Reading the Monty Hall problem thread reminded me of a problem i once saw where you have a family with two children, where you know one is a boy, and you are supposed to find the probability for the other child being either a boy or a girl. For some reason the probability is given as 1/3 for it being a boy and 2/3 for being a girl.

The logic seems to go something like that you have four possible combinations, GG BB GB BG, since you know one is a boy the GG is out, and so you have BB GB and BG as possibilities.

Now I never figured out why this is supposed to be right. The problem, as far as I see it, is that the combination of BG and GB is the same in this problem, since the sequence is totally irrelevant for the question of whether the other child is a boy or girl(as it is only the probability of the gender of the sibling that you are supposed to compute). As I see it there is really only three possible combinations in the group of two children, either (both are boys), (both are girls) or (one is a boy and one is a girl). we know at least one is a boy, so the (both are girls) are out, leaving two possible answers, or 1/2 that the one you choose is correct.

Now, if the difference between BG and GB denotes something, for example age; the first of the pair being the oldest, there is an error here, that being that you in fact have more possibilities; Gg, gG, Bb, bB, Bg, gB, Gb and bG(capital denotes the child of known gender). Now the known child is a boy, so this excludes Gg, gG, Gb and bG, but you are still left with Bb, bB, Bg and gB, and so you have 2/4 for the sibling being a boy, so that is still 1/2(but if you were supposed to guess both the gender and whether the child is younger or older, you have a 1/4 probability)

So, where is it my logic is wrong, or have I perhaps missed/forgot some additional restriction that the puzzle imposes?

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But if you have BB, BG & GB

arn't the BG and GB senarios just the same so it should realy just be 1 in 2 chance that it is a girl because without the order all you have is BB &BG?

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For each kid, there's a 1/2 chance of that kid being a boy or a girl. With two kids, there's thus 1/4 chance of both girls and 1/4 chance of both boys. That leaves 1/2 chance of one of each.

Thus one of each is twice as likely as two boys.

However, if your first child is a boy, the odds that your second child will be a girl are 1/2, not 2/3.

Ain't probability fun?

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SeanF

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But in real life, the sex of the first child appears to have an influence on the sex of the second child. Couples who give birth to a boy seem to have a higher probability of having a boy as the second child. Couples who give birth to a girl seem to have a higher probability of having a girl as a second child.

There may be more at play here than statistical probability.

Single (type of) bullet theory????

If (I have not yet completely finished digesting the news) the answer to the Monty Hall problem is correct: then I suspect that the 2/3 answer is correct.

Although the order doese not seem relevant, that does not seem to let you collapse the chances.

It looks like in the Monty Hall problem: how you can't ignore the case where you picked the wrong door in your first guess, so Monty only has one choice of wrong door to then open for you. (These were the "repeated rows" in my last table).

If you picked a sample of 2 child families, you would "expect":
1/4 GG
1/4 BG
1/4 GB
1/4 BB

Now, in this question, order is irrelevant, so you could say it is:
1/4 GG
1/4 BG
1/4 BG
1/4 BB

Which collapses to:
1/4 GG
1/2 BG
1/4 BB

(i.e. BG is 1/2 not 1/4: the "other" 1/4 GB can't be thrown out)

Now, since you know one is a boy, the GG is out, and you get:
2/3 BG
1/3 BB

So it's 1/3 likely that you have the BB case, and that the other is a boy.

I sincerely hope I'm not wrong this time, and I've finally come up with a signature (TBA).

Cheers,

That difference is small, though, and so we ignore it--for purposes of shortening the debate

But, one boy and one girl in the long run happens twice as often as two girls and twice as often as two boys. It is twice as likely.

Just because something is one of 3 possibilities, does not mean it has a probability of 1/3.

It could have that probability, but it doesn't have to. It depends on the distribution. It could have any proability -- and in the case of 2 children being boy and girl, that happens half the time.

(Just like having 2 doors to shoose from does not necessarily make them equally likely to hide the prize.)

