#weird probability
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That’s how it should work.
Since it would either be a 3 on the first die or a 3 on the second.
Are you familiar with Bayes's theorem?
Conditional probability.
P(both 3 | at least one 3) = P(both 3 ⋂ at least one 3)/P(at least one 3) = P(both 3)/P(at least one 3)
Oh, yeah, that's a nice approach, too.
nah i've been meaning to learn it tho
they don't teach this notation in the uk until 17
What notation?
What kind of combinations would result from this?
I mean, I'm a fan of Boolean algebra notation.
Also, you can think about it like this: the permutations “reflect”, i.e., (1,3) is different from (3,1). Thus we get 2*6-1=11 possible probabilities, and only one of these is (3,3), since reflexivity in this case would not work on it.
There's two fundamental rules of probability theory you need to know to basically be able to do anything possible. They are: P(AB) = P(A|B) * P(B) = P(B|A) * P(A) P(A + B) = P(A) + P(B) - P(AB)
@ebon dew You still with us?
at gcse in england they only teach probabiltiy trees so i just tried to represnet it using that and failed
idk wht the `¦ means
It means "given".
Tables suffice here. They’re useful regardless, but it’s good to know the notation.
P(A|B) means "the probability of A given B" or "the probability of A if B is true."
@ebon dew
Oh, wait, so this is supposed to be solved just using the definition of probability?
Well, that kinda sucks...
That’s what it looks like.
I mean, there's no indication of how it's "supposed" to be solved.
And we're not getting a lot of feedback from the helpee about what exactly confused them.
Yeah, true, but seeing as what Max said about notation and stuff, that's what I thought.
Yeah, there isn’t necessarily an off limits method. It is only a question with no confined boundaries, i.e., in a textbook.
I mean, it's possible to do it like that, of course, but that's so inefficient...
What if the dice had 20 sides?
sorry
it's just the new vocab and stuff, i understand now why my initial reasoning was wrong, but now it's about coming up with an expression for the probabilty
I'm not sure what you mean by that.
Not a single word was understood
I’ll rephrade
Rephrase
I see why I cannot just assume the first dice is the one that rolls a 3, so i get that my idea of 1*(1/6) was wrong
But idk how to form an expression for the probability now
If you don’t know the notation, create a table.
I don't know what you mean by that.
It follows that you have, from the first roll of a 3, the ordered pairs (3,1),(3,2)...(3,6).
Ye I mean I’m preparing for an Olympiad style thing tomorrow so I’ll know now to just make a table if something like this occurs but now it’s just interest
Where the (x,y)->(first roll, second roll).
Create the listing for the 2nd rolls, and make sure you don’t overcount.
You don’t know that the first roll is 3 though?
This is casework.
Look, I think I can help correct whatever your mistake was in your probability tree.
Why?
Look, you need to tell us how you're thinking about the problem in order for us to help you.
You need to work with us.
Ok.
this was my original idea, but i understand how it's wrong n ow
Why is it wrong?
cause it could be the second dice that is 3
this is wht i meant by probability tree
Hmm.
Okay, so now which branches contain at least one 3?
top 3
So we only want to pay attention to those branches.
Oh? What's this?
its intended for like 14/16 year olds
Hm... I don't think I've ever seen this method.
Look, probability theory isn't that bad.
idk this might be an unconventional problem for it
It all comes down to these two rules.
ik but i hve had no formal education on it aside from "probability trees", i just self study everything, and usually it does not include probabilty, and more so calc and lin algebra
yea ik but this olympiad thing is done by 13 year olds and i dont think they learn this
They do if they self-study it.
the test is based on things they would learn tschool,
...these rules are probability theory. Like, they're exactly the same thing that you're doing in your little probability trees, but, like, algebraically, and so way more convenient.
i dont think the averge person sitting this test would understnd any of those symbols, but i know
That's a bid odd, though. The notation isn't difficult, why not have it in school? You are going to use it later in university, anyway, so then there wouldn't be a need to relearn it.
similra to how we were taught the power rule without knowing that it follows first principles or what any of it actually means on the intuitive level
Odd.
uk school has basiaclly like 11-16 gcses, 16-18 a levels, then uni
at a levels, i think they teach you it
but im 15 so i have to learn it all myself
But, like, it’s universal notation, and also is pretty easy to interpret.
Sorry, I don't really know what those mean.
tbf we learn union and the upside down union in sets
$\cap$ and $\cup$ in set theory?
Temu Mod
as in, set1 = [1,2,3,4,5], set2=[1,3], nd they woiuld make u draw a venn diagfram and find the union
yea
i disagree with it
Oh, that's even more odd, then... Usually you learn them at the same time, roughly.
oy
Yeah.
it's hardly set theory i mean
I learned mine last year.
idk set theory but i assume its more than
"put these numbers in this circle, and the other numbers in the other circle"
It still is, though. It's operations on sets, so, set theory.
It’s some part of it.
You disagree with it?
with the order we learn things
Look.
Here's Boolean algebra.
Or, wait, before that.
You do understand, like, Boolean logic, right?
ytea i can code
Okay, so.
In Boolean algebra, the Boolean product corresponds to "and", and the Boolean sum corresponds to "or".
Oh, actually, sorry to divert the conversation, but I'm just really curious.
If you want to write the probability of thins A and B happening at the same time, would you write it as P(A ⋂ B) or as P(A ∧ B)? As in, do you view events more like sets or more like statements about things?
Just curious.
I prefer the set approach, by the way.
@ebon dew
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