Boy Or Girl paradox

Question

I want to ask question about famous Boy or Girl paradox. I completely understand, why the probability is 1/3 in the most famous variation of this paradox, but why the probability is 1/2 in the following variation?

Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?

Let's compare children by their, for example, weight (not age). First child weights more. So we have 4 variants:BB, GB, BG, GG. Variant BB isn't suitable, so we have 3 variants with equal probability and answer is 1/3. But if we make such a comparison by age, we suddenly get other answer, which is 1/2. Wiki states, that this answer is correct. Where's the mistake in my solution?

Answer 1

Remember probability is about information (or rather, a lack thereof). That is, we calculate probabilities on the basis of given information (or again, lack thereof). So, when you try to answer this problem by only considering weight, you are not taking into account the information that was given to you about the age, and hence you get the wrong answer.

As an analogy, suppose we had the following problem:

You have 2 coins, a red one and a blue one. I flip both coins, and the red comes up heads. What is the chance the blue one comes up heads?

And now suppose your answer is: Well, let's take the two coins and sort them by weight. Now, I don't know whether the red or the blue one is heavier, but given that one of the coins came up heads, the probability space is (heavier one first): HH, HT, TH. So, the probability of the other one coming up heads is $\frac{1}{3}$.

Would you consider this a good answer? Of course not! The answer should be $\frac{1}{2}$, and the mistake in the 'solution' is that you didn't take into account all the information, which is that a specific coin came up heads, and you are asked about the other coin coming up heads.

Or, as an even more extreme analogy. Suppose I say: I flip 1 coin, and it comes up heads. What is the probability it came up heads? And you say: well, it's $50$%, because I assume it is a fair coin. No, once you are told it came up heads, the probability it came up heads is $100$%. Again, when asked about probability, it's a question that says "given everything you do know, what can you say about some event happening?".

Now, back to your specific case. Given that we do know the red one comes up heads, notice that the probability space cannot be (heavier one first): HH, HT, TH. Why not? Well, if the red coin was the heavier coin, then TH is ruled out, and you are left with HH, HT. And if the red coin is the lighter one, then the HT is rules out, and you are left with HH, TH. And notice that in both case, the probability of the other coin coming up heads is now indeed $\frac{1}{2}$, as it should be.

Answer 2

The point here is that there is only one oldest child. Therefore that one is fixed and we only end up with two possibilities: G-B or G-G.

Answer 3

Label the children by both sex and relative age: "OBYB" for "older boy, younger boy", "OBYG" for "older boy, younger girl", and so on. Then there are eight events, of which four are ruled out by the condition that the older child is a girl, leaving the four events {YBOG, OGYB, OGYG, YGOG}; and in two of those four events, both children are girls.

Dealing with the interdependence of relative age, sex, and relative mass is left as an exercise for the reader.

Answer 4

You have created a sample space encoding the weight order and sex of the siblings, ignoring the critical information on age (ie: the elder sibling is female): $$\rm \{F_HF_L, F_HM_L, M_HF_L, M_HM_L\}$$

Then you treat the four outcomes as being equally probable.   However, weight is not independent of either age or sex, particularly under the condition that the eldest is female.

Answer 5

With the children sorted by weight (but knowing the older child is a girl), there are actually four cases (which we are, somewhat unrealistically, treating as equally likely):

Case 1: $B_Y, G_O$ (Boy (heavier but younger), Girl (lighter but older))

Case 2: $G_O, B_Y$

Case 3: $G_O, G_Y$ (older girl is heavier in this case).

Case4: $G_Y, G_O$ (younger girl is heavier in this case).

So if we treat the cases as equally likely, the chances of two girls is $\frac12$.