Please don't hate me for this puzzle

[Az]

You are going to your friends house. You know your friend has two children ages 4 and 6. You do not know what gender they are. When you get there a boy answers the door. What are the odds that your friends other child is a boy as well? (Yes the boy answering the door is her son, its not that sort of a trick problem).

[AngryYoungMan]

1. 50-50

Right? Unless I'm missing something.

[Az]

2. Wrong

This puzzle has been the bane of statisticians and mathematicians around the world.

[AngryYoungMan]

3. Can I discern whether the kid at the door is the older or younger sibling?

eom

[Az]

4. Uknowable from the evidence provided

But ultimately irrelevant to the problem.

[AngryYoungMan]

11. Okay...

So, among families with two kids, there is a greater (or lesser) likelihood that the kids are the same gender, as compared to the likelihood that they're of different genders. This is the only conclusion I can draw from my first answer (50-50 odds) being incorrect.

Since this flies in the face of what one would expect---which is 50-50 odds on ANY kid's gender---there has to be facts not in evidence, or, colloquially, "something you're not telling us."

In other words, the puzzle can't be solved by means of the facts provided. The answer has to be evidence of a larger statistical truth that is not contained in the puzzle.

Right?

[Az]

16. Correct the question is flawed

We do not know what is being asked. We don't know if its a individual odds problem or a population problem. If it was an isolated case then it would be 50-50. But since we do not live in isolated cases the best answer is derived by considering what the odds are on a communal scale.

[AngryYoungMan]

19. So do I "win"?

:)

[Ohio Dem]

23. Okay, I'll buy that.

Not a math question at all, really, is it?

[Rabrrrrrr]

21. I think that...

Initially, I was thinking that the chance of any child being male, but then realized the question is really asking, "Whats the chance of having two male children out of two children". It's not deceptively worded, though it looks like it, since it isn't like the "If I flip a head on a coin, what's the chance the next flip will be heads?" which is always 50%; it's asking, "what's the chance that if I flip 2 coins, they will both land head's up?"

[xray s]

5. 25%

[Az]

6. Must of taken a stats class

Now tell why that is the right answer.

[xray s]

14. I think it is the same as tossing heads twice

Each toss is 50/50. But the odds of tossing heads twice is only 1 in 4. The possibilities are Boy boy, boy girl, girl boy and girl girl.

[Against ME]

17. Wait, if it is 50 50 everytime,

then their is just as good a chance for you to flip heads again as their is for you to flip tails.
The previous flip is irrelevant, is it not?

[xray s]

24. For each individual flip, yes, but...

If a woman has two kids, and the odds of having a boy or girl is 50%, the only possible outcomes are boy/boy, girl/boy, boy/girl and girl/girl. hence, the probabilty of boy/boy is one in four.

[xray s]

26. ...unless Barbara Bush is the mom

then the kid is probably a chimp

:7

[Against ME]

28. But we already know one is a boy, right so

the next "flip" is 50-50, no? and the girl/girl is out the window

[Ohio Dem]

That wasn't the question, though.

If the question was, "What's the probability that a family with two children will have two boys?" then you would be correct.

But the question was "What's the probability that a family with two children will have two boys, given that the eldest (or youngest - it doesn't matter) is a boy?" The probability that each one is a boy is ½. If you know that one of them is male, that leaves 50-50 for the second.

Except, from what I've read elsewhere in this thread, it doesn't work that way in families. I guess children aren't coins, and there are other factors at work.

[Bake]

70. Not if these are independent events.

Your premise is wrong. It's not, what are the odds of having two sons, it's either (a) what are the odds of having a second son given that one of them is already a son, or (b) the two are totally independent so it's what are the odds of having a son, regardless of what the other child is.

Bake

[Az]

18. Bingo

From a selection of population the odds are 25% that a woman will have 2 boys.

[FatbackSlim]

20. Eh?

But because a boy opened the door, the possibility of two girls has been eliminated. So the choice is really from only three possibilities (b-b, b-g, g-b), which is why I believe the correct answer is 33%, not 25%.

[Rabrrrrrr]

25. ooh, probabilistically, that doesn't work

that way.

[FatbackSlim]

29. You're going to have to explain...

I'm trying, but I can't see how the odds don't change by eliminating one of the possibilities.

[Rabrrrrrr]

31. well, now that I've been thinking about it,

maybe you could be correct.

