help-13

[unborn cradle]

yeah, haven't coded the calculator yet

[vague valve]

alr, no problem

it doesnt matter what the cards outside of the combo we want are right

[unborn cradle]

correct

[cedar kiln]

@vague valve Has your question been resolved?

[unborn cradle]

I'm working on coding a solution to this, I just got distribution to work:

(1∪2∪3)∩4∩5∩6
(4∩5∩6∩1)∪(4∩5∩6∩2)∪(4∩5∩6∩3)

distribute test 2
(1∪2)∩(3∪4)
(1∩3)∪(1∩4)∪(2∩3)∪(2∩4)

distribute test 3
(1∪2)∩(3∪4)∩(5∪6)
(1∩3∩5)∪(1∩3∩6)∪(1∩4∩5)∪(1∩4∩6)∪(2∩3∩5)∪(2∩3∩6)∪(2∩4∩5)∪(2∩4∩6)

distribute test 4
((1∪2∪3)∩4∩5∩6)∩((1∪2)∩(3∪4))
(4∩5∩6∩1∩1∩3)∪(4∩5∩6∩1∩1∩4)∪(4∩5∩6∩1∩2∩3)∪(4∩5∩6∩1∩2∩4)∪(4∩5∩6∩2∩1∩3)∪(4∩5∩6∩2∩1∩4)∪(4∩5∩6∩2∩2∩3)∪(4∩5∩6∩2∩2∩4)∪(4∩5∩6∩3∩1∩3)∪(4∩5∩6∩3∩1∩4)∪(4∩5∩6∩3∩2∩3)∪(4∩5∩6∩3∩2∩4)```

it transforms an arbitrary algebraic set expression into a union of intersections so that inclusion-exclusion can be applied

[neat raven]

!iamnotstudying

!iamnot

[vague valve]

thats great, whats left to do?

[unborn cradle]

Fixing an edge case with empty sets

I have this so far, the results agree with my previous calculations:

hit_cards: [11, 3, 7], missed_cards: []
Draws with no missed cards: 91390
Draws with no missed cards and not all hit cards: 84922
Draws with no missed cards and all hit cards: 91390 - 84922 = 6468
Total draws: 91390
draw chance before mulligan: 6468/91390 = 3234/45695


hit_cards: [3, 7], missed_cards: [11]
Draws with no missed cards: 23751
Draws with no missed cards and not all hit cards: 22195
Draws with no missed cards and all hit cards: 23751 - 22195 = 1556
Total draws: 91390
draw chance before mulligan: 1556/91390 = 778/45695


hit_cards: [11, 7], missed_cards: [3]
Draws with no missed cards: 66045
Draws with no missed cards and not all hit cards: 40535
Draws with no missed cards and all hit cards: 66045 - 40535 = 25510
Total draws: 91390
draw chance before mulligan: 25510/91390 = 2551/9139


hit_cards: [11, 3], missed_cards: [7]
Draws with no missed cards: 40920
Draws with no missed cards and not all hit cards: 34225
Draws with no missed cards and all hit cards: 40920 - 34225 = 6695
Total draws: 91390
draw chance before mulligan: 6695/91390 = 103/1406


hit_cards: [7], missed_cards: [11, 3]
Draws with no missed cards: 14950
Draws with no missed cards and not all hit cards: 3876
Draws with no missed cards and all hit cards: 14950 - 3876 = 11074
Total draws: 91390
draw chance before mulligan: 11074/91390 = 5537/45695


hit_cards: [3], missed_cards: [11, 7]
Draws with no missed cards: 7315
Draws with no missed cards and not all hit cards: 3876
Draws with no missed cards and all hit cards: 7315 - 3876 = 3439
Total draws: 91390
draw chance before mulligan: 3439/91390 = 181/4810


hit_cards: [11], missed_cards: [3, 7]
Draws with no missed cards: 27405
Draws with no missed cards and not all hit cards: 3876
Draws with no missed cards and all hit cards: 27405 - 3876 = 23529
Total draws: 91390
draw chance before mulligan: 23529/91390 = 23529/91390


hit_cards: [], missed_cards: [11, 3, 7]
Draws with no missed cards: 3876
Draws with no missed cards and not all hit cards: 0
Draws with no missed cards and all hit cards: 3876 - 0 = 3876
Total draws: 91390
draw chance before mulligan: 3876/91390 = 102/2405

Process finished with exit code 0

[unborn cradle]

@vague valve

about 19.23%

[vague valve]

for the original example?

[unborn cradle]

yes

[vague valve]

wow nice

so this works with any input?

[unborn cradle]

The inputs are deck_size, hand_size, and combo_counts

The only thing that doesn't work is if you need to draw duplicate cards for the combo

I used

DECK_SIZE = 40
HAND_SIZE = 4
COMBO_CARDS = [11, 3, 7]

as inputs

[vague valve]

i see, whats the issue with duplicate cards? need a seperate case for those i assume

[unborn cradle]

yeah

the issue is that if there are duplicates the results aren't independent

The other issue is that the code works by subtracting the combo cards we didn't draw from the deck size, so if we tried to use [10, 10, 10] to represent drawing 3 of the same card with 10 in the deck, it would subtract 30 from the deck size instead of 10

I can send you the code but it's in 3 separate files

I'm not sure if it would run in an online interpretter

[vague valve]

would it be hard to get duplicates working? theres a 4 card combo im trying to calculate

[unborn cradle]

I'm not sure

I'd have to work out the math first

before I could start coding it

[vague valve]

alr, gl with that. could you run it for
DECK_SIZE = 40
HAND_SIZE = 4
COMBO_CARDS = [4, 4, 4, 4]

[unborn cradle]

About 8.81%

wait

it's negative wtf

[vague valve]

thats not supposed to happen i assume lol

[unborn cradle]

nope

something's wrong lol

[unborn cradle]

ok I did something wrong with the math

I'm not sure how to fix it though

[unborn cradle]

ok I think I know how to fix it

the issue is it's calculating
|(A∩B)∪(A∩C)|
= |(A∩B)| + |(A∩C)| - |(A∩B)∩(A∩C)|

but it doesn't realize |(A∩B)∩(A∩C)| = |A∩B∩C|

It thinks the two A's are different sets

so I'll have to create an object to keep track of them so I can compare them

[vague valve]

alr

[vague valve]

did you get it working?

[unborn cradle]

no not yet it might take a bit because I basically have to refactor everything

[vague valve]

did you get around to doing this?

[unborn cradle]

no sorry I should have time today though

[unborn cradle]

@vague valve 48437824/2088033025 ~ 2.32% for

DECK_SIZE = 40
HAND_SIZE = 4
COMBO_CARD_COUNTS = [Card(4), Card(4), Card(4), Card(4)]

(assuming I did it right this time)

~~I was able to consolidate all the code into one file if you want to run it yourself here: https://www.online-python.com/~~ Edit: bugged, fixed version below

You can change the variables

DECK_SIZE = 40
HAND_SIZE = 4
COMBO_CARD_COUNTS = [Card(4), Card(4), Card(4), Card(4)]

at the top to test different results

[unborn cradle]

oh I found a bug, the deck_size was incorrectly decreased for the mulligan, I fixed it now

[vague valve]

nice

send updated code? i didnt know about that wesbite, seems very useful

[unborn cradle]

here you go

Hopefully there's no more bugs 😅

Ideally we could test it against a simulation for accuracy, if we really wanted

[vague valve]

how exactly does the "card combo count" variable work? how do i differentiate between different cards

[unborn cradle]

so like

for the ace, queen, jack example

it would be

COMBO_CARD_COUNTS = [Card(7), Card(11), Card(3)]

7 Aces, 11 Jacks, 3 Queens

COMBO_CARD_COUNTS = [Card(4), Card(4), Card(4), Card(4)]
is for a 4-card combo with 4 copies of each card

[vague valve]

yeah, but what about 2 of the same card for a combo?

[unborn cradle]

oh it doesn't work for duplicates unfortunately

I'm not sure what the math for duplicates looks like

It might be worth asking in #combinatorial-structures , if someone could help me with the math I could try to code it depending on how difficult it is

[upper laurel]

you here?

