this post was submitted on 04 Feb 2025
129 points (99.2% liked)

Asklemmy

44804 readers
868 users here now

A loosely moderated place to ask open-ended questions

Search asklemmy 🔍

If your post meets the following criteria, it's welcome here!

  1. Open-ended question
  2. Not offensive: at this point, we do not have the bandwidth to moderate overtly political discussions. Assume best intent and be excellent to each other.
  3. Not regarding using or support for Lemmy: context, see the list of support communities and tools for finding communities below
  4. Not ad nauseam inducing: please make sure it is a question that would be new to most members
  5. An actual topic of discussion

Looking for support?

Looking for a community?

~Icon~ ~by~ ~@Double_[email protected]~

founded 5 years ago
MODERATORS
[email protected]
[email protected]
[email protected]
[email protected]
129
What are some examples of 'common sense' which are nonsense? (lemmy.ml)
submitted 3 days ago by [email protected] to c/[email protected]
146 comments fedilink hide all child comments
 

Wikipedia defines common sense as "knowledge, judgement, and taste which is more or less universal and which is held more or less without reflection or argument"

Try to avoid using this topic to express niche or unpopular opinions (they're a dime a dozen) but instead consider provable intuitive facts.

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 15 points 3 days ago (3 children)

The gambler’s fallacy is pretty easy to get, as is the Monty Hall problem if you restate the question as having 100 doors instead of 3. But for the life of me I don’t think I’ll ever have an intuitive understanding of the birthday problem. That one just boggles my mind constantly.

[–] [email protected] 9 points 3 days ago

Lemme try my favorite way to explain the birthday problem without getting too mathy:

If you take 23 people, that's 253 pairs of people to compare (23 people x22 others to pair them with/2 people per pair). That's a lot of pairs to check and get only unique answers

[–] [email protected] 12 points 3 days ago (3 children)

Really? The birthday problem is a super simple multiplication, you can do it on paper. The only thing you really need to understand is the inversion of probability (P(A) = 1 - P(not A)).

The Monty hall problem... I've understood it at times, but every time I come back to it I have to figure it out again, usually with help. That shit is unintuitive.

[–] [email protected] 1 points 1 day ago

Adding my own explanation, because I think it clicks better for me (especially when I write it down):

  1. Pick a door. You have a 66% chance of picking a wrong door, and a 33% of picking the right door.
  2. Monty excludes a door with 100% certainty
  3. IF you picked a wrong door, then there's a 100% chance the remaining door is correct (so the contingent probability is p(switch|picked wrong) = 100%), so the total chance of the remaining door being correct is p(switch|picked wrong)* p(picked wrong) = 66%.
  4. IF you picked the right door, then Monty's reveal gives you no new information, because both the other doors were wrong, so p(switch|picked right) = 50%, which means that p(switch|picked right) * p(picked right) = 50% * 33% = 17%.
  5. p(don't switch|picked wrong) * p(picked wrong) = 50% * 66% = 33% (because of the remaining doors including the one you picked, you have no more information)
  6. p(don't switch|picked right) * p(picked right) = 50% * 33% = 17% (because both of the unpicked doors are wrong, Monty didn't give you more information)

So there's a strong benefit of switching (66% to 33%) if you picked wrong, and even odds of switching if you picked right (17% in both cases).

Please feel free to correct me if I'm wrong here.

[–] [email protected] 1 points 1 day ago

My explanation is better:

There's three doors, of which one is the winner.

First, pick a door to exclude. You have a 66% chance of correctly excluding a non-winning door.

Next, Monty excludes a non- winning door with certainty.

Finally, open the remaining door and take the prize!

[–] [email protected] 4 points 2 days ago (1 children)

My favourite explanation of the Monty hall problem is that you probably picked the wrong door as your first choice (because there’s 2/3 chance of it being wrong). Therefore once the third door is removed and you’re given the option to switch you should, because assuming you did pick the wrong door first then the other door has to be the right one

[–] [email protected] 2 points 1 day ago

Thanks for the help, it was easier this time 😅

[–] [email protected] 1 points 3 days ago

The birthday problem is super easy to understand with puzzles! For example, how does laying out the edges increase the likelihood of a random piece to fit.