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A stats Q. / Brain teaser

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So I heard this on the radio, as well as a potential answer, but I seem to be able to come up with two very different answers which are causing a bit of argument with my mum the maths teacher... 

So you meet up with a couple of old friends, you know they've had two children and they've brought one with them. The child they have brought with them is a boy. What are the chances that the other child is also a boy ? 

 

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Interesting question. You could say 50% as they are independent events, but then if you combine the probabilities you get 25% that they're both boys right?

e.g. there are four combinations:

Boy + Boy
Boy + Girl
Girl + Boy
Girl + Girl

So there is a 1 in 4 chance that both are boys.

EDIT: But we know that one of them is a boy, so maybe it's 1 in 2?

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Knowing all that prior to meeting them would stand you at 50%?

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Ok, so one of the answers I have is 1/2 the other is 1/3 - from the four possible families of two children one is ruled out by knowing they have at least one boy - so in the other three choices only one is two boys so 1/3 ? This is the one my mum is going with, but I'm fairly sure the answer is 1/2. 

What do people think of that?

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Boy + Girl and Girl + Boy is the same thing. So only three possibilities, right? And one definitely is a boy. The boy is either going to have a sister or a brother. 50/50 surely?

Edited by Mikee

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couldn't the answer for 1/3 go something like this: 

1 boy is present, so girl+girl is off the table. 
that leaves you with 3 options: girl+boy, boy+girl, boy+boy
each one has a probability of 1/3, so you'd end up with a 2/3 chance for mixed genders and 1/3 for equal. 

Edited by jeff costello

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This is a great brain teaser! Hmm. I don't think it's as simple as 50:50, because you know there are only two data sets and you know they aren't exclusive of each other due to your prior knowledge.

So........... Thinking through this out loud... there are 2 possible sets of data:

Set a - 1 boy (item 1) + 1 boy (item 2)
Set b - 1 girl (item 3) + 1 boy (item 4)

They have brought one boy with them, so either they have brought item 1, item 2 or item 4.

If they brought item 1, that leaves item 2 at home.

If they brought item 2, that leaves item 1 at home.

If they brought item 4, that leaves item 3 at home.

So that leaves 3 scenarios for the remaining item - boy, boy, girl. So I'd say the possibility of the remaining kid being a boy is 2 out of 3. Wow, my gut feel was 1/3 but my (probably flawed :P ) maths said otherwise!

 

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If they had 100 kids and they brought 99 of them and they were all boys what would be the chances the unseen one is a boy? I'd say 50/50.

I don't get how the answer isn't 50/50.

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15 hours ago, jeff costello said:

1 boy is present, so girl+girl is off the table. 
that leaves you with 3 options: girl+boy, boy+girl, boy+boy
[each combination is equally likely so, ] each one has a probability of 1/3, so you'd end up with a 2/3 chance for mixed genders and 1/3 for equal. 

ha! totally by chance i just listened to the same radio show. ("infinite monkey cage" bbc3) 
and this is actually the correct explanation ( with a little addition in brackets)

the 50/50 explanation does not include that piece of knowledge you already have ( "one is a boy" )

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But girl+boy is the same as boy+girl, particularly considering you already know that the boy exists and that only leaves you with that combination once and the boy+boy once so the probability is 1/2 of each event... 

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15 minutes ago, monkeyseemonkeydo said:

But girl+boy is the same as boy+girl, particularly considering you already know that the boy exists and that only leaves you with that combination once and the boy+boy once so the probability is 1/2 of each event... 

They may be the same in that there is one girl and one boy, but they represent two different scenarios as far as calculating the probability is concerned.

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10 minutes ago, monkeyseemonkeydo said:

But girl+boy is the same as boy+girl, particularly considering you already know that the boy exists and that only leaves you with that combination once and the boy+boy once so the probability is 1/2 of each event... 

it is not the same in this context, isn't it? you only know one is a boy, not which one it is.
with boy/boy the one you know about could be either one, in boy/girl you know about the first one, in girl/boy you know about the second one. 

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1 hour ago, Tony Harrison said:

They may be the same in that there is one girl and one boy, but they represent two different scenarios as far as calculating the probability is concerned.

But not in a logical sense surely. In the knowledge that there is definitely one boy the 'two' outcomes represent the same answer so should be treated as a single scenario.

1 hour ago, jeff costello said:

it is not the same in this context, isn't it? you only know one is a boy, not which one it is.
with boy/boy the one you know about could be either one, in boy/girl you know about the first one, in girl/boy you know about the second one. 

You know one is a boy, it makes no difference which one it is. The two possibilities of the second child is it's either a boy or it's a girl. Again, the boy/girl vs. girl boy is irrelevant since you know the boy exists therefore the order is already determined/irrelevant.

Edit: And the plane takes off, damnit.

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That's my problem with the whole thing- if you use the logic that gets you to 1/3 it's because you say the boy you know could be the first or second child (in terms of age) and gives you the three possible family outcomes, but as soon as you say it IS either the first OR the second you remove either boy/girl or girl/boy from the equation and it becomes 1/2 again. 

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I think that the human mind likes to try to simplify things, and sees girl + boy the same as boy + girl. If you imagine that having the two children are separate events (which they are) then that should show how the two are distinct. Whilst it may seem like the same outcome ("the couple have a boy and a girl") in mathematical and logic terms they're still different scenarios.

Let's look at it this way - here are the combinations of those two events:

Parents have a boy, then subsequently have a girl
Parents have a boy, then another boy
Parents have a girl, then another girl
Parents have a girl, then have a boy

This means there are four different scenarios, with a 25% each of occurring. Since to us two of those scenarios are practically the same, we can sum those probabilities, e.g. there's a 50% chance that the parents end up with a girl and boy born in unspecified order. This doesn't answer the original question - I'm just trying to show how those two seemingly identical scenarios shouldn't be seen as the same thing.

The plane definitely takes off.

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Yeah I get that, so that's why I see 1/3 as a genuine solution, but my problem is once you actually test it out by saying, ok it's the first child they had that you meet- in that scenario you can rule out girl boy and girl girl to leave boy boy or boy girl so 50%. If you then test out and say it's the second born - you get exactly the same thing - so it doesn't matter which child - first or second and the probability is 50% whereas if you just say you don't know it's 1/3. This is my problem, I can't see a legitimate flaw with either method (I'm hoping it's my old dulling mind), but I definitely want one to be wrong.

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i'm not sure, but i think what you calculate here is only half the scenario: if you know the sex of the first child, what's the chances of the sex of the second child - which would be 50/50 (and a boring riddle)

maybe it helps if you re-phrase the original problem like this: 
the parents have 50/50 chances of having either same-sex children (boyboy, girlgirl) or mixed-sex children (boygirl, girlboy). knowing that one child is male takes only one option off the table (the girlgirl one), which takes away half the chance of them having same-sex children. 

Edited by jeff costello

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Please. The plane doesn't take off. Friction in the wheel bearings.

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