[vague valve]

yep im here

[upper laurel]

whats stopping you from just redrawing cards until you get the hand you want

[vague valve]

you cant do it multiple times

[upper laurel]

you can only do it once per card you redraw?

[vague valve]

i think yes

[upper laurel]

"i think"?

[vague valve]

im not sure what you mean

[upper laurel]

so you need to edit your post to refer to the first 4 cards you pull out as "starting cards" instead of "any of the 4 cards you drew one time"

[unborn cradle]

it's a mulligan

you only get one mulligan in most card games

[upper laurel]

you can also edit the post to have the word mulligan in it

either way it needs to be changed

[unborn cradle]

magic/hearthstone/LoR etc.

[unborn cradle]

You can only redraw one time, is that not clear from the original question?

[upper laurel]

"you drew one time" is vague because it doesnt have a "that" or "the first time" or "starting time" or anything like that

[unborn cradle]

fair

[upper laurel]

if the question is going to stay open for this long it should ideally have no catches, even minor ones

[vague valve]

you draw 4 cards

[upper laurel]

axerity bear with me

[vague valve]

cards 1 2 3 and 4

[upper laurel]

no no you dont have to say anything

you can edit your post by clicking the ... option in the right

then clicking "edit"

[vague valve]

stop condescending to me

i dont appreciate it

[unborn cradle]

?

[upper laurel]

see here you chose to re-explain the same idea instead of agreeing to edit the post

[vague valve]

sorry im gamer raging rn

[upper laurel]

I am simply telling you to not do that

[unborn cradle]

we just want you to edit the original question lol

so someone who reads the pinned message can understand it better

[upper laurel]

yea the message gets pinned when you ask it, its been up there for a week

[vague valve]

ok i labeled the 4 cards you draw at the start the starting cards

[upper laurel]

yea thats good

[unborn cradle]

I found this https://deckulator.appspot.com/

doesn't work with mulligans though

[vague valve]

that would have been too convenient to exist for sure

[dense ledge]

There is something that makes me very uneasy with these calculations...

say we have deck 5q 1k 1j 1a and
want to draw 3q 1k (and assume we have hand size of 4 and we can redraw any of cards exactly once)
after first draw we get: 2q 1j 1a
We have two strategies we can follow we can redraw 1j and 1a (only the cards we didn't draw correctly), and we will either end up with the 3q 1k or we might also end up with 4q
The other strategy is to redraw all the cards since there are only 4 cards in the deck left we are guaranteed to get exactly them, which would be 3q 1k.

So by discarding cards you wanted to get you can actually improve your chances

[unborn cradle]

That's a good point, I was assuming the first strategy was always (at least weakly) dominant

so really I would have to calculate the probabilty for every possible mulligan

[vague valve]

yeah i thought about that but i wasn't sure if it'd apply since i couldnt find a situation where it would, but it makes sense though

[unborn cradle]

I would recommend asking this in #combinatorial-structures if you haven't already

[vague valve]

alright

although i thought you first: mulligan all cards that aren't part of the combo,
then second: if you have more than enough of a card for the combo, mulligan those

doesn't that work?

[vague valve]

should i just copy and paste the original message and explain where we're currently at?

[unborn cradle]

I would rewrite to be a bit more concise and more general

[upper laurel]

you should edit the original problem to be how it is so far (and say you edited the message to be updated)

[vague valve]

alright i posted it. not sure if its good but i tried

[spice rose]

2d

answer sheet says 2.8mto3.2m

.closed

[cedar kiln]

@vague valve Has your question been resolved?

[vague valve]

trying to use this to calculate getting at least one specific card (4 copies) but i don't think the percentage is right, im doing something wrong

DECK_SIZE = 40
HAND_SIZE = 4
COMBO_CARD_COUNTS = [Card(4)]

[unborn cradle]

I think that's right

[vague valve]

it says 60% chance

[unborn cradle]

yeah for 1 card with 4 copies in the deck?

sounds about right

[vague valve]

thinking about it now i guess it does make since, its just unintuitive with the mulligans

unrelated but combinatorial structures hasn't made much progress regarding this, is there somewhere else where it would be better to ask?

[unborn cradle]

not sure lol

[cedar kiln]

@vague valve Has your question been resolved?

[narrow wind]

I don't know if it's been stated already, but are we assuming that discarding is done optimally? (i.e. always discard a card if it's not in the target hand and never discard otherwise?)

[vague valve]

yes

[narrow wind]

This assumes that we can't make a decision after each redraw. If we can make a decision after each redraw then we can simply discard the 2 we don't want and then discard the other 2 one at a time if we still don't have our target hand

the deck can contain 40 of the same card, 1 of 40 different cards, and everything in between
I suspect that this means that the probability of drawing any given card is always just 1/40 if the deck has an equal probability of being any possible deck

If that's true then the question is easy

Do we have any constraints on how the deck is constructed?

[vague valve]

im not sure what you mean

also yes, you can make a decision after each redraw

[narrow wind]

Does the deck have an equal probability of being any one of the 40^40 possible decks that we can construct in the obvious way? Is it that except without duplicates? Is it that but without duplicate sets but randomly shuffled?

If it's the first one then the next card in a deck always has a 1/40 chance of being any given card, so the question becomes easy. If it's one of the other two cases then I think the same still applies but I'd have to think about it more

[drifting storm]

$\oint\frac{1}{z}{d}z=i\tau$

[vague valve]

its been a month since the original question has been asked lol

[crisp halo]

https://tenor.com/view/grave-dead-die-run-tour-gif-22471974

[normal thunder]

Maybe no one knows how to solve?

[vague valve]

we need Albert Einstein to be resurrected from his grave for this one for sure

[dense ledge]

did he know probabalistic game theory or whatever this is?

[vague valve]

im sure he could figure it out

[unborn cradle]

I could probably solve this if I had someone else to collaborate with

so far my attempted solution failed because of a false assumption

[vale ore]

<@&286206848099549185>

[weary ingot]

@vale ore this channel is occupied, please look in #❓how-to-get-help

[random shale]

That's a IMC Question lol

[random shale]

#competition-math

[nocturne gazelle]

wut

that is?

an IMC question?

[random shale]

For anyone trying to find the question

[random shale]

International Mathematics competition

Basically the IMO for university students

[nocturne gazelle]

ik what it stands for i just didnt know that the question was from IMC

[random shale]

This doesn't seem too bad for each set of cards tbh

[nocturne gazelle]

theres supposed to be 52 cards

[random shale]

Yeah 2024 IMC, literally released like 2 days ago?

[nocturne gazelle]

oh i havent checked that one yet

[random shale]

The difficulty for the first 3 on both days was lower than previous years tbh

[nocturne gazelle]

oh right

[random shale]

But the final 2 on both days were harder, good questions tho

[nocturne gazelle]

they seperate 6 questions in 2 days and give like 3 hours or smth

each day

[random shale]

10 questions in 2 days and 4 hours each? I think

[void spoke]

Limit sum?

[vague valve]

math collab?

[random shale]

@vague valve

The question is probably not easy to generalise into a nice form, like doing it for a regular deck of cards (excluding J,Q,K) it doesn't seem too bad but for generic cards it'll depend on many factors

[void spoke]

U all talking bout this question?

[random shale]

Nah this wasn't his channel

This

[random shale]

This question

||Spam log rules to simplify as a sum minus a log||

[void spoke]

The pic one right?

Oh this

[random shale]

Yea

[unborn cradle]

I just mean if someone else could work with me on it

[vague valve]

sorry that i cant be much of help

[unborn cradle]

We don't need a nice form. We just need a form and an algorithm can compute it.

[worldly chasm]

There is a very large amount of scroll back on this thread, making it extremely daunting and time consuming to jump in and help.

Can someone (@vague valve ?) give a summary of the work done so far and what open questions are lingering, and perhaps edit them (or a link to the post clarifying everything) into the pinned post? As well as any clarifications to the original question to make it better formed.

