Wonderfully logical illogical probability problem

[mks2005]

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.

Superb explanation. This paragraph (in particular the bit I underlined) really highlihgts the key concept behind this way of thinking. Thankyou, div0.

 

[OSB]

Superb explanation. This paragraph (in particular the bit I underlined) really highlihgts the key concept behind this way of thinking. Thankyou, div0.

Yeah seconded, div0's hit the mark (my explanatory skills are somewhat less über!!).

 

[div0]

Great explanation, and this is exactly why this problem is like the Monty Hall problem.

Superb explanation. This paragraph (in particular the bit I underlined) really highlihgts the key concept behind this way of thinking. Thankyou, div0.

Yeah seconded, div0's hit the mark (my explanatory skills are somewhat less über!!).

Thanks

Probability often isn't intuitive and so it's easy to get confused in cases like this and think that the 'obvious' answer must be correct. I did this sort of thing as part of one of my courses at university and REALLY struggled to get to grips with it at first. It's so easy to create a counter-argument that 'feels' right and comes to the more 'intuitive' answer. But I do believe that my explaination gives the correct answer, and hopefully it's clear where it comes from.

 

[OSB]

Probability often isn't intuitive and so it's easy to get confused in cases like this and think that the 'obvious' answer must be correct. I did this sort of thing as part of one of my courses at university and REALLY struggled to get to grips with it at first. It's so easy to create a counter-argument that 'feels' right and comes to the more 'intuitive' answer. But I do believe that my explaination gives the correct answer, and hopefully it's clear where it comes from.

Yeah thats' where i'm at now, another two terms of it at least! I kind of enjoy it really!

 

[Haircut]

Threads like this reinvigorate my love of maths, kind of wish I'd carried on a bit further with maths related things now.

 

[Psyk]

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?

Yes, but if you roll the dice 10 times there's a much greater chance of at least one of them being a 6 than if you just rolled once.

 

[Snakedoctor]

Makes perdect sence them knowing. One thing is, there are very slightly more women in the world than men. So the chances that they are also a girl, do increase.

To address this situation, I understand, there are already "man sharing" organisations in the US.

 

[div0]

Yeah thats' where i'm at now, another two terms of it at least! I kind of enjoy it really!

What course is it that you're doing? Is it a maths based degree?

My probability course came in as part of a physics course, which was rock solid! I ended up having to drop the physics for 3rd year as a joint degree was just too hard.

 

[Bill Door]

It's 50/50 regardless of if you know if the child is the younger or not. By virtue of knowing that one child is female the probability of the elder child being female is no longer 50%.

For those argueing the 2/3 case, please point out any logical flaws in the following...

There are 2 children.
You know that one is a girl.
There is a 50% chance that she has an older sibling, and a 50% chance that she has a younger sibling.
Apply the logic used in the example where the age order of the girl is known for both of these scenarios.

 

[div0]

It's 50/50 regardless of if you know if the child is the younger or not. By virtue of knowing that one child is female the probability of the elder child being female is no longer 50%.

For those argueing the 2/3 case, please point out any logical flaws in the following...

There are 2 children.
You know that one is a girl.
There is a 50% chance that she has an older sibling, and a 50% chance that she has a younger sibling.
Apply the logic used in the example where the age order of the girl is known for both of these scenarios.

Did you read my last 3 or 4 posts?

I tried to make it as clear as possible

Your statements are correct, but they are not addressing the question.

You are looking at it as TWO completely INDEPENDENT EVENTS (the birth of one child and the birth of another).

I have explained that this is NOT the way to approach the problem. The father ALREADY KNOWS the OUTCOME of BOTH EVENTS (both births) BEFORE he tells you the information.

Of the 4 ways he COULD have had his two children - he can ONLY have had them in one of the 3 (EQUALLY LIKELY) ways that allow at least 1 girl. TWO of these THREE ways result in a boy being the other child, and only 1 of them results in it being a girl.

As each of the 3 are EQUALLY likely, this is what causes the probability to be 2/3.

If he ONLY knew the sex of ONE child (his daughter), then it is true the the probability of the other child being a boy is 50/50. But this is NOT the case, he has PRIOR KNOWLEDGE, that affects the probability.

I know its not intuitive and I know your logic seems solid, but I've done my best to explain it in previous posts. If theres anything specific that you want explained, then I'll try and answer it as best I can.

 

[PhillyDee]

Its probability, not chance, or logic.

 

[OSB]

Its probability, not chance, or logic.

Exactly!

What course is it that you're doing? Is it a maths based degree?

I'm doing a Natural Sciences course (Cambridge), so this year i'm doing a module of Maths, one of Physics one Chemistry and one Geology.

 

[Jokester]

Seriously read this:- http://mathforum.org/library/drmath/view/52186.html

Both the OPs cases refer to picking a child from a family, not the case of randomly picking a family and then being told that at least one is a girl.

In both cases in the OP, we are being specifically told that the child with the father is a girl and this is the only way we have established that the family has at least one girl, by effectively taking a sample to find out. In both OPs cases the sample space contains only the other child, whilst in the other case both children remain in the sample space.

