Wonderfully logical illogical probability problem

  • Thread starter Thread starter OSB
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OSB said:
Yeah of course that's true, i know that if A and B are independent events P(A|B)=P(A). And IF event A was birth of a girl and B the birth of a boy, but that's not what they are. A is of a two child set there is a girl and B of a two child set there is a boy.

Now due the 4 different ways of having girls and boys in a two child set A and B are not independent. The 4 ways are BB, BG, GB and GG.

Therefore as you can see if event A is present (there is a girl) then P(B) is different than were we to just look at P(B)

Doing the maths P(B|A)=2/3
where-as P(B)=3/4

You haven't defined events A and B clearly enough to reflect the situation.
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I have defined A and B clearly enough, A is the birth of one child, B is the birth of an other child, doesn't matter the order. The fact is A has a 50/50 chance of being a boy, so does b, knowing for certain that A is actually a girl has zero impact on the probability of child B, regardless of A being the younger or older child.

Edit:
probability.jpg
 
Ok i'm going to do the problem in a mathematical notation to get rid of all the problems of linguistics and definitions (once they have been dealt with).

I shall define my events as follows:

Event A - the first (therefore the eldest) child (in a set of two) is a girl; P(A)=1/2
Event B - the first child is a boy; P(B)=1/2
Event C - the second child is a girl; P(C)=1/2
Event D - the second child is a boy; P(D)=1/2

Obviously events A and B and events C and D are mutually exclusive.

A tree diagram to help visualise the problem is as follows:

treeyg3.jpg


I will use the following notation:

AںB – meaning the union of A and B, that is A or B

A∩B – meaning the intersection of A and B, that is A and B


From the OP, situation one is "I have two children, this is my daughter Lisa" and we want the probability that his other child is a boy. This equates to P((BںD)|((AںC)), solving:

P((BںD)|((AںC)) = P((BںD)∩(AںC)) / P(AںC)

= [P(B∩A)+P(B∩C)+ P(D∩A)+P(D∩C)] / P(AںC)

= [0+¼+¼+0] / ¾

= ½ / ¾

= 2/3

The second situation is "I have two children, this is my youngest child, Lisa". Here the problem is P((BںD)|A), solving:

P((BںD)|A) = P(B∩A)+P(D∩A) / P(A)

= 0+¼ / ½

= 1/2


Hopefully that should be clear enough (though maybe a bit complicated) and should be rigorous enough.
 
OSB, even if the proof is completely rigorous there will always be some that doubt you!

I remember in the 0.999...=1 thread I posted up a watertight proof using the continuum property of the real line and still people wouldn't have it.
 
damaster275 said:
Are those statements not the same?
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In the second case BG and GB are both possibilities, but by specifically identifying one child as female in case 1 removes the possibilty that the other child is female and that child is actually the boy (hence the paradox). In this case the other child can be either male or female and is independent of the other child, giving 50/50.

I think the confusion most people seem to have is because wiki refers to order of birth, but this could be anything, even the order the children are introduced to you (ie "this is my daughter"), or the order that the coins are flipped etc.
 
Haircut said:
OSB, even if the proof is completely rigorous there will always be some that doubt you!

I remember in the 0.999...=1 thread I posted up a watertight proof using the continuum property of the real line and still people wouldn't have it.
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True-dat!!

1/3 = 0.333...

3 * 1/3 = 1,

job done :p:D
 
"I have two children, this is my daughter Lisa"
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"I have two children, at least one is female"
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These two statements are identical and provide exactly the same information and the same answer - the chance of the other child being a boy is 2/3.

Identifying the child makes no difference - you neither know whether it is the older child or younger child - all you know is that one of his children is a girl. The fact that she's standing in front of you provides you with no further information than the second statement.

Jokester said:
In the second case BG and GB are both possibilities, but by specifically identifying one child as female in case 1 removes the possibilty that the other child is female and that child is actually the boy (hence the paradox). In this case the other child can be either male or female and is independent of the other child, giving 50/50.
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No it doesn't - it still only removes the possibility of a BB combination. Leaving the other possibilities as GG, GB, BG each with equal probability of occuring.