[vague valve]

alr i did it

thanks chatgpt

well, i edited the original message

[worldly chasm]

As in you used ChatGPT, or you're suggesting I used ChatGPT?

[vague valve]

nah i used it

[remote nacelle]

had some free time and tried your problem for a toy deck of numbers instead of cards. for clarity, i will state my comprehension of the problem statement. we have a shuffled deck of cards of size $n$ composed of $n_d$ distinct cards each of which appear $n_i$ times in the deck so $\sum_{i=1}^{n_d}n_i=n$. the player is aware of the $n_i$ i.e. the composition of the deck. the player gets a random sample without replacement of $k \leq n$ cards from the deck to form his hand. he can then choose to redraw or not redraw from the deck each card in his hand once one by one (this mean he can adapt his strategy after each single redrawn card). we want the probability of getting a certain desired hand after all the redraws have been done. the player is assumed to play optimally choosing the redrawing strategy that maximise the probability of obtaining his desired hand after each redraw. here is a table for the deck 122333 when the desired hand is 123. as existentialistic found, the distribution of the distinct hands before the redrawing process is a multivariate hypergeometric distribution. then i assumed the redraws were done simultaneously but as was clarified by troposphere and confirmed by axerity in #combinatorial-structures the redraws are made one card at a time making my little table obsolete or at best an underestimate of the probability. nonetheless here it is. feel free to verify it.

[wraith dagger]

pola_touche
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[unborn cradle]

@vague valve To clarify we can mulligan cards one at a time right? And decide whether or not to discard based on what we redraw?

[unborn cradle]

Your understanding seems correct, I have checked the picture yet

[unborn cradle]

My thought was we could adapt the formula used on this website to account for redraws: https://deckulator.appspot.com/

Here's a sample calculation:

The bottom card should be called "card 4" but we get the same formula

[vague valve]

yes

[remote nacelle]

and when a redraw is done, is the replacement card and the rejected card drawn and put back in randomly or is replacement card drawm from the top of the remaining deck and the rejected card put at the bottom? this can complexify the problem even more because you could deduce information on the next drawn card if the cards are not redrawn and put back randomly from the remaining deck.

[vague valve]

the cards you reject do not go back into the deck, and the redrawn cards are drawn from the top of the remaining deck

[remote nacelle]

ultimate pain

[vague valve]

why would that be an issue?

[remote nacelle]

because i just did an example with a bad assumption again

[vague valve]

whoops

well i updated the original question again

[unborn cradle]

These details are important man lol

[vague valve]

i didnt know they would be lol, i was trying my best to keep it from being an essay

[upper laurel]

minor typo in there, try rewording the first two sentences to "In a randomly shuffled 40-card deck, you draw 4 cards at the top to be your starting cards. You have the option to use a mulligan, where you discard a starting card in your hand and redraw another card. Mulligans do not need to be done all at once."
that way you include the option to do another mulligan of a different starting card if your first mulligan fails to get the card you want

[vague valve]

alright edited

thanks

[unborn cradle]

@remote nacelle I think this is the general formula. If someone could confirm and possibly help me type in LaTeX that would be great.

where

n = deck size
k = hand size
ni = #copies of card i
ki = #minimum desired copies of card i
(only copies with ki > 0 are indexed)

p = #distinct desired cards

Should be Cp = kp in the left sum

[unborn cradle]

$\sum_{C_p = k_p}^{n_p}$...$\sum_{C_2 = k_2}^{min(n_2, k - \sum C_{i > 2})} \sum_{C_1 = k_1}^{min(n_1, k - \sum C_{i > 1})}$ $\frac{\binom{n_1}{C_1} \binom{n_2}{C_2} \dots \binom{n_p}{C_p} \binom{n - \sum n_i}{k - \sum C_i}}{\binom{n}{k}}$

[wraith dagger]

Existentialistic

[unborn cradle]

Could someone confirm whether this is the correct probability or not? (without mulligans)

n = deck size
k = hand size
ni = #copies of card i
ki = #minimum desired copies of card i
p = #distinct desired cards

[worldly chasm]

I am probably missing something subtle, but if we are able to mulligan only a set number of times, and we do not need to mulligan if we do not want to, and the rules are accommodating enough that we don't ever need to make a suboptimal draw, and you draw n cards with m mulligans, then wouldn't the question reduce down to "do the top m+n cards contain a set of the desired cards?"

[unborn cradle]

No

#combinatorial-structures message

[unborn cradle]

Do you know if my formula is correct?

[worldly chasm]

Ah, the subtlety I was missing is the rule of the mulligan is you can only redraw a specific card in your hand only once. It's as if you have n slots for cards, and if you have m mulligans, m ≤ n, and once you replace a card in a slot via a mulligan that slot is now locked.

[upper laurel]

each mulligan is per-card, the instrctuions were subtly hinting to this by implying that mulligans only apply to starting cards

axerity has had to edit the question more times than there are valid approaches to the problem

[worldly chasm]

Are you double counting here? For instance the hand 122 where the desired cards are 12 would be counted in two ways, first by counting the unitalicized 2 in the sum with C2, and the italicized 2 in the sum with the leftovers, and then vice versa?

If you aren't overcounting it though, then you might not be counting it at all. This seems like it needs inclusion exclusion.

[unborn cradle]

what are n, k, ni, ki in this example?

[worldly chasm]

I'm trying to conform to your use of variables

Oh

[unborn cradle]

what's the difference between 2 and 2 lol

[worldly chasm]

n is larger than 3, k is 3, p is at least 2.

They're normally indistinguishable 2s that I made distinguishable for the purpose of talking about them individually

[unborn cradle]

I think this term guarantees we don't overcount

it's basically how many different ways we can draw the cards we don't care about

where k - sum Ci is the remaining hand size

[worldly chasm]

Yes but in the hand 122 where we need 12 then the extra 2 isn't really a number we don't care about

[unborn cradle]

There will be separate cases where C2 = 1 and C2 = 2

[worldly chasm]

I get that, what I'm fearing might happen is in the case C2 = 1, we will also count the second 2 in that final term

And I'm not seeing how that is avoided?

[unborn cradle]

we won't, because we're doing n2 choose C2

for that card type

we're not doing n2 choose k2

we're doing n2 choose C2

[worldly chasm]

The final term

The "leftover" one

[unborn cradle]

the final term subtracts all the 2's

n - sum n_i

[worldly chasm]

Ok, I see it now

[unborn cradle]

Sorry I'm not quite understanding your counter-example

ok cool

[worldly chasm]

In that case I do not have a quibble

And I thought the use of min was clever. Though I probably would have ordered the summations the other way around

[unborn cradle]

I stole it from the website to be fair

Just had to reverse engineer the formula and I changed the notation slightly

[unborn cradle]

It's funny you mention inclusion-exclusion because that was my first approach and I think I significantly overcomplicated it

[worldly chasm]

I'm happy to be wrong. I frequently overcomplicate things

[unborn cradle]

I wrote code to put an arbitrary expression of unions and intersections into disjunctive normal form and then count the size via inclusion-exclusion

@worldly chasm I think there is a problem with my formula though. It doesn't work when we care about all the cards and the combo size is less than our hand size. I think we need Cp = max(kp, k - sum(ki))

[worldly chasm]

Oh true

Good point and good find

[unborn cradle]

but we only need the max when we have no extra cards...