Alas, however, I must get to bed and be ready for work. :-)

But I can see your logic now, and you might have the right one, since this is neither "flip one coin, what's the chance of the next flip" or "What's the chance two flips coming up the same", it's more like "you have two flips, one flip you know is heads, what's the chance the other flip is also heads?"

I'm not sure if those last two are different or not, but they might be.

Been, what, 15 years since my prob and stats class...

[Ohio Dem]

32. Here ya go.

Suppose you flip a coin. It's either heads or tails. It doesn't matter what the result of the first flip (or the first thousand flips, for that matter) was. The next flip is independent from the other flip. Coins don't have memories. The probability of a head (or tail) is 50%. Period.

If you flip a coin and get a head, the probability that the next one is a head is 50%. Same for the next flip, and the rest of the flips. Individually, the probability of getting a head is 50%.

Now, if you ask what the probability of getting two heads in a row without knowing what the first flip yielded, then it would be 25%. But since you know the result of the first flip, you can forget about it.

[Against ME]

35. Thank god,

that's exactly how i saw it.

The ratio of head to tails is not necessarily 1 to 1, because each flip is independent, and the ratio of boys to girls isn't 1 to 1 because each pregnancy is independent of the last.

If I have 490 boys in a row, the chances of the next one being a boy is a 50 percent chance. Much like hot streaks in basketball, if their's a 50 chance that MJ will make the shot, then it is not weird to see him make 5 or 6 shots in a row, because the dice reset after every shot.

[Rabrrrrrr]

38. But we're talking collectives here

Not independant actions, but the SUMMATION of independant actions.

Though I believe the correct answer cold very well be 33%, it might be 25%, but certainly is not 50%.

[Ohio Dem]

41. No, we're not.

Once you know the gender of the first child, there is only one left to predict. And they're quite independent events. The second one is either a boy or a girl, at least on this planet. 50-50 in other words.

[Against ME]

45. If, we were trying to figure out what gender,

the children were, and in what order they came, then the answer would be 33%, but since we are just trying to deduce the gender, it is 25%

If we have a boy, then the next child has a equal chance at a boy or a girl.

But if we were adding into this equation order, then it would have the possiblities of b-g g-b b-b

[Ohio Dem]

40. Sports streaks make people forget that the dice "forget."

You can get on a roll in sports. Good days and bad days. Coins and dice aren't like that.

And the prob that a good basketball player will make a foul shot isn't 50%, anyway. They are better than that. Much like loaded dice or weighted coins.

[Az]

27. Correct

My bad on the previous answer. Its been awhile since I trotted this old puzzle out. It is 33%.

[FatbackSlim]

30. Woo-hoo!

My stats prof can sleep well tonight!

[FatbackSlim]

37. Explanation

You know from the question that mom has two kids. There are four possible combinations (boy-boy, boy-girl, girl-boy, girl-girl) considering both gender and order of birth. As has been pointed out, each birth is independent, has a 50% chance of either gender, and therefore each of the four possibilities has an equal 25% (50% * 50%) chance of occurring.

HOWEVER!

A boy opens the door. As a result, the girl-girl combination is impossible, meaning that there are three remaining possible combinations. Since they were all originally equally likely, they still are, but are now each 33% likely.

So...of the three combinations, only one (boy-boy) allows a boy to answer the door while allowing another boy to be wandering around. The other two cases (boy-girl, girl-boy) each leave a remaining girl after the door is open.

Q.E.D.

[Ohio Dem]

43. You may be right.

I think a lot of it depends on how you read the question. Since the family already has two children, and the one that answers the door is a boy, you have a point.

I was looking at it like this: A family has a boy. Now the woman is pregnant again. The odds of a boy (or girl) is 50%.

I see your point. I'm gonna have to think about this for awhile.

[WeaselRippedMyFlesh]

50. It is 50/50 and heres why

I never posted here before but i cant let this pass.

You were right before Ohio Dem it is 50/50

All prove it using some laws i learned in stats class (which i'm taking now and have a exam in next week)

The probability of a having a one boy with one child is 0.5

Now the probability of having a two boys is 0.25
But we know that one child is a boy
so were looking for the probability of having two boys given that one is a boy.

