Wonderfully logical illogical probability problem

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I had to work this one out this morning, i first thought wtf!? Then i worked it out, see what you think:

You meet your new tutor in town accompanied by a young girl. He says to you "I have two children, this is my daughter Lisa". What is the probability that his other child is a boy. Would it make a difference if he had said "I have two children, this is my youngest child, Lisa"? (You may assume a 50% chance of any one birth being a boy or a girl.


Have fun!

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There are four configurations of children:
Youngest first the possibilities are GG, GB, BG, BB (G - Girl, B - Boy) and all are equally likely.

First situation.

BB of them is ruled out by knowing that one is a girl.
Thus, we only have GG, GB and BG left.
Knowing that the child with him is a girl gives a 1/3 chance that he has another girl and 2/3 chance that the other child is a boy.

Second situation.
We now know that the youngest child is a girl, this leaves only GG and GB.
Thus the probability is 1/2 that the other child is a boy.

Hope that's right!

Spot on!!!! Crazy how looking at it you would think 50-50 both times, stupid question! But actually when you do the maths its 2/3 first time and only 1/2 the second!!!

 

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This chick had balls!?!?!

 

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If you don't know the gender of either of the children, the tree looks like this:

if you know that one of the children was a girl, (let's say that the girl was the oldest child (it doesn't matter which way around it is for the maths)) the tree diagram looks like this:

As you can see, there's a 50% chance of girl+ boy, and there's a 50% chance of girl+ girl.

Thanks.

But in the first case you still canoot have BB therefore your diagram is flawed, trust markyp23 he has the correct answer as counter intuitive as it seems!

 

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I should probably point out I've got an SYS in Statistics and Probability .

The two questions the OP has asked are exactly the same from the point of view of probability, and are not the same as the question that yields 2/3.

I'd like to point out that doesn't mean you're views are correct. The two cases ARE different. Clarification:

When having two children the following combinations are possible (note order is important):

BB
BG
GB
GG

Where B=Boy and G=Girl.


Now in the first case the information you are given is that he has two children and one is a girl (ignoring any linguistical nuances where he might phrase it this way or that way because of whatever). So this rules out the BB option, leaving only BG, GB and GG. Out of these 3 remaining possibilities 2 contain a boy, therefore the probability his other child is a boy is 2/3.

In the second case the information we have is not only that one child is a girl, but that his second child (the youngest) is a girl. Now the only remaining possibilities are BG and GG. Here out f these two options only 1 contains a boy, therefore the probability his other child is a boy in this case (given this information) is 1/2.


Hope that is clear enough for everyone?

 

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Seriously, this is WRONG! for the questions posted by the OP.

The 2 cases from the OP are both differently worded versions of case 1 from this page here:-

No they're not. I'm the OP, the first case i stated is analogous to case two on your link (in that it is only the gender of one child is stated, not whether that is the older or younger child). Where as the second case in the OP both the gender and whether that child is the older or the younger is also stated, analogous to case 1 on that link. I don't see how you think they're both case 1 from your link?!

 

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Ah sorry didn't realise you were the OP, no they are both case 1, because you have stated "this is my daughter" in both cases, the age of the child is irrelevant to the probability. By stating "this is my daughter" removes the opposite case of being possible and it's irrelevant what one was born first. If you KNOW (ie probability = 1, not 0.5) which child is female there are only two options possible, BG, or GG. the alternative boy girl combination of GB is NOT possible. Younger or older makes no difference as all it does is change GB to BG and BG to GB, it's restricted the sample space by removing the third option that gives the 2/3.

The correct question you need to ask is "I have two children, if at least one is female (this allows for both combinations of BG and GB) what is the probability that the other is male"

The probabilities are linked to what options (from the four:BB BG GB GG (with order mattering, ie BG and GB are different)) are left once the information given in the statement is taken into account.
For this case let the first child be the youngest ie in the option GB the girl is younger than the boy.

Copied from the OP, the first statement is "I have two children, this is my daughter Lisa".

Here we know at least one is a girl so the BB option is out. I think you're happy here that the prob therefore or the other being a boy is 2/3.

The second statement is "I have two children, this is my youngest child, Lisa". This obviously rules out the options BB and BG (as we know the first child is a girl) This is where this second statement differs from the first, you are given more information. You are told that not only is at least one of his two children a girl, but also that it is the youngest child which is a girl. So with optios BG abd BB rules out the only remaining are GB and GG which gives the probability of 1/2.