"I have two children, this is my youngest child, Lisa"?
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This is the information that removes the BG option - as you now know that his youngest child is a girl.
 
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timdog said:
BUT THE FIRST QUESTION DOESNT ASK ABOUT AGE
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And?

Age is irrelevant. Replace older/younger for childA and childB, it still follows that all 4 combinations must be considered. The only combination that can be ruled out is ChildA+ChildB cannot both be Boys.
 
div0 said:
Identifying the child makes no difference - you neither know whether it is the older child or younger child - all you know is that one of his children is a girl. The fact that she's standing in front of you provides you with no further information than the second statement.
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But it does, compare these two here:-

A random two-child family with at least one girl is chosen. What is the probability that it has a boy? - An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two boys, what is the probability that two random children are of different gender?"
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And gives a tree like this:-

probability2.jpg

Once you remove the BB case, gives you the 2/3 chance that one sibling is a boy.

is a totally different question to:-

"I have two children, this is my daughter. What is the probability that his other child is a boy?"
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In which case the tree is (adding in age relationship to show it makes no odds):-

probability.jpg

By specifying the daughter is older/younger merely removes one side of the tree and ultimately still landing you with 50/50 odds.

OSB's proof only applies to the first case and I don't dispute that, but it certainly doesn't apply to the second as he has ultimately reduced the problem to a single event (since age doesn't matter), either that the other kid is a boy or it is a girl.

If you look at the reworded first question, it is clearly referring to two different events. the chance one child is a boy/girl and the chance that the second child is a boy/girl, but without the chance that both are boys.
 
Ok think of it this way....

The first question in the OP is this...

'I have 2 children' (childA and childB)

what is that chance that childA OR childB is a boy?

well we know that childA has a 50% chance of being a boy or a girl
and likewise that childB has a 50% chance of being a boy or a girl

There is now a 25% chance of childA being a boy AND childB being a boy
And also a 25% chance of childA being a boy AND childB being a girl
And also a 25% chance of childA being a girl AND childB being a boy
And also a 25% chance of childA being a girl AND childB being a girl

The man now says - 'this is my daughter'

The ONLY option you can rule out is the first one Boy+Boy. There out of 3 remaining EQUALLY likely outcomes, 2 of them would result in him having a boy and only 1 would result in the other child being another girl.

In the second part of the OP, the man effectively says 'this is my daughter and she is childA'

This removes the first 2 possibilities, leaving 2 EQUALLY likely outcomes. Of those two outcomes, 1 results in the other child being a boy, and the other outcome is a girl - hence 1/2 chance of a boy.
 
Jokester said:
But it does, compare these two here:-
And gives a tree like this:-

http://premium1.uploadit.org/jokester//probability2.jpg

Once you remove the BB case, gives you the 2/3 chance that one sibling is a boy.

is a totally different question to:-


In which case the tree is (adding in age relationship to show it makes no odds):-

probability.jpg

By specifying the daughter is older/younger merely removes one side of the tree and ultimately still landing you with 50/50 odds.
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Ok I thought about this last night and I'm fairly sure I can explain it and leave little or no doubt :)

The tree above only works in the event that the father introduces you to his daughter without HIM knowing the sex of his other child. If he does NOT know the sex of the other child, then that tree works perfectly and the chance of his other child being a boy or a girl is 1/2.

It may not seem intuitive, but because he KNOWS about both children BEFORE he tells you about (introduces you to) ONE of them, then it is this PRIOR knowledge of HIS that affects the probability.

For example if in true OcUK style a guy goes out and pulls a girl, sleeps with her and to show that he is truley Alpha he gets out of bed and pulls her flat mate and sleeps with her too.

Both girls later have a child, but don't tell the father.

Later in life one of the children (a girl) tracks her dad down and says hi.

The probability of the other child (could be older / could be younger) being a boy or a girl is still 50/50.

However, if he had met BOTH children and THEN introduces you to a (this is crucial) daughter. The probability of his other child being a boy is now 2/3.

Hard to believe I know, but I'll try to be as clear as I can....

The birth of any child is like the toss of a coin and can equally result in a 50/50 outcome (boy/girl, or heads/tails).