This would be the probability of having two boys divided by the probability of having a boy with just one child.
so 0.25/0.5= 0.5 or 50/50 (This is a not something a just made up is a theorem from stats)

In other words..
There are 4 combinations BB BG GB GG
But once you know the first one is a boy you can throw out the GG and the GB combinatiion leaving just 2.

[Against ME]

52. YOu are right,

but the order is irrelevant in this problem. The first one or the second one could be a boy, and the anwser would be the same. You assumed that first was boy, but it could very likely be that the second kid is a boy.

Since we do not care about order, G-b and b-g are the same thing. you can neither throw out g-b or g-b, but yo ucan add them together.

[Ohio Dem]

59. That makes sense, too.

I'm gonna have to look at this when it isn't so late.

Welcome to DU, by the way! It wasn't one of the "shit weasels" from Stephen King's "Dreamcatcher" that ate your flesh, was it?

[Ohio Dem]

65. Okay, here we go. This is the answer.

If a family has two children, there are four possibilities.

Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl

These have been listed in order of birth, older then younger.

The child that opens the door is a boy, so that means Girl/Girl is surely not the answer. That leaves three. But assuming the boy that answered the door is the first born, you can throw out the Girl/Boy option. Similarly, if the boy that answered the door is the second born, you can throw out the Boy/Girl possibility. With only two children, the boy that answered the door is either the older or the younger of the two. That means that one of the Boy/Girl-Girl/Boy options disappear. That leaves two options, Boy/Boy and the remaining mixed gender option.

50% it's a boy.

[WeaselRippedMyFlesh]

67. There you GO

Thats what i was trying to say but you explained it in english much
better than i could with math

[Ohio Dem]

34. Hey Az, aren't there really four possibilities?

bb, bg, gb, gg?

[lfairban]

44. Yes, and when the boy answers the door

it eliminates the gg possibility and either the bg or the gb depending on whether the boy that answers the door is older or younger of the two children.

I still think the correct answer is 50%.

Consider this:

Same situation, except when you go to the door, a child answers, but the child says nothing and you can't really tell if it is a boy or a girl. What are the odds that the other child is a boy? A: 50% Anybody disagree?

So why do the odds on the gender of the second child change all of a sudden when you recognize the gender of the first? It doesn't make sense. :shrug:

[Ohio Dem]

46. I'm sticking with 50%, too.

But I think it depends on how one interprets the question, and that depends on the wording.

[FatbackSlim]

47. Dependent vs Independent

The trick is that when you link the two births together (by counting up the total number of boys across multiple births) you've taken independent probabilities and turned them into dependent probabilities.

[Ohio Dem]

49. I tend to agree. It depends on how one asks (or reads) the question.

I read it the wrong way, I think. I just went and read it again, and I think you may be right. It's not a "what will the next child born be" question.

We know the family has two children. One's a boy. That means both children aren't female. That leaves three possibilities.

[FatbackSlim]

51. Bingo!

And on that note...g'night!

[WeaselRippedMyFlesh]

53. NOO

It only leaves two

There are 4 combinations gg bb gb bg

assuming the older answers the door and is a male
you throw out the GG AND the gb

if the younger was there you throw
out the GG and the bg

They are never dependent.

[Ohio Dem]

56. Who said it was the older child that answered the door.

We don't know that.

[lfairban]

61. They didn't.

But you still need to throw out one or the other, and either way, it ends up as 50%.

The questions still comes down to, if a woman has two children and one is a boy what are the odds the other is a boy. A: 50%

It is pretty much the same as asking if your first child is a boy, what are the odds the second one is also.

[Against ME]

55. But order is irrelevant.

b-g and g-b are one in the same.

[Ohio Dem]

57. Mathematically, they're not the same.

When you have two children, there are four ways it can go.

BB

BG

GB

GG

Those are the possibilities.

[Against ME]

62. yes, but if you their are two children, and age is irrelevant,

like it is in this, the BG and GB become one in the same, and in this case GG is out the window, so BG and GB = 1, and BB = 1, 1+1 =2

Two possibilities, 50 percent chance.

If you have one boy, then their are only two more possibilities, a boy or a girl.

But if you want to know the order in which the came, then G-B and B-G become two different answers.

This case is like 4+5, it's the same as 5+4
If order was relevant, then it would be -, 4-5 and 5-4 are diffent.