[worldly chasm]

And I agree that's probably the most straightforward way to repair it

[unborn cradle]

uhhh

or we might have some extra cards but not enough to fill the hand 😡

Cp = max(kp, k - sum(ki) - (n - sum(ni))

I think

[worldly chasm]

So if we have 1, 2, 3 of cards of type a, b, and c, and we want 1 of each, and we have 1 card of type d, and our hand is 5 big, then we have a situation where C1 = C2 = 1, then when we get to C3 we will have max(1, 5 - 3 - (7 - 6)) = max(1, 1)

Instead of sum(ki) I think you mean sum(k(i<3))

@unborn cradle ^

[unborn cradle]

yes I think so

that's right

k_(i < p)

or I could write it like this

Cp = kp + max(0, k - sum(ki) - (n - sum(ni))

@worldly chasm I think that's more clear, the second term is the correction term

[worldly chasm]

That does make sense

[unborn cradle]

$\sum_{C_p = k_p + max(0, k - \sum k_i - (n - \sum n_i))}^{n_p}$...$\sum_{C_2 = k_2}^{min(n_2, k - \sum C_{i > 2})} \sum_{C_1 = k_1}^{min(n_1, k - \sum C_{i > 1})}$ $\frac{\binom{n_1}{C_1} \binom{n_2}{C_2} \dots \binom{n_p}{C_p} \binom{n - \sum n_i}{k - \sum C_i}}{\binom{n}{k}}$

that's right but it doesn't fit on one line lmfao

[wraith dagger]

Existentialistic

[eager trail]

oh ye so like 1+1 squared is 4

im very smart

how do isolve this

[unborn cradle]

Please use an unoccupied channel

[eager trail]

oh yea my bad

[remote nacelle]

idk if this is correct, i would need to parse it for longer, but from what i understand the complex sum on top of this expression is due to the case when the player want a certain combination of cards $\sum_{i=1}^p k_i$ in his hand less than the hand’s total size $k$. when i said that the distribution for the hands is a mvt hypergeometric distribution, i assumed that the player wants exactly $k$ cards i.e a specific hand not a certain subset of desired cards in the drawn hand.

[wraith dagger]

pola_touche

[remote nacelle]

my question is if this complexity is really required. what does the player really want. an exact 4 cards hand or certain desired multiset of cards in his hand? @vague valve

[vague valve]

im not sure what you mean by certain desired multiset

[remote nacelle]

well let’s say you have the hand 1223 from some deck containing the cards 1,2, and 3. does the player wants 4 specific cards with repeats or can he wants to have only subsets with repeats for example 122, 223, 123 or 111 in his 4 card hand. a multiset is a mathematical set with repeats basically.

[vague valve]

so you want to calculate if a player has multiple desired hands?

[remote nacelle]

no i want to clarify if the cards the player wants at the end of the mulligans are a subset with repeats of the full hand

like can the player want at the end to have the cards 1 2 2 in his hand when his hand will contain 4 cards

[vague valve]

if im understanding it correctly then it depends on how many cards are in the combo

i think the answer is yes

like if a player wants 122 and they draw 1322 then the combo has been achieved

[remote nacelle]

said another way, the player’s goal is to get a certain target “sub-hand” in his full 4 card hand after the mulligans

this makes the problem yet again more complex

[vague valve]

if the hand is 4 cards and the combo is 4 cards then it wont matter right

[remote nacelle]

yeah this extra complexity happens if the combo is less than 4 cards with the 4 card hand

[unborn cradle]

We don’t need it to be simple, we just need it to be computable in a reasonable amount of time

[remote nacelle]

then we could simulate it with a computer to estimate the probability like in R or python. that wouldn’t be a formula or the exact probability, but it could approach it to a given precision with enough simulations.

[unborn cradle]

Yeah that might be good for verifying if we have the correct formula or not

Or even just a good approximation

[remote nacelle]

though the simulation will probably assume the player is using a certain redrawing strategy, not THE optimal redrawing strategy. unless there is a way to computationally search for the optimal redrawing strat.

[unborn cradle]

We can redraw one at a time though

And decide based on what we receive

[remote nacelle]

yeah i guess, my worry is that the optimal strategy might not be unique and that it will be something too complex (a lot of cases) for a normal monkey to execute. like something akin to a chess engine. it would be nice if a simple optimal strategy exist. for example something similar the one proposed by troposphere in combinatorial structure seem pretty good. only discard for redraw the cards that are not in your desired sub-hand or that are in your subhand but more than the number of times you want that card.

[unborn cradle]

Yeah I suppose we don’t know which card to redraw first

Wait no it doesn’t matter

Just redraw what we don’t need

Until we can’t redraw anymore

[vague valve]

i imagine it would get complicated if you had multiple desired hands though

[remote nacelle]

we can test it out with a simation but we have no proof it’s optimal

[unborn cradle]

I’m pretty sure the proof is trivial

I can try to articulate it

[unborn cradle]

Let x be the size of the desired multiset - the size of the largest subset of the multiset in our current hand. If there are multiple largest subsets, cards which cannot be discarded (i.e. were not in our starting hand) will be prioritized over ones that can. Clearly x ≥ 0, and our goal is to get x = 0. Discarding a card not in the subset can never increase x, so it will only benefit us. Discarding a card in the subset can never decrease x, so it will never benefit us. So the strategy is optimal.

[remote nacelle]

can you apply this to the deck 122333 with target 123 when the hand that was drawn before the redraws was 223? i just want to see the proof’s logic in a concrete example.

[unborn cradle]

We discard one of the 2's. If we get a 1 we're done. If we get a 2 we discard the other 2. If we ever get a 3 we discard the 3 in our starting hand.

It's important that cards not in our starting hand are prioritized in the largest subset

which frees up some cards to be discarded

[unborn cradle]

Does that clear things up?

[remote nacelle]

i’m unsure but yeah that seem to work you basically show you can’t do better than that strat

what was making me unsure is the fact that the strat can fail, but i guess it fail less time or as frequently as any other strat.

[unborn cradle]

Isn't that the definition of "optimal"?

[remote nacelle]

it made me think of an upper bound on the probability of getting the desired subhand. once you draw your hand, then you basically have to find the probability that the k top card of the remaining deck contains at least the card(s) you still need for your desired subhand.

then it’s only a matter of having a strategy that can always get these cards in your hand to show that it’s optimal but your proof seems better

or determining when an optimal strategy can get the remaining cards to compute the exact prob

[unborn cradle]

Yeah I think this approach is backwards. Usually we need to know the optimal strategy to determine the probability of success.

[neat raven]

axerity vs gwad 0 ravivar

[vague valve]

the what now

[raven shard]

they were making a joke
both you and the person they mentioned have help channels older than a month

[vague valve]

oh lmao

[upper laurel]

this can go through smaller cases but is too slow to work for n=40 and k=4 above approximating

[tacit cedar]

Guys

i have problem. I have 4 numbers 0,2,10,42. thats a series. what is a pattern here?

[weary ingot]

please open a new channel! #❓how-to-get-help

[fossil lava]

i’m guessing it’s that every term is an odd power of 2 plus the previous term. so:
0
2 = 0 + 2^1
10 = 2 + 2^3
42 = 10 + 2^5

and i guess that would make the next term
42 + 2^7 = 170

[amber holly]

Could someone explain this?

Why did x + 30/a get removed?

[pallid moon]

is there more context

if this is all the context they probably just set k_1 to be this constant

so plugging in gives $y=k_1x-k_1/2$ which is much simpler

[wraith dagger]

buboblakistoni

[inner garden]

I agree that more context is needed, but it looks like a derivative problem. The way I learned it is you kind of reduce the variables. -60/a x is actually -60/a x^1, and 30/a is actually 30/a x^0 (which is 1), it's just usually implied and written the way you've always seen for simplification. When taking a derivative the exponents get subtracted by 1, and I can't quite remember the reason why, but when you take the derivative of a variable that has an exponent of 0 it's like your multiplying the constant by 0. So 30/a (x^0 which is 1) turns into 30/a * 0, which effectively "cancels it out". Then -60/a x^1 becomes -60/a x^0 (again which is 1) therefore -60/a. Thinking about it like this usually works for simple problems like yours. Not sure why they changed y to k1 though, usually when doing derivatives it becomes "prime = ' " if I remember correctly, which would be written y' = -60/a

[remote nacelle]

!occupied

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[remote nacelle]

yeah, if @vague valve agrees, this demo imitating the deck-o-lator site is an appropriate solution to the question. i just want more details on what’s happening in there. like do you check all possible redrawing strats for heach possible drawn hands?