But the OP is not about simply saying that you know the outcome of ONE birth and so what is the probability of the OTHER birth being a boy! It is about the father KNOWING the outcome of BOTH births and then telling you some information about ONE of them! It is the fact that he KNOWS BOTH outcomes BEFORE telling you something about it that changes the probability - because the question is not just simply about the 'next'/'previous' birth, but actually about him KNOWING BOTH children and revealing information about only ONE.

It is subtly different and I'll try to give you an example, which I hope will remove any doubt!

If I throw TWO coins and hide them in a cup (without seeing them). Just by looking at ONE of the coins I can make no prediction about the OTHER. The chance that the other coin is a head or a tail, does not change whether I show a head or a tail for the uncovered coin.

However if I get to see BOTH coins and then tell you that one of them is a TAIL, then the chances that the other coin is a HEAD are now 2/3. This is simply based on the fact that I have seen BOTH coins BEFORE telling you about them.

I'm sure you agree that if I throw 2 coins (can be simultaneously, or can be one after the other, it doesn't matter), then on average I will get:

HH
TH
HT
TT

Each coin results in a 50/50 split between heads and tails (2 heads and 2 tails each).

If I DON'T see either coin and then uncover only 'Coin1', then whether it is a head or a tail, has no effect on the chances of the other being a head or a tail.

For example, if I uncover coin1 in the example above, then on 1 occasion coin2 is a head and in the other coin2 is a tail (so a 50/50 outcome for the other coin)

HOWEVER, if I uncover BOTH coins and look at them and then tell YOU that ONE of the coins is a tail. Now you can see that there are 3 ways (of equal probability) in which I would be able to tell you I have 1 tail. Of those 3 ways, 2 of them result in the other coin being a head. So the chances are 2/3.

It all revolves around KNOWING BOTH outcomes and then RETROSPECTIVELY giving information about ONE of those outcomes.
 
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Jokester said:
OSB's proof only applies to the first case and I don't dispute that, but it certainly doesn't apply to the second as he has ultimately reduced the problem to a single event (since age doesn't matter), either that the other kid is a boy or it is a girl.
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It applies to both cases, the second case is in essence a single event as div0 explains in the post above. By fixing the age of one child we are no longer looking at possible 2 child combinations but instead just the outcome of one birth. The proof is rigorous and applies to both cases, how can it not.
 
OSB said:
It applies to both cases, the second case is in essence a single event as div0 explains in the post above. By fixing the age of one child we are no longer looking at possible 2 child combinations but instead just the outcome of one birth. The proof is rigorous and applies to both cases, how can it not.
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I'm not sure that statement quite goes hand in hand with the point I'm making.

It is the fact that the father KNOWS the outcome of BOTH EVENTS (births) BEFORE he provides any information about ONE of these EVENTS.

He is telling you that ONE of TWO EVENTS (births) resulted in a daughter. He is NOT telling you the result of EITHER INDIVIDUAL EVENT.

It is subtly different, but his KNOWLEDGE of BOTH outcomes has an affect on the probabilities.

If he only knows about the sex of one child, or if he specifically tells you the sex of one child, then the remaining EVENT (other birth) is STATISTICALLY INDEPENDANT and is a 50/50 chance. But by KNOWING BOTH BEFORE he tells you the sex of ONE of them, then it affects the probability of the sex of the 'other' child (as described in my previous post).
 
if i keep rolling a 6 sided dice the greater the chances are that the number 6 will come up.

even though each time i roll the dice its 1/6 that the number 6 will come up.

The effect is cumulative.
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If you roll a dice ten times the odds on getting ten sixes are the same as getting 1,2,3,4,5,6,1,2,3,4 or any other sequence, isnt it?
 
BullBoyShoes said:
If you roll a dice ten times the odds on getting ten sixes are the same as getting 1,2,3,4,5,6,1,2,3,4 or any other sequence, isnt it?
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Correct, but that's to do with EVENTS happening in a particular ORDER/SEQUENCE.

The OP is not about ORDER or SEQUENCE (oldest/youngest is irrelevant) - it is simply about the fact that the man knows about BOTH his children before he reveals any information.
 
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