[WeaselRippedMyFlesh]

64. But

The probability of a girl and a boy is NOT
equal to the probability of a girl and a boy GIVEN that one is a boy

[Against ME]

68. If I have a two bags of marbles,

and each one has a blue and green marble in it, And I reach in the first bag and pull out a blue Marble(boy), then if I reach in the second bag, their are only two possible things i can pull out, a blue(boy) or a green(girl).

But if I have two bags, bag 1 and 2, each having two marbles, and I don't know which bag is wich, then I could pull out a blue or a green marble out of the second bag, not know wich bag i was pulling from.

that would leave me with 3 possibilities,

blue bag 1/green bag two
blue bag 1/green bag two
green bag 1/blue bag two

[WeaselRippedMyFlesh]

60. Then look at it this way

Say there only 3 choices GG BB and GB

P(GG)=.25 P(BB)=.25 and P(GB)=.5

you know that one of them is a boy

the probablility of GB given that one of them is a boy
would be P(GB | B) which is equal to P(GB Union B)/P(B)

which is 0.25*0.5/0.25
=.125/.25 = .5
Therefore the probability of GB given one boy is 0.5
And because of this the probability of BB must also be 0.5
because the sum must be 1 and GG given one B is 0.

I known this might look at little confusing do those that arent taking stats right now but trust me im not making this up.

[Against ME]

63. Thanks,

was this meant for another post, I agreed that it was 50%, I was one of the first to argue it.

[are_we_united_yet]

15. because

there is a 50% chance of having a boy.

another child would be 50% of 50%

[Ohio Dem]

22. It seems to me that the two children are independent events.

The gender of the first child has no influence on the gender of the second child.

If you flip a coin and it's heads, the probability of the second coin being a head is ½.

The probability of flipping two heads in a row is ¼, but that goes away once the first coin is flipped. The events are independent.

[MisterP]

42. the operative is AS WELL

I get it!

[supernova]

54. Anybody who

paid attention in High School biology when genetics came up can get this. There's only 4 possible combinations for child pairings.

[Ohio Dem]

7. 51-49 it's a female.

The ratio of males to females isn't 50-50.

[Az]

12. We can allow for an even spread of male female in the population

Even with the slight difference its not 49%.

[FatbackSlim]

8. Two thirds

67% likely the other is a girl.

On edit: confused boy and girl. One third likely it's a boy.

[BigMcLargehuge]

9. three to one

.

[Against ME]

10. 100% chance

eom

[Rabrrrrrr]

13. 1:4 based on genetics

and being two kids, yes?

Though if the dad only has male making sperm, it's 100% chance. :-)

[lynndew2]

33. 100% boy

you said so in your qustion...lol other child is boy as well.

[Rabrrrrrr]

36. Boy, you're pulling a semantic technicality on that one

But yes, it does imply the second child IS a boy, doesn't it?

[lynndew2]

39. just a reminder to all how easily words can be

taken out of context. sorry but that was fun :P

[Ohio Dem]

48. I hate you for this puzzle.

Just kidding, of course. It's good to think on the weekends. I guess.

[Sperk]

58. Here are all the possible outcomes of having child 1 and 2

Boy/Girl
Boy/Boy
Girl/Girl
Girl/Boy

four possible outcomes, only one meeting the criteria of the story...there is a one in four chance of the next child being a boy.

That's my story and I'm sticking to it.
:P

[Against ME]

66. Girl/Girl is out the window,

so that takes it down to 1 in 3, and since we are not looking for order, G/B and B/G are the same thing. One is a boy, the other is a girl, in both of them, and their is no given order, it's just a different way of saying the same thing in this case. So that takes it down to a 1 in 2 chance. 50/50

[TreasonousBastard]

69. This one seems too simple...

even after reading all of the other replies so far.

The probabilities originally are:

BB= 1/3
BG= 1/3
GG= 1/3

Note that the ages are irrelevent in this case, so we don't need the extra BG or GB, which would change the odds to 1/4.

So, after we see that there is one boy, the odds are now:

BB- 1/2
BG= 1/2

It works out the same whether dependant or independant in this case.

It would be a little trickier question if there were 6 girls, 2 boys, and a transgender in the house.

[Az]

71. Bzzzt

The initial odds are

BB = 1/4
BG = 1/4
GB = 1/4
GG = 1/4


Since we know there is a boy this removes GG leaving

BB = 1/3
BG = 1/3
GB = 1/3