4 ms·
I 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 :-)
by reitanqild 8y ago
I 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.