[vague valve]

how does this work

not sure how to use it

[remote nacelle]

you open it on a browser on a computer, works for me

[vague valve]

ok yeah i got that part but like the result isnt useful

3 desired cards 40 in deck no matter what i do it just gives me the max trial number/ max trial number

[crimson sedge]

why dont peopleclose the chats

[crimson sedge]

Maby so people can learn from others questions

[prisma pike]

Because bro isn't done yet💀

[normal thunder]

Yeah... Let them cook...

Hmmm... Is this meme correct?

[vague cove]

https://tenor.com/view/dog-gets-flung-gif-25527373

[restive oriole]

💀

[nova aurora]

this is not a math help anymoree

[nimble basin]

.close

!close

[pulsar pine]

.close

[remote nacelle]

you sure you use it right like it works like the dec-o-lator website. you specify the composition of the desired hand and the composition of the deck. In mtt's implementation it look like this for a deck of 4 cards 10 each with a desired hand of 1 Q 1 j and 1 A

[remote nacelle]

but you can't specify the number of card you draw so yeah it's still useless

in the general case but he can fix it easy i'm pretty sure

[torpid abyss]

.done

[upper laurel]

has the original problem where youre looking for a specific hand instead of a specific subhand already answered

[remote nacelle]

not exactly in that case, but we have a candidate optimal strategy for the redrawing and we can use simulations for estimating the prob. it’s a matter of implementation since the prob of a certain hand before the redraws is a multivariate hypergeometric dist if we are looking for a specific 4 cards hands with 4 cards drawn. it’s not a weird sum of probabilities from that dist as with the subhand version of the problem. from my perspective, what is left to do is to compute for each possible hand the prob that an optimal strat gets the hand you want.

i am hoping this is what you did with your html code, but i did not put the time in to understand/parse it. also i’m not familiar with html. a bit of explaining of the code would be appreciated.

[upper laurel]

you realize the code didnt fall out of the sky, right

I didnt intend to stay in this channel, so the code only answers the original question and nothing else specific to the latest axerity news (nor too close a resemblence to the other calculator website)

the code is only built to answer the original question where you draw your starter cards, take starter mulligans if you want, and ask for the specific probability that the entire hand you got is the hand you wanted
your hand size is from adding up the # desired

to test this, (hand size) cards are drawn for the initial starter deck, then another (hand size) cards are drawn to prepare for the possible mulligans
if the hand matches what's desired, or neither the hand or the mulligans contains enough cards for the desired hand, then the trial ends

if not, then the code brute-forces through all possible ways to mulligan the starting hand
when (hand size) = 4, there are 4 + 4 * 3 + 4 * 3 * 2 + 4 * 3 * 2 * 1 ways or in general < e * (hand size)! ways to go through

depending on how many trials you want, the code will either:

• randomly draw (hand size) * 2 cards to test as above, then give you an approximate number using that number of trials, or
• go through all possible cases and give you an exact fraction if you have enough trials to allow it
  you need nPr(deck size, 2 * hand size) / (hand size)! trials to go through all possible cases

other than bounding the numbers, I didnt soup up the website to do much more than what it appears to do
it doesnt reduce its trials nor introduce any strategy to really help the original problem

for deck size = 40, hand size = 4, this is 129 billion cases
if the code ran through a million cases per second, it would take almost 36 hours to complete

but writing it off as useless without even reading the code? bold prediction, and its correct

[hot oriole]

https://youtu.be/26ZMgnVKS0c?si=KNTIVzdYNkvCDKNo

[upper ruin]

Why are you spamming this everywhere??

[unborn cradle]

wtf why are there so many hard coded values for n <= 40n

[upper laurel]

correction: those lists contain every hardcoded value for n! and nCr(n, k) so that theres at least one optimized part of the code, at negligible benefit

[unborn cradle]

lmfao

[unborn cradle]

I mean it's not the worst idea it's just funny to look at

[upper laurel]

funny to look at is sometimes why I write code that way

I had to generate that list with a separate js script

[unborn cradle]

Could you give a brief overview on how it works?

[upper laurel]

on what?

[unborn cradle]

the code

I don't know html too well

I wrote my attempted solution in python

[upper laurel]

it was already given here, this is the brief overview

do you want a detailed overview?

[unborn cradle]

oh

ok one moment

[upper laurel]

you didnt read it I gather

[unborn cradle]

It's not general enough imo

[upper laurel]

you realize Im not going to work on the code? I just posted it to emulate the original question

just seems a bit off where I post the code and the first thing people say is "it needs to do more"

[unborn cradle]

can you chill lol

I'm not attacking you

it's a good start at least

[upper laurel]

you gotta think through what this means in response to a post like that next time

[unborn cradle]

🆗

[upper laurel]

not going to say sorry or anything?

[unborn cradle]

are you alright buddy?

[upper laurel]

what is that question supposed to mean

[whole totem]

...

uhm

when your done

I need help.. please?

[upper laurel]

you gotta go post in the available channel to get help, this one's dedicated to axerity's question

[vague valve]

i appreciate your help and im assuming its not intentional but you're definitely straddling the line of being condescending

[upper laurel]

how so

[vague valve]

its just that you come off condescending/combative in your messages sometimes

i can reference your messages if you think it would help illustrate my point but i'd rather not

[upper laurel]

go for it, youre repeating yourself

[vague valve]

accidental reply hold on

[upper laurel]

not a big deal, keep going

[vague valve]

#help-13 message
its pretty clear here that existentialistic was just trying to give feedback. i can understand being annoyed about people asking you to do more coding when you don't want to, but saying things along the lines of "you should've thought about what you were saying before you said this" / asking for an apology over something like this is an overreaction

#help-13 message
i misunderstood what you were asking of me here, but asking to "bear with me" when explaining to me how to edit a message (which is very simple, so saying this is implying i'm stupid), assuming i'm refusing to do what you're asking first instead of me just misunderstanding, just the general verbiage of the messages here convey that you don't respect me

[upper laurel]

neither does this convey respect

we agree that you should reword the message over the course of 5 minutes, so when you then say that, it conveys you didnt read it too closely

so when you say it like that, it sounds like you were getting cocky(?) as if I misunderstood what you were saying

so here I was thinking you were trying to justify your original post instead of agreeing to edit it, said in a cocky manner
thats why I respond back like that, its not meant for if you were just misunderstanding

[vague valve]

that's fair enough as well, and i appreciate you being willing to listen to what i said

[upper laurel]

its also good that even though you said you were just having a Heated Gamer Moment, you still bothered to bring it up here to make sure its 100% resolved

[prisma pike]

Miracle this didn't escalate

[unborn cradle]

@vague valve I wrote some code that approximates the probability via simulation. It won't give the exact answer but it should be close enough for practical purposes.

import random

DECK_SIZE = 40
HAND_SIZE = 4
COMBO_CARDS_TOTAL_COMMA_DESIRED = [(4, 1), (4, 1), (5, 2)]
SIMULATION_COUNT = 10**6

def main():
    deck = create_deck()
    success_count = 0
    fail_count = 0
    for simulation in range(SIMULATION_COUNT):
        current_deck = deck.copy()
        hand = []
        for draw in range(HAND_SIZE):
            draw_card(hand, current_deck)

        if does_hand_contain_combo(hand):
            success_count += 1
            continue

        for card in hand.copy():
            if card >= len(COMBO_CARDS_TOTAL_COMMA_DESIRED) \
                    or hand.count(card) > COMBO_CARDS_TOTAL_COMMA_DESIRED[card][1]:
                draw_card(hand, current_deck)
                hand.remove(card)

        if does_hand_contain_combo(hand):
            success_count += 1
        else:
            fail_count += 1

    print(f"Successes: {success_count}, Failures: {fail_count}")
    print(f"Success rate {success_count/SIMULATION_COUNT:.4%}.")
    print(f"Standard deviation: {math.sqrt(success_count*fail_count/(SIMULATION_COUNT ** 3)):.4%}")

def create_deck():
    deck = [[index for _ in range(pair[0])]
            for index, pair in enumerate(COMBO_CARDS_TOTAL_COMMA_DESIRED)]
    deck = sum(deck, [])  #flatten

    non_combo_cards_label = deck[-1] + 1
    for card in range(DECK_SIZE - len(deck)):
        deck.append(non_combo_cards_label)

    return deck

def draw_card(hand, deck):
    choice = random.choice(deck)
    hand.append(choice)
    deck.remove(choice)

def does_hand_contain_combo(hand):
    for index, pair in enumerate(COMBO_CARDS_TOTAL_COMMA_DESIRED):
        if hand.count(index) < pair[1]:
            return False
    return True


if __name__ == '__main__':
    main()```

Currently it runs a million simulations then finds the average success rate, you could lower it to 100,000 if you want it to run faster (replace the 10 ** 6 with 10 ** 5)

oh I made a mistake

ok fixed

[proud hemlock]

.close

[snow escarp]

fr

[vague valve]

thanks a bunch! just to confirm you put (number of cards in deck, card type) for the COMBO_CARDS_TOTAL_COMMA_DESIRED variable?