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It's interesting though. Because if it is only slightly over 50% that you're second kid will be the same as the first, then if you asked people to bet on randomly picked two-child families after having the sex of one kid revealed then this fact might lead them to go for the same sex, and on average they'd still lose

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If logic doesn't work, then surely it does.

If you take the GG,GB,BG &BB combinations they could be said to represent the order in which they are born so as you already have the first being a boy then all you are left with is BG&BB, so the chance of next being a boy is 1/2.

the same can be said in flipping a coin, even if you have 49 heads then the 50th still has a 50:50 chance of being another head.

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The problem doesn't say that the first-born is a boy, it only says that at least one of the children is a boy without specifying order.

but the boy is the first one to be counted.

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I think that in the original puzzle, being told that at least one is a boy is the same probablistically as this variation:

Imagine that we get a load of two child families and remove all the ones that have two girls. If we pick one at random there is a 3 in 3 chance that they have at least one boy, and a 2 in 3 chance that they have exactly one girl. So being told that at least one is a boy doesn't tell us anything or change any of the odds. Put another way, you could be told that there is at least one boy before the family is chosen.

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There are 10 types of people in the world. Those who understand ternary, those who don't, and those waiting for a bus.
If logic doesn't work, then surely it does.

the chance that the other will be a boy is 1/2
the chance that the other will be a girl is 1/2

ie the chances of it being a girl or a boy is 1/2+1/2=1(certainty)

what other options are their, a goblin?

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the chance that the other will be a boy is 1/3
the chance that the other will be a girl is 2/3.

ie the chances of it being a girl or a boy is 1/3+2/3=1(certainty)

their are no other options


If you just picked a (2 child) family at random there is 1/4 for each combination: BB BG GB GG, right? If all you are told is that it is not GG, the other three combinations are still equally likely.

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There are 10 types of people in the world. Those who understand ternary, those who don't, and those waiting for a bus.
If logic doesn't work, then surely it does.

by the same logic if you spin 9 tails with a coin then the chance of getting a tail for the tenth spin is 1/(2^10) there for you could place your bets on getting a head.

This can't be how it is.

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It's not. They had even odds for getting a boy or girl each time. It's the fact that only one of the possible outcomes has been eliminated, not two, that makes the difference.

If you flipped 10 coins, and for the outcomes where at least 9 of them were tails you aske the question "9 are tails, what's the unrevealed oen" (and discarded all other outcomes) then what is the chance that the unrevealed one is a tail? Remember that there is only one way that you could have flipped 10 tails, but 10 ways that you could have flipped 9 tails and 1 head.

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There are 10 types of people in the world. Those who understand ternary, those who don't, and those waiting for a bus.
If logic doesn't work, then surely it does.

If you have a family with with two children and you know one is a boy then either the other one is a

Boy

or it is a

Girl

you have only two choices therefore the chance of getting another boy is 1 in 2(1/2).

if the chances of the child being a girl=2/3 and the boy=1/3

then you would have the possibilities of

a)1 boy(that you know of) and 1 girl

b)1 boy(that you know of) and 1 girl

c)1 boy(that you know of) and 1 boy

so aren't (a) and (b) just the same out comes?

so you only have 2 out comes 1boy and 1 girl or 2boys.

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Yes, that may be what the puzzle was hinting at, but still I feel its missing something… hmmm… How about this point of view:

Say that you have a list of the children of the four two child families from the registry of inhabitants in a small town(that makes 8 children), you know that there is one of each of BB, GG, GB, BG.

You are instructed to choose one boy from the list at random, and give the probability for the sibling being male? I would say it is 1/2, since the BB family has two boys on the list, so that family has a higher chance of being selected than one of the one boy families individually. I think this is likely to happen if you choose a child at random from a larger group of two child families too.

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That's true if you choose a child at random, but not if you choose a family at random.

Frog march, like the Monty Haul question, I think this one is best answered by experiencing it. Make up about 50 cards or slips of paper (you can make fewer, but as always, statistics gives the best predictions when your sample is large). For each one, flip two coins, and based on their outcome label each card. I don't care who's heads and who's tails; if you get two boys, or a boy and a girl, write it down. But if you get two girls, forget about it, and flip the coins again (we're only counting families who have at least one boy). Now, go through the cards and tally whether you have two boys or a boy and a girl, and see what result you get.