You're link has some interesting stuff about what (in a real life situation) the person (tutor in this case) would actually say depending on how many daughters/sons they had. In this example this has to be ignored, i copied it verbatim from the sheet of problems i was set, personally i would phrase it slightly clearer.

Hopefully you can see that being told whether she is the younger or older does matter (but not which one, merely that you are told one, seems bizarre to me even now). Heh, what a great question 'eh!!

 

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Quoting for proof for the naysayers.

I'm presuming those boy-girl combinations were generated randomly? Good effort though and it shows exactly what need to be shown. That is when the position of the child is fixed (child1 orchild2 or youngest or oldest) as well as the gender, the probability drop from 2/3 to 1/2 (or in that model 1/3 to 1/2).

Nice one Zogger

 

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From http://mathforum.org/library/drmath/view/52186.html

Case 1 in OP, except with girl swapped with boy.

The key part is that you're specifically picking out a child (a girl in the OP). But in the other case you're not. As you see it makes no difference whether the kid is younger or older.

Edit: The 2/3 case totally relies on you not specifically identifying one child as being female as it removes the possibility of that child being a boy!

This is actually a different situation and a different question is answered, this siblings problem is not the same as the two daughter problem i posted.

To remove the ambiguity from the OP:

Situation one:
You know - Man has two children
You know - One is a girl


Situation two:
You know - Man has two children
You know - One is a girl
You know - It is the youngest which is the girl


Now situation one answered:
4 combos {BB, BG, GB, GG} (let the first child in the combo be the eldest).
We know one is a girl therefore BB is out, we are left with {BG, GB, GG}.
Out of these three possible situations two contain boys, giving rise to the 2/3 answer.

Now situation two answered:
We know not only that one child is a girl (ruling BB out) but also that it is the youngest which is a girl which therefore also rules out GB.
We are therefore left with {BG, GG}.
Now you can see that the prob of the other child being a boy is 1/2.


This has to be clear enough, i can't see there's any flaw in that and this removes all linguistical ambiguity from the OP (which i admit was confusing but that was only as i copied it verbatim).

 

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But surely if a woman has 5 boys in a row the odds on the 6th being a boy are still 50% ?

Zogger you are correct.

The probability that the 6th child will be a boy is 50%. As this is an independent event.

With your wife about to give birth the prob that its a boy is still 50%. However given that you (one) have (has) two children at least one of which is a girl, the prob that the other is a boy is 2/3.

This is due to the combinations of B and G possible and the fact that BB is obv not poss as you (one) already have (has) a girl.

Out of {BB, BG, GB, GG} BB is not poss which leaves {BG, GB, GG} where two out of those three contain boys.

Tis just the strange counter intuitive workings of probability.

 

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True, but if you know a woman has 6 children and that at least 5 are female the chances of the 6th being male are not 50%.

This is true, as long as order does not matter, ie you don't know which of the 6 is the male (not in some weird hermaphroditic way) but in the ordering sense, So you can't say the boy is the oldest, otherwise the probability of him being a boy will again drop to 1/2. Strange i know!

 

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You meet your new tutor in town accompanied by a young girl. He says to you "I have two children, this is my daughter Lisa". What is the probability that his other child is a boy. Would it make a difference if he had said "I have two children, this is my youngest child, Lisa"? (You may assume a 50% chance of any one birth being a boy or a girl.


The fact that you meet your tutor determines the family, you then find out that one of that family's children is a girl.

In the OP these are two distinct cases, one where we are told whether she is the youngest or the oldest (which it actually is is not important) where in this case the prob is 2/3. Then the other case where we are not told whether she is the elder or the younger, where the prob is then 1/2.

 

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Well, yeah I was just looking at the first part as that's the one that seems to be causing the issues!

Ah ok

Conclusion: Statistics sucks

Seconded

Actually if you say the boy is oldest the probability rises to 100%

Bah!

 

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But hopefully people can see that is all about ordering. If you have 6 children there are many possible combinations eg:

GGGGGG

GGGGGB
GGGGBG
GGGBGG
GGBGGG
GBGGGG
BGGGGG

GGGGBB
GGGBBG
etc etc

Looking at those top two, if you say ok the first 5 children are girls (as in the top two) the prob of the 6th being a boy is 1/2 (as a boy or a girl birth are equally likely).