[unborn cradle]

(number in deck, number you want)

[vague valve]

ah ok

should i keep the channel up? it would be nice to get an actual formula for this instead of approximations but as you said this should be good enough for what i'll use it for

[unborn cradle]

import random
from collections import namedtuple

DECK_SIZE = 40
HAND_SIZE = 4
COMBO_CARDS_TOTAL_COMMA_DESIRED = [(4, 1), (4, 1), (5, 2)]
SAMPLE_SIZE = 10**6

COMBO_CARD = namedtuple("COMBO_CARD", "total, desired")
COMBO_CARDS = [COMBO_CARD(total, desired) for total, desired in COMBO_CARDS_TOTAL_COMMA_DESIRED]

def main():
    starting_deck = create_deck()
    success_count = 0
    fail_count = 0
    for sample in range(SAMPLE_SIZE):
        current_deck = starting_deck.copy()
        hand = []
        for draw in range(HAND_SIZE):
            draw_card(hand, current_deck)

        for card_number in hand.copy():
            if card_number >= len(COMBO_CARDS) \
                    or hand.count(card_number) > COMBO_CARDS[card_number].desired:
                hand.remove(card_number)
                draw_card(hand, current_deck)

        if hand_contains_combo(hand):
            success_count += 1
        else:
            fail_count += 1

    print(f"Successes: {success_count}, Failures: {fail_count}")
    print(f"Success rate {success_count/SAMPLE_SIZE:.4%}.")
    print(f"Standard deviation: {math.sqrt(success_count*fail_count/(SAMPLE_SIZE ** 3)):.4%}")

def create_deck():
    deck = [[card_number for _ in range(combo_card.total)]
            for card_number, combo_card in enumerate(COMBO_CARDS)]
    deck = sum(deck, [])  #flatten

    non_combo_card_number = len(COMBO_CARDS)
    for card in range(DECK_SIZE - len(deck)):
        deck.append(non_combo_card_number)

    return deck

def draw_card(hand, deck):
    choice = random.choice(deck)
    deck.remove(choice)
    hand.append(choice)

def hand_contains_combo(hand):
    for card_number, combo_card in enumerate(COMBO_CARDS):
        if hand.count(card_number) < combo_card.desired:
            return False
    return True


if __name__ == '__main__':
    main()

This is a slightly cleaner version

It should give the same output it's just organized a bit better

[vague valve]

unrelated but i made some likely gut wrenching code in terms of readability but its the first time i coded anything if you want to see it

[unborn cradle]

sure, for this problem?

[vague valve]

no, for something else

[vague valve]

?

[unborn cradle]

Not sure

[unborn cradle]

sure I can take a look

[slim anchor]

hello

byebye

[crimson sedge]

how has axerity come above palaeeudypte?

[void sand]

I wonder why too

[wet lynx]

Hey can someone hop on a call with me and help me with my geomtry howmework

pls

[wary iron]

Do you have a specific problem you're stuck on?

[wet lynx]

@wary iron Can you help please

[wary iron]

Think about what RT is in terms of RS and ST

[wet lynx]

I got this already

can you help with this

[wary iron]

One sec

[wet lynx]

k

[wary iron]

collinear means that all the points are on the same line

we are given that PQ is congruent to RS which in simple terms just means they are the same

we also know that PS = 21 and PR = 17

[wet lynx]

ok

what is qr

[wary iron]

my drawing is kinda bad but what additional information do you think u can figure out from the givens

[wet lynx]

what is the value of QR

[wary iron]

i can help you figure it out but im not gonna give you the answer

[wet lynx]

they already gave us the anwsers for ps and pr

ok

[wary iron]

what other line segments do you think you can solve for?

[wet lynx]

Idk

this is my first week in geomtry

[wary iron]

well if we have PS and PR we can probably figure out RS right

[wet lynx]

ok but we are looking for QR

[wary iron]

we cant solve for QR quite yet

[wet lynx]

ok

[wary iron]

if we know other lengths we can figure out QR

[wet lynx]

how do we start

[wary iron]

can you agree that PQ + QR + RS = PS

[wet lynx]

yes

[wary iron]

and we are given that PS = 21

so PQ + QR + RS = 21

[wet lynx]

yes

[wary iron]

and we are given that PQ = RS

so PQ + QR + PQ = 21

aka 2*PQ + QR = 21

[wet lynx]

ok

[wary iron]

are you still with me

[wet lynx]

no

😦

[wary iron]

since PQ = RS we can just replace RS with PQ

[wet lynx]

ok

[wary iron]

2PQ + QR = 21
and
2RS + QR = 21

both work

i just chose PQ cause why not

so if we can figure out PQ or RS then we can figure out QR right

[wet lynx]

right

[wary iron]

do you have any ideas on how we can solve for PQ or RS?

[wet lynx]

No

[wary iron]

well can you agree with me that PR + RS = PS

[wet lynx]

yes

[wary iron]

and we are given that PR = 17 and PS = 21

so do you think you can solve for RS?

[wet lynx]

No I dont get this im getting frustrated

[wary iron]

what are you confused about

[wet lynx]

Idk I dont understand anything

[wary iron]

you're fine up to here?

or are there previous steps you didnt quite understand

dont be afraid to ask questions even if they sound dumb

[wet lynx]

I understood for a sec and now I dont

[wary iron]

my drawing is pretty bad to be fair

lemme try drawing it a bit better

[wet lynx]

how does PR+RS=PS

[wary iron]

all of these points are collinear

which means that you can draw a line and it will pass through all of them

[wet lynx]

ok like a square with a x in the middle

[wary iron]

based off my drawing R is clearly between points P and S

[wary iron]

what do you mean by this

here i think this is a better visual representation

[wet lynx]

oh ok

So a straight line with multiple points

[wary iron]

im not really sure what you're trying to say but i dont think its correct

[wary iron]

yes

based on the drawing do you think you can find RS

[wet lynx]

yes

its 4

Right?

[wary iron]

yes

[wet lynx]

finally

[wary iron]

and we are given that PQ = RS

so what does PQ equal

[wet lynx]

4

[wary iron]

so now do you think u can solve for QR?

[wet lynx]

13

[wary iron]

nice

[wet lynx]

yessssssssssss

That feels so good to understand you are great

[wary iron]

no problem

.close

[wet lynx]

What about this

[wary iron]

i gotta do my own homework man sorry

[wet lynx]

oh ok

no problem I got 2 problems left I think I can do it myself

thank you so much

[vague valve]

my poor channel

[void sand]

[fallow apex]

What's the ques ?

Also channel can't be poor or rich

[vague valve]

pinned

[fallow apex]

nah i ain't doing any more combi enough combi for 2 months already

[dire geode]

@vague valve can you just move your question to stack exchange. Help channels aren't meant to be open for this long

[coral jewel]

[warm island]

hi

[dire geode]

.close

[cedar kiln]

Channel closed

Closed by @dire geode

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[steel cave]

can someone walk me through this?

[steel cave]

I know it’s simple but I’m getting hella brain fog and need some sense of direction

[cedar kiln]

@steel cave Has your question been resolved?