As an aside, one of the reasons this is confusing is because of the way the sample is selected. Generally, if you know of a family who has two children, at least one of whom is a boy, it might often be because you've met them. Suppose you meet a mother out walking with her son, and in the conversation, you find out that the father is somewhere else with their other child. What's the probability that the other child is a boy? It's 1 in 2, but this is not the same selection rule as the one described, even though it may seem like it if you aren't paying close attention.

Suppose we make this a coin-flip puzzle.

Before you start, you’ve got these chances (boy = Heads, girl = Tails):

T then H

T then T

H then T

H then H

So you flip the first coin and you get Heads. What are the probabilities of what might result in the next flip?

Looks to me like you would have an equal chance of either an H or a T.

If your first flip results in Tails, seems to me that you would have an equal chance of either a Tails or Heads in your second flip.

In the original puzzle, in figuring the probabilities, you have to throw out the two possibilities of T-H and T-T, since your first toss resulted in an H. That leaves you with only the H-T or H-H possibilities, which seems to me to be a 50-50 chance for each one.

Hmmm, Yes, I see it, we are creating a group that is in favor of pairs of siblings of different gender if we pick our random child by family, even if there are only 1/2 the number of girls in the group, but if we do it by individual child without filtering by family we will have an equal amount of boys with brothers as boys with sisters since both boys in the BB families will be each others brother.

Heh, I guess this shows that one needs to be careful with what figures one plot into ones computations when doing statistics and probabilities...

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Nope. In the original puzzle, you throw out only the T-T result. You don't know that the first throw was heads, you just know that one or the other of them was heads. That leaves both T-H and H-T in your possible results.

The original puzzle said, "...you have a family with two children, where you know one is a boy..." Since T-H meets the criteria, you can't throw it out.

But as Grey pointed out, it is imperative that you know only that one of the children is a boy, not that you know a specific child is a boy.

If you meet a boy who says, "I have one sibling, is it a boy or a girl?" the odds are 1/2.

If you meet a mother with her son and the mother says, "I have one other child, is it a boy or a girl?" the odds are 1/2.

If you meet a mother and she says, "Our first child was a boy, is our second a boy or a girl?" the odds are 1/2.

If you meet a mother and she says, "One of our two children is a boy, is the other a boy or a girl?" the odds are 1/3-boy, 2/3-girl.

Probability is all in what you know or don't know. If I draw a card and ask somebody what the odds are it's a spade, they'll say 1/4. If I show somebody else the color of the card then ask what the odds are it's a spade, they'll say either 1/2 or 0. If I show somebody the entire card and ask what the odds are it's a spade, they'll say either 1/1 or 0.

So who's right?

They all are. Because they all know different things about the card.

Same with this case.

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2009 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

Ok, I see. I mis-remembered the original terms and I didn’t bother to re-read them.

This is the same error that people make with the Monty Hall problem. If it hasn't been named already it should be, maybe the "fallacy of equal probabilities."

If I buy a lottery ticket, either it is the winning one, or it isn't. You only have two choices - are they equally likely?

There is only one ticket that is right, there are many that are wrong. There is only one way to have two boys, there are two ways to have a boy and a girl.

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There are 10 types of people in the world. Those who understand ternary, those who don't, and those waiting for a bus.
If logic doesn't work, then surely it does.

Yeah, what's up with that? Why do some people assert that two possibilities must have equal proability.

In the next hour, you will die or you will not. Two possibilities.

Are any of the 50-50 people willing to explain why they are not in a constant state of despair?

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Maybe that’s why so many people worry about giant earthquakes, volcano caldera eruptions, asteroid collisions, comet crashes, and atomic wars.

You mean if the boy isn’t with the mother on that particular day, and she mentions him, then there is a 2/3 chance the other one is a girl?

If the boy is with her, there is 1/2 a chance the other is a girl, but if he stays home that day and she refers to him, there is a 2/3 chance the other is a girl? What if the boy is hiding behind a tree?