4 ms·
Here's the math, I will break it down based on Boy/Girl and seen. For the first case here are the different combinations for two children based on sex (let's s
by pg_bot 8y ago
Here's the math, I will break it down based on Boy/Girl and seen.
For the first case here are the different combinations for two children based on sex (let's say left denotes older and right denotes younger to distinguish the difference between girl and a boy and boy and a girl.)
B B
B G
G B
G G
These outcomes are each equally likely, but since we know that one has to be a boy we can eliminate the (G G) case leaving us with two chances out of three that the other child is a girl.
The other case can be viewed as follows (S) denotes seen.
B(S) B
B B(S)
B(S) G
B G(S)
G(S) B
G B(S)
G(S) G
G G(S)
I can eliminate all cases where I see a girl out the window, leaving me with the following equally likely cases:
The neighbor has two boys and I see the older boy. The neighbor has two boys and I see the younger boy. The neighbor has an older boy whom I saw and younger girl. The neighbor has an older girl and a younger boy whom I saw. The neighbor has a girl in 50% of the remaining possible cases.
- deleted 8y ago[deleted]
- perl4ever 8y agoSuppose they are twins and therefore there is no older and younger. Does that change the probability from 2/3 in the first case?
- reitanqild 8y agoI guess this is one of a few place to answer mu. One of the twins is still considered to be born first and therefore older :-)
- perl4ever 8y agoI wrote a program and successfully got the right answers experimentally for each case, so I'm not doubting the answers, but I don't buy the explanation. While you may argue that one twin is always born first, this is an idealized mathematical problem, so it can't depend on particular details of physical reality in that way. There is nothing logically preventing them from being the same age, so it can't be pivotal to solving the problem. And in fact, I experimentally got the right result without any reference to ages.
- zazen 8y agoSo, you don't actually have to regard BG and GB as separate cases if you don't want to. People like to write the outcome space as ordered pairs BB, BG, GB, GG (ordered by age or whatever else), because this has the advantage that all four outcomes are equiprobable, so you can calculate probabilities just by counting. If you prefer to regard the outcome space as unordered pairs {B,B}, {B,G}, {G,G} this is absolutely fine: you just have to bear in mind that the prior probability of {B,G} is 1/2, whereas it's 1/4 for the other pairs, so a tiny bit more calculation might be required. P({B,G}| not {G, G}) = P({B,G})/(P{B,B} + P{B,G}) = (1/2)/((1/2) + (1/4)) = 2/3 as with the other method.
- rcxdude 8y agoIgnoring the increased probability that twins are of the same gender, no. Regardless, the chance of having two children of different genders is twice that of having two boys. If you split the set into 'two boys, a boy and a girl, and two girls', you must accept that the 'a boy and a girl' case is twice as likely as the first and last options.
- szemet 8y agoIf you assume that the siblings can be differentiated (thus ordered), it makes no difference. You can for example order them taking their names alphabetically, or by weight, or by location at a given time, etc... If differentiation is not possible, then they are the exact same person [1], so the parents have one child and it must be boy then ;) [1] https://en.wikipedia.org/wiki/Identity_of_indiscernibles https://en.wikipedia.org/wiki/Identity_of_indiscernibles