[soft rock]

just plug the points into the graph

and if its true then the point lies on it

[steel cave]

okay thank you I figured it out

.close

[cedar kiln]

Channel closed

Closed by @steel cave

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
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[vocal drum]

Can someone help me with my assignment?

[vocal drum]

<@&286206848099549185>

[slate lintel]

cmon you could have waited for at least a few minutes

[vocal drum]

I'm sorry, it won't happen again.

[void spoke]

Fear shall be known -hayley💀

[fallen moat]

Hello there Trash Boat!
Do you know how to start with Q1?
like have you done anything on the question?

[raven shard]

(fire mordecai & rigby reference btw)

[vocal drum]

Yeah I did

[vocal drum]

Thank bro

[fallen moat]

reading

[vocal drum]

And I don't know how to go further

[fallen moat]

i see,
if it's (-4)+5 that means
negative 4 plus 5

but not(-4)×(5)

[vocal drum]

Ohhhhh I thought im going to multiply them

Alr hang on

[fallen moat]

got it?

[vocal drum]

Like this? And continue it on the rest?

[fallen moat]

yea, and then the same problem with -3+(-5)

[vocal drum]

Ohhh alr alr ok brb

[cedar kiln]

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@vocal drum Has your question been resolved?

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[lost thunder]

why detA=-2
A is nxn, n is natural

[lost thunder]

.close

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[spring coyote]

guys im not sure if this correct

chatgpt say 1 thing, textbook notes other thing

i guess its not and its actually 0/6

= 0

[vagrant elbow]

It is 0 yeah

Be careful when using ChatGPT, it is general not very good at this sort of stuff

Use WolframAlpha if you just need an answer

[spring coyote]

wow

its anotehr AI?

[vagrant elbow]

Eh I don't think it's AI

But it's useful

,w limit as x goes to infinity sqrt(6x^4 - 6x^2 + 2022)/(6x^4 + 6x^2 + 2022)

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by neonperseus

[vagrant elbow]

There might be some sort of clever factorisation that can solve this but what you can also do is simply analyze the dominant terms in all the polynomials

[wraith dagger]

ananas

[spring coyote]

yes

very nice

[vagrant elbow]

Some people aren't comfortable with it

But since you are that's the best way to solve this

[spring coyote]

Woah

it actually kinda

turns text into

but not as precise as chatgpt

[vagrant elbow]

Yeah it's more traditional, you can't really talk to it like a human

Try to be as simple and precise as possible

[spring coyote]

yes yes verynice

[honest zodiac]

who can help to find the determinant?

[cedar kiln]

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[crude gale]

I'm not sure how this shows that Z[w] is an integral domain, could anyone help

[crimson delta]

what are the properties of an integral domain

note that Z[omega] is a subset of C

[crude gale]

no zero divisors so (a+bw)(c+dw)=0 implies one of them must be 0

[crimson delta]

as those are both also complex numbers what do we know

[crude gale]

Oh it follows

Because complex numbers are a field

[crimson delta]

yes

[crude gale]

Which imply integral domain

[crude gale]

Ohh so they are really just trying to show closure under multiplication here?

[crimson delta]

yes. the other properties either follow from C or are trivial

like additive closure/inverse

[crude gale]

Ahh okay get it get it, thanks alot

[crimson delta]

would be good to make a complete list

and check each of them of as trivial or following from C

just as an exercise

[crude gale]

For subring part, I just have to show

Closure under subtraction, closure under multi and non-zero right

[crimson delta]

nonempty. yes

[crude gale]

Oh yes nonempty*

Aight aight thanks alot got it

.close

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[coarse blaze]

THanks :D

Hello, my professor went really quick over this exercise and this is what I understood

I am not sure this is enough proof

can anyone explain why or why not this is enough?

[obtuse mango]

roughly speaking that's the idea but the thing u've written in the middle doesn't make sense

(lim x->infinity of x)

[coarse blaze]

I meant n, not x

in both three limits

[obtuse mango]

ah right

[crimson delta]

I hope its supposed to be an intersection sign

[obtuse mango]

you could write it like that but i wouldn't

[coarse blaze]

It is :D

[obtuse mango]

oh yeah whoops i just glanced over that lol

[coarse blaze]

yeah it was

it is an intersection

hopefully no more mistakes :)

[obtuse mango]

it's more that let's say there was a number x in our intersection

then -1/n <= x <= 1/n for all pos. ints. n

so we can take the limit of the LHS and RHS

(in some sense you could say we're taking the limit of x as n tends to infinity but this is kinda meaningless so i really wouldn't)

[coarse blaze]

So we would say

for every x inside of the interval it must be true that

[crimson delta]

I would phrase it from the other perspective

[coarse blaze]

-1/n <= x <= 1/n for all natural n

[crimson delta]

let x be nonzero. then there exists n with 1/n<|x|, so x is not in [-1/n,1/n]. and then also not in the intersection

[coarse blaze]

we take limits and we arrive to the conclusion that 0 <= x <= 0 so the only x in the interval is 0

[coarse blaze]

That's how I had done it, actually. She said it was too much work

[crimson delta]

lol

[coarse blaze]

I have a slight problem with this

let's say x <= 1/n for every n>0

Are we sure we can keep the <= while doing the limit?

My intuition says we shouldn't, but 0 is a correct value of x

[crimson delta]

its the squeeze theorem

[coarse blaze]

Ok! I'll read about it

thanks @obtuse mango and @crimson delta

:D

.close

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[coarse blaze]

.reopen

[cedar kiln]

✅

[coarse blaze]

,rotate

[coarse blaze]

In this case, it's obvious that we can't use <=

how would I justify the fact that we can't use it here but we could previously?

(I understand this isn't the squeeze theorem since we aren't squeezing towards x)

I'm missing the big U before the intervals, its

$$ \bigcup_{n\ge 2} [1/n, 1 - 1/n] $$

[wraith dagger]

dp

[crimson delta]

well the point was, before it needed to be in the interval for all n

now just for a single one

[coarse blaze]

Yeah, but if

$$ 1/n \le x \le 1-1/n $$

[wraith dagger]

dp

[coarse blaze]

then that x belongs

why can't we carry the <= over when we do the limit

[cedar kiln]

@coarse blaze Has your question been resolved?

[coarse blaze]

<@&286206848099549185>

(sorry for accidental ping, @ helper)

Hi, you have to use another channel

for example, #help-0

this one is still in use

no worries :D

[cedar kiln]

@coarse blaze Has your question been resolved?

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[calm wharf]

hello

[lusty pawn]

yeah question?

[idle gulch]

Hi

[calm wharf]

one sec

posting

the question

and the fig.

how tf do i go this?

[idle gulch]

Consider angles 🙂

[calm wharf]

i get that

but how to formally write the answers

in step by step process

[idle gulch]

Like this

[lusty pawn]

sum of opp angles are 180 in cyclic quad

[idle gulch]

Now you will get sum of three angles of triangle 180 ° thing

So this way figure out , F, H,G,E

You could get the desired thing

[calm wharf]

ooooo

thankkkkssss

[idle gulch]

Remember that the

[lusty pawn]

is that ncert?

[idle gulch]

Four angle internals you consider

Will sum to 180 °

[calm wharf]

yeahh

ig my friend send it

[lusty pawn]

which class?

[calm wharf]

9

[lusty pawn]

oh

9 ncert was a disaster

[calm wharf]

thankssss

[calm wharf]

yeah ni shit

[calm wharf]

u indian

[lusty pawn]

yeah

[calm wharf]

:))

[lusty pawn]

11th

[calm wharf]

niceee

[lusty pawn]

not - so - nice

[calm wharf]

now to wrap this up

thanks @lusty pawn @idle gulch

.close

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[gilded pilot]

Hi. Can someone please help me prove this?

[gilded pilot]

This is what I've done so far.

Usually I'm able to find such N for a given epsilon, but I am a bit confused by r^n here.

[flint plinth]

what sorts of theorems do you know? for example, do you know that a_n --> 0 if and only if |a_n| --> 0? and do you know that a decreasing sequence of positive numbers has a limit?

