4 ms·

>I tell you I have two children, and (at least) one of them is a boy born on April 1. What probability should you assign to the event that I have two boys? If y

by sagaro 3y ago

>I tell you I have two children, and (at least) one of them is a boy born on April 1. What probability should you assign to the event that I have two boys? If you think that is going to be too cumbersome, simply tell me whether the probability is close to 1/2 or to 1/3, or to some other simple fraction, and provide an estimate as to how close. I am going to not heed the author's suggestion of setting up the problem correctly and use intuition: For the week problem it was 14+13 = (7 * 4 -1) in the bottom and 13 (7*2-1) on the top. So for a normal year (365 days), it would be (365*2 - 1)/(365*4 - 1). Close to 1/2.

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 ago
  But 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 ago
  Now 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 ago
    And what do you do with non-discrete attributes?
- phalf 3y ago
  But 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.
FabHK 3y ago
Good intuition. The answer is, indeed, (2-p)/(4-p), where p is the probability that the kid in question is born in the time period in question, ie 1/7 or 1/365, as the case may be. So, you’re correct. One can pump the intuition further: If you only know that there are no girls, the result is 1/3. If a boy opens the door, or you know that the older one is a boy, the result is 1/2. So identifying one of them as the boy increases the result from 1/3 to 1/2. By saying “one is a boy born on Tuesday”, you give some information, thus nudge the result a bit toward 1/2. Saying “one is born on 12/30”, say, you identify the kid more and nudge the result nearly all the way towards 1/2. And indeed, if (in the formula above) we set p=1 (no identification), 1/3 results, and if we set p=0 (full identification), 1/2 results. Quite neat.

by sagaro 3y ago

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 ago
But 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 ago
Now 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 ago
  And what do you do with non-discrete attributes?
phalf 3y ago
But 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.

FabHK 3y ago

Good intuition. The answer is, indeed, (2-p)/(4-p), where p is the probability that the kid in question is born in the time period in question, ie 1/7 or 1/365, as the case may be. So, you’re correct. One can pump the intuition further: If you only know that there are no girls, the result is 1/3. If a boy opens the door, or you know that the older one is a boy, the result is 1/2. So identifying one of them as the boy increases the result from 1/3 to 1/2. By saying “one is a boy born on Tuesday”, you give some information, thus nudge the result a bit toward 1/2. Saying “one is born on 12/30”, say, you identify the kid more and nudge the result nearly all the way towards 1/2. And indeed, if (in the formula above) we set p=1 (no identification), 1/3 results, and if we set p=0 (full identification), 1/2 results. Quite neat.