3 ms·
I was lazy so I just did: import itertools as itools xs = [f'{i}-{j}' for i in range(30) for j in range(12)] ys = ['b', 'g'] zs = list(itools.product(xs, ys
by ksaho 3y ago
I was lazy so I just did:
import itertools as itools
xs = [f'{i}-{j}' for i in range(30) for j in range(12)]
ys = ['b', 'g']
zs = list(itools.product(xs, ys))
boys = [z for z in zs if z[1] == 'b']
len(boys)/len(zs)
----
gives me 0.5, or 1/2.
The moral of the article is to set up your universal set correctly and do not throw out any information, now matter how irrelevant.
- p1esk 3y agoBut in this case the moral of the article is exactly the opposite - if you throw out the irrelevant information the answer becomes trivial (no need for your calculations).
- wruza 3y agoNow I tell you I have two children, and (at least) one of them is a boy born on April 1st, has medium-length black hair, is 4'2", interested in 5 out of 12000 available rpg games, has 8 close friends and sword-shaped fingernails. I believe your script may require a datacenter and a scalable architecture now.
- phalf 3y agoAnd what do you do with non-discrete attributes?
- phalf 3y agoBut that's exactly what you did! You just counted which fraction of all kids is boys. You threw away that this is about pairs of kids, about april 1 on which one of them has their birthday and is a boy, and we want to know the sex of the other one. You already simplified by applying symmetry reductions. You could have just done print(1/2) and proclaimed that the output was 0.5 which therefore must be the solution.