[gilded pilot]

Yeah, I would like to get rid of the absolute value signs, am I on the right track?
Yes, a decreasing sequence has a limit in R (if all the terms are positive of course).

I was thinking, maybe I can use the fact that r is less than 1 and more than 0

[flint plinth]

show that |r^n| is a decreasing sequence

and then there's a nice trick

[gilded pilot]

so like, its powers are going to be less than r for sure

[flint plinth]

the sequences |r|^n and |r|^(n+1) have the same limit

call it L

[gilded pilot]

ok

[flint plinth]

so |r|^n -> L

and |r|^(n+1) -> L

but notice that |r|^(n+1) = |r| |r|^n

so |r|^(n+1) also converges to |r| L

which shows that L = |r| L

see what you can do with that

[gilded pilot]

Can we write this in epsilon notation? So , prove that r^n is convergent using epsilon, considering that it is in (-1, 1)
I understood the part where |r^n| and |r^(n+1)| converge to the same L))

Sorry if I am asking dumb questions I just got started with this topic

basically I would have to also prove this in that case

But intuitively I do get how it must have a limit, cuz r^n > r^(n+1) > r(n+1+1) ... etc.
is the limit 0 btw?

yeah it is

how do we prove it though? we need to find at least one natural N for every epsilon, am I right?

can I express N in terms of epsilon? if we do that the proof will be completed I think @flint plinth

<@&286206848099549185>

N must be more than log(r)(epsilon)?

and we need to ensure it is positive

sth like this?

so if I got * at least one N that works for all epsilon, the proof is complete right?

also idk to get the smallest possible natural N for which all a(n), n>=N are in the epsilon neighborhood, should I round it down? yeah I think I should

anyone?🥹

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[slim mountain]

dont redeem it

[cedar kiln]

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[restive oriole]

my math teacher gave me the definition that "A cylinder is a surface that consists of all lines parallel to a given line and passing through a given plane curve". Does this mean that, if I say draw an "s" shape or some other squiggle on the xy-plane and extended that through the z-axis the resulting surface would be considered a cylinder?

[hollow trail]

under that definition something like this is considered a cylinder, yes

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[errant marten]

Please help me to factor this polynomial.

[rugged pebble]

You can think a*b = -3 and a + b = -5

Sorry 2(a + b) = -5

[errant marten]

.close

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[young coyote]

Im struggling with this notation problem, I know its (-inf,-1)U____ but I dont understand how to put the other part in

[plain ridge]

(-1, 2)?

which literally just means "from -1 to 2"

[plain ridge]

oh i should ping

[young coyote]

thank you so much, I was hard overthinking that

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[ruby beacon]

Help my teacher has created some concept on creating nth step diagrams for visual patterns and I'm so confused

[ruby beacon]

i can't find any information on nth step diagrams online

[digital cliff]

would you mind showing an example

[ruby beacon]

umm so basically he draws like the core of a pattern so like the constant

then he visually adds where the pattern adds from

i understand basic algebera with patterns and stuff but i dont understand the diagrams

!close

nvm

ill figure it out

/close

!end

.close

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[digital cliff]

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[bold pond]

$4x + y^2 = 12, x = y$ how do I set up the integral for the area between the curves?

[wraith dagger]

wakamole

[dusky peak]

find their intercepts first

that gives you the bounds

as you can see here, the area highlighted here is exactly the area below curve 1 minus area under curve 2

this is true when (in the relevant interval), one function is always >= the other (which it is here)

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[sudden mirage]

Can someone help me with this

[cedar kiln]

@sudden mirage Has your question been resolved?

[sudden mirage]

<@&286206848099549185> (sorry for the ping)

[lavish void]

‼️

wdym you need help

vertical stretch multiply x values by recirpical of 3/2

then reflect and translate

[sudden mirage]

Why reciprocal

Nvm dumb question

[lavish void]

bc its a vertical stretch

[sudden mirage]

Uhh

[lavish void]

all good

you need the equation

[sudden mirage]

Oh I see

[lavish void]

wait just makung sure if the negative is attached to the number its reflect y right?

[sudden mirage]

Yes

It’s reflected

If it’s attached to b value it’s a horizontal reflect and if it’s attached to the a value it’s a vertical one

If that was what you were confirming

[lavish void]

yes

ok so do you still need help?

[sudden mirage]

Wait it’s

A verticle

Vertical.

Factor

Across the x axis

So it wouldn’t be reciprocaled

That’s only for horizontal

I cant spell

It’s okay, I think I got it tho! Thanks

.close

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[vernal lichen]

can anyone help me

[vernal lichen]

with this

bc i think my teachers wrong

[dire geode]

Where is it wrong

[vernal lichen]

y int

[dire geode]

Yea should be -6

[vernal lichen]

yessir

[dire geode]

Good find

[vernal lichen]

watch me get a point back tmrw

trust

idk how she got that

easiest part

[cedar kiln]

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[winter lynx]

( 1 0 1 )
( 0 1 1 )
( 1 1 0 )

[winter lynx]

im going to find the inverse of this matrix

I get a good result of a 3x3 matrix

but the answe sheet multiply this 3x3 matrix by 1/2

and I dont know where they get 1/2 from

[odd verge]

Determinant

[winter lynx]

how do you get the determinant?

and whats that

[odd verge]

Hmm you will have to learn determinant before inverse

Let's say A is a matrix the determinant of A is written as |A|

[winter lynx]

my lectures havent mentioned them yet

okey, what does the determinant tell me?

[odd verge]

Have you seen something like this before

[winter lynx]

not when it comes to matrices

just vectors

I believe

[odd verge]

Oh ok

You should learn the determinant of matrices and adjoint

[winter lynx]

ive solved inverse of matrices without the determinant so far

why do I need it this time?

or has the determinant been 1 for the other examples?

[odd verge]

Did you do inverse for 3x3

[odd verge]

Probably

Because you multiply by 1/determinant

[winter lynx]

this is how I solved this one

isnt this inverse correct?

even though I dont have a determinant

[odd verge]

What method is this

[winter lynx]

gaussjordan

trying to get the reduced echolon form

if thats what its called

the pyramid with zeros

[odd verge]

I haven't learned this one yet 😔

I only know the inverse=adjoint/determinant

[winter lynx]

ah okey

so how do you get the determinant

[odd verge]

https://youtu.be/3ROzG6n4yMc?si=FItytiUgihk2Oq3e

[winter lynx]

nice

thats a lot of writing to be done

[winter lynx]

I still get the wrong determinant

how do you get 1/2 with just 1's and 0's

[cedar kiln]

@winter lynx Has your question been resolved?

[desert atlas]

If I'm not mistaking the determinant is -1 here

You don't need a determinant for the inverse

[desert atlas]

The first step isn't correct

The matrix should be
1 0 1
0 1 1
0 0 -2

[winter lynx]

I see thanks

[winter lynx]

but the answe is 1/2 * [1 -1 1]
[-1 1 1]
[1 1 -1]

[desert atlas]

And that is not the determinant nor the adj matrix

[winter lynx]

its the inverse

right?

[desert atlas]

The whole thing yes

When you have a number * matrix
It just tells you that each element of the matrix is to be multiplied with that number

[winter lynx]

yeah but why would you need the 1/2

only the last row needs to be divied by 2

unless you multiply row 1 and 2 with 2, then it makes sense

[desert atlas]

1 2
3 4
5 6
7

I did it in that order

[cedar kiln]

@winter lynx Has your question been resolved?

[winter lynx]

I see, thanks for the help

.close

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[pliant apex]

What does TrendLine mean in charts?

[cedar kiln]

@pliant apex Has your question been resolved?

[dire geode]

depends on context, but could be
https://www.mathsisfun.com/data/least-squares-regression.html

[pliant apex]

?

[dire geode]

what's your confusion about least squares

[pliant apex]

???

[dire geode]

can't help if you don't use words. good luck

[pliant apex]

fuck you

[hot crag]

um

he's
only trying to help

like he said, we can't exactly help if we don't know what your doubt is

[pastel vault]

<@&268886789983436800>

[wicked mantle]

Why

[limber marsh]

please don't be rude to other users