However if you only say at least 5 are girls, what is the prob the other one is a boy, the prob is 6/7. This is because there are more ways you can have 5 girls and a boy than there are to have 6 girls.

Is this making sense?

 

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I don't think it is clear enough mate no. Your query was on the probability of the sex of 2nd child being a boy or girl in both situations. The confusion is arising as folk are thinking that because it is possible in situation 1 that the 2nd child could be older or younger has any bearing on its sex.

I know in probability you need to break down into possible solutions like BG GB etc...but when the question is concerned purely with the sex of the 2nd child, whether it is older or younger can't have any bearing - you can say you have more possible options BG and GB and GG, which you do, but I think it is a mistake to say (in this instance) that because you have one more option (BG) this is relevant to probability of the sex.

Ok i see what you're saying, hopefully my post above with the 6 children shows the situation a little better, i'll do the same for just two.

The 4 possible combinations or boy and girl are:

BB
BG
GB
GG

Now, if we are told that at least one is a girl and then we are asked what is the probability the other is a boy, it is 2/3. Because of the three options containing a girl two also contain a boy. I think this part is clear.

However if we are told that the first child is a girl and then asked what is the probability the other is a boy, it is only 1/2. This is because there are only two options in which the first child is a girl, and then in only one of these is the other child a boy.

Again it is all to do with ordering, combinations and permutations.

 

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Yep answer to that question is 2/3.

Are you happy with the answers to the two different statements (poorly worded) in the OP now?

That if we are merely asked "given a girl, what is prob the other a boy" it is 2/3.

But that is we were asked "given the youngest is a girl, what is prob the other is boy" it is 1/2.

 

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I think I'm coming round to that way of thinking now

Still not 100% sure

Hehe, i still don't accept the answers willingly myself, avoid stats if at all possible!!

 

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Here's where I get confused. I'm not sure that's correct. It's still a two-sibling set. Still the only information we are given is that at least one of the two is a girl.[/i]

Not true, you are told two things:
1. At least one is a girl
2. (perhaps more importantly) it is the youngest which is the girl (could have been the oldest though, which it actually is makes no difference, only that one of the two is defined.)

I can't see how that the question is any different to:

"Of all two-sibling sets, what is the probability that a given two-sibling set which contains at least one girl, also contains a boy?"[/i]

It's different in that this situation (where the girl is the youngest) is the same as saying:
"Of all two-sibling sets, what is the probability that a given two-sibling set in which the first child is a girl, also contains a boy?"

I believe the answer is only 0.5 in the following case:

"Of all two-sibling sets, what is the probability that a person in a given two-sibling set with at least one girl, is a boy?"

Heh in this case the prob is not 0.5 its 0.67 (2/3) as the set could be GG, GB or BG; giving rise to the two out of three probability.

hope this helps

div0's post clarifies things nicely i believe.

 

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Thanks, this phrase got my head around it. Definition.

I enjoy thinking about this kind of stuff, gotta be good for the brain.

Heh no worries, is good for me as well because to be able to explain it one has to really understand it (excuse the posh turn of phrase using 'one' but it removes confusion!). Also people come up with variations on the problem which a good to look at. I have to say i thought it may cause a stir but not this much cafuffle!!

 

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Dont get how it can be anything other than 50/50 given there are only 2 possible outcomes from the question "what is the probablity that the child is a boy or a girl?

The child is a boy
The child is a girl.




Also dont get how "If person A has 9 children all girls what is the probablity of the next child being a boy or a girl".

isnt it still 50/50? Regardless of there being 9 or 99 or 999 girls beforehand. Because in each individual separate child they can only be one or the other.

Thought Jokester was right.

I suck at statistics though

Taken as an individual case "what is the prob this next child is a boy" the answer is of course 50/50 as whether you have a boy of a girl is not dependent on how many other boys or girls you've had.

However if you look at the situation as a whole (without knowing in which order the children were born) the situation drastically changes. With 9 girls and 1 boy (without the order being known) they could have been born as follows:

GGGGGGGGGB
GGGGGGGGBG
GGGGGGGBGG
GGGGGGBGGG
GGGGGBGGGG
GGGGBGGGGG
GGGBGGGGGG
GGBGGGGGGG
GBGGGGGGGG
BGGGGGGGGG

There there are nine possible ways of having 9Gs and 1B. However there is obv only 1 way of have 10Gs

Therefore if you know that in a group of 10 children that at least 9 are Gs the prob the other child is a B is not 1/2 (50/50) its actually 9/10