 

[Haircut]

Seriously read this:- http://mathforum.org/library/drmath/view/52186.html

Both the OPs cases refer to picking a child from a family, not the case of randomly picking a family and then being told that at least one is a girl.

In both cases in the OP, we are being specifically told that the child with the father is a girl and this is the only way we have established that the family has at least one girl, by effectively taking a sample to find out. In both OPs cases the sample space contains only the other child, whilst in the other case both children remain in the sample space.

No, that example is picking a child from a group of families.
In the OP the family is chosen first by way of bumping into the tutor. It is then this family and only this family we are dealing with.

 

[divine`]

Question:

Does it really matter?

 

[div0]

Seriously read this:- http://mathforum.org/library/drmath/view/52186.html

But that backs up my argument completely.

In the example where the answer is 1/2 - YOU pick the child (to be identified) at random, with NO knowledge BEFORE YOU pick

Supposing that we randomly pick a _child_ from a two-child family. We
see that he is a boy, and want to find out whether his sibling is a
brother or a sister. (For example, from all the children of two-child
families, we select a child at random who happens to be a boy.) In
this case, an unambiguous statement of the question could be:

From the set of all families with two children, a child is
selected at random and is found to be a boy. What is the
probability that the other child of the family is a girl?

Note that here we have a pool of kids (all of whom are from two-child
families) and we're pulling one kid out of the pool. This is like the
problem you're talking about. The child selected could have an older
brother, an older sister, a younger brother or a younger sister.

Let's look at the possible combinations of two children. We'll use B
for Boy and G for girl, and for each combination we'll list the older
child first, so GB means older sister while BG means younger sister.
There are 4 possible combinations:

{BB, BG, GB, GG}

From these possible combinations, we can eliminate the GG combination
since we know that one child is a boy. The three remaining possible
combinations are:

{BB, BG, GB}

In these combinations there are four boys, of whom we have chosen one.
Let's identify them from left to right as B1, B2, B3 and B4. So we
have:

{B1B2, B3G, GB4}

Of these four boys, only B3 and B4 have a sister, so our chance of
randomly picking one of these boys is 2 in 4, and the probability is
1/2 - as you have indicated.

In the example the gives the answer 2/3 - the family is selected first (the TWO children are KNOWN) and then information is given about ONE of them.

But now let's look at a different way of selecting the "boy" in the
problem. Suppose we randomly choose the two-child _family_ first. Once
the family has been selected, we determine that at least one child is
a boy. (For example, from all the mothers with two children, we select
one and ask her whether she has at least one son.) In this case, an
unambiguous statement of the question could be:

From the set of all families with two children, a family is
selected at random and is found to have a boy. What is the
probability that the other child of the family is a girl?

Note that here we have a pool of families (all of whom are two-child
families) and we're pulling one family out of the pool. Once we've
selected the family, we determine that there is, in fact, at least one
boy.

Since we're told that one child (we don't know which) is a boy, we can
eliminate the GG combination. Thus, our remaining possible
combinations are:

{BB, BG, GB}

Each of these combinations is still equally likely because we picked
one of the four families.

Now we want to count the combinations in which the "other" child is a
girl. There are two such combinations: BG and GB.

Since there are three combinations of possible families, and in two of
them one child is a girl, the probability is 2/3.

Both the OPs cases refer to picking a child from a family, not the case of randomly picking a family and then being told that at least one is a girl.

No they don't refer to just picking a child from a family! They refer to a father (of a random 2child family) telling us that he has a daughter.

In both cases in the OP, we are being specifically told that the child with the father is a girl and this is the only way we have established that the family has at least one girl, by effectively taking a sample to find out. In both OPs cases the sample space contains only the other child, whilst in the other case both children remain in the sample space.

No. The father, KNOWING what TWO children he has, is TELLING/SHOWING us that he has a daughter. It is distinctly different.

 

[Jokester]

No, that example is picking a child from a group of families.
In the OP the family is chosen first by way of bumping into the tutor. It is then this family and only this family we are dealing with.

Note that here we have a pool of families (all of whom are two-child
families) and we're pulling one family out of the pool. Once we've
selected the family, we determine that there is, in fact, at least one
boy.

Since we're told that one child (we don't know which) is a boy, we can
eliminate the GG combination. Thus, our remaining possible
combinations are:

But in both cases, we do know which, the girl is standing in front of us and we're being asked what the probability of the other child being male is, which is now a single event.

 

[Haircut]

But in both cases, we do know which, the girl is standing in front of us and we're being asked what the probability of the other child being male is, which is now a single event.

But we don't though.
The fact that the girl is standing in front of us only tells us that there is at least one girl in the family, nothing more.

 

[div0]

But in both cases, we do know which, the girl is standing in front of us and we're being asked what the probability of the other child being male is, which is now a single event.

You're looking at it from the wrong angle.

If the child that is revealed to us is chosen at RANDOM, then the answer is 1/2.

But a child is chosen/revealed, by someone already KNOWING the sex of BOTH children, then the probability changes, because of their prior knowledge of the situation.