a boy, Child2 is eldest and a boy
Child1 is youngest and a boy, Child2 is eldest and a girl
Child1 is youngest and a girl, Child2 is eldest and a boy
Child1 is youngest and a girl, Child2 is eldest and a girl

(Sorry DonaldD, we both missed some yesterday.)

Now because we have introduced age/birth order we have doubled the size of the problem space but we haven't actually changed anything. The probability of both children being boys is now 2/8 = 1/4.

And if we collapse the problem space by saying Child1 is a boy we get:

Child1 is eldest and a boy, Child2 is youngest and a boy
Child1 is eldest and a boy, Child2 is youngest and a girl
Child1 is youngest and a boy, Child2 is eldest and a boy
Child1 is youngest and a boy, Child2 is eldest and a girl

So the probability of both children being boys is 2/4 = 1/2.

The key thing to remember is that there are two children and they are distinct and separate entities. One child's gender is not linked in any way to their sibling's gender (in this mathematical problem space anyway, even if in real-life there are quite significant links).

Q.E.D.

[ July 27, 2010, 05:41 AM: Message edited by: Badvok ]

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Aris Katsaris
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quote:
Originally posted by JoshuaD:
[QB]
Now, same scenario. A room full of 1,000,000 dads who each have two children and at least one is a son. Instead I first ask, because I'm trying to kill time, "What day of the week was your son born on?" and receive an answer.

Really? What the hell of an answer will people with two sons give you? This question is meaningless and unanswerable for people with two sons.

quote:
I then continue to ask "Is your other child a boy?" How many will answer yes?
Other of what? If someone with two sons says "I have at least one son", then asking if your other child is a boy is meaningless question because there's no "other". He wasn't referring to any particular boy.

quote:
Scenario 5: I have a room of 1,000,000 fathers who have 2 children where one child is a boy. I stand them in a line and ask them each "What day was your son born on?" If answer any day other than Tuesday, I ask them to leave.
Again a meaningless question for fathers with two sons.

quote:
the specific boy mentioned is older, then the odds are 50:50. If the specific boy mentioned is younger, then the odds are 50:50.
If the father is referring to a specific child, then the odds are 50:50.
If the father isn't referring to a specific child, then the odds are 1/3.

That was what my whole contribution to the thread was about.

quote:
If the only two choices lead to a 50:50 chance, why would the intermediate odds be different just because we don't know which one it is yet?
Because when a father says "atleast one of my sons" is a boy, the possibilities don't collapse to
a- I chose to mention younger son.
b- I chose to mention older son.

they collapse to this:
a- I fathered one boy then one girl.
b- I fathered one girl then one boy.
c- I fathered two boys.

I urge you to look at my own earlier contribution to this thread where I argue that the different reasons for the father mentioning the info he mentioned, affects the probability.

-----

This thread has derailed back into an earlier stage, with people forgetting the fact we've mentioned several times already: "atleast one son" doesn't pinpoint to one particular child. You can't ask a father the day his "atleast one son" was born, because they may be more than one.

Read the thread before you post, people.

[ July 27, 2010, 06:34 AM: Message edited by: Aris Katsaris ]

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Aris Katsaris
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Look at conditional probability please.

Probability(A given B) = (Probability of A AND B) / Probability (B)

Probability(two boys, given atleast one boy) = Probability(two boys) / probability(atleast one boy) = (1/4) / (3/4) = 33%

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