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It might not be the lack of hats in such a situation that's hazardous, but the lack of clothing well south of the hat region. Or if not hazardous, perhaps adventurous. e ...xactly!
I have a feeling the men without hats are happy to be partying in a dark room. After all, they can dance if they want to, and they can leave their friends behind.
I'm a game developer, and a strikingly similar scenario - and result - came up awhile back when doing a deep dive on some item drop rate adjustments. Imagine you have a monster that drops an item when defeated at a rate of 1 in 100 times and then you defeat 100 of that monster. What's the chance you've gotten at least 1 of that item? Due to the "at least 1", this is easier to count the inverted result of "how many times did you fail to get the item" and repeat 100 times. So: (1-1/100)^100. And then invert that result: 1 - (1-1/100)^100. Giving a result of approximately ~63.4% chance of getting at least once. Generalizing this as n instead of 100, and then letting an n approach infinity, we get the result: Lim n->inf [1 - (1 - 1/n)^n] = 1 - (1 / e)
I remember mentally calculating critical chance for multiple critical chances way back in the Warcraft 3 days and my rough math always seemed to end up near either 33% or 66%
It does indeed sound like a Berlin nightclub, but the calculation was not gender specific, and it could therefore also be applied to any dark room where people remove their hats on entering.
1/e also shows up in another famous math problem, which I'll poorly paraphrase: When dating, what % of the total pool should you check out before committing to one? The answer is 37% of the pool, 1/e.
11:00 just a note: I think here either the sum should be marked with i < j (not i =/= j), or, if written in this way, there should be 1/2 in front of it. You don't want to repeat elements: A1 intersection A2 and then A2 intersection A1. But the result is correct, the non-repeating sum over i < j is equal to n choose 2.
Came here to say this. The n choose 2 fixes this, since it ignores permutations of the choosing, which is why even though the left hand side is wrong, the right hand side is correct. This also means the higher order indices in the summation later on for the inclusion exclusion formula need to be written as i < j < k ... for all free indices from the set, or put another way 1 =< i < j < k ... =< n.
In my head I thought "its definitely something like e, or 1/e", and imagine my surprise when I saw the result! Although, not much of a surprise, whenever probability is involved, e will show up.
Yeah, for the mathematically initiated, it may be far less surprising. but it’s so amusing to take a silly word problem like that regarding hats, and the answer is 1/e 😂
I remember finishing our subfactorials a year ago and I loved them so much. I made a scratch project to plug a number in to give the sub factorial of the number.
I think that if we asked 10000 people the question (for some arbitrary number of hats, like 20) to give a percentage from 0-100 that no-one gets their hat back there would be quite a peak at 37%. Yeah, there would likely be one at 73% too.
Fixed points in this sense are cycles of length 1. The obvious generalisation is to permutations with other minimum (and indeed maximum) cycle lengths. These would be practical things to know.
e and pi are the most interesting numbers like “oh you have a weird value for this problem which no field of mathematics even comes close to?” Plug in e or 1/e or e^2 or the eth root of e or e^pi just keep plugging variations of e and pi and it’ll probably work and if it doesn’t even Euler can’t help you
So, if you need to calculate a subfactorial for some reason and you wish to save a lot of time, just divide the factorial by the number e then round to the nearest integer. It works every time.
Seems like a useful way to calculate the secret santa arrangements where no-one gets its own gift. Thanks for the video, I wasn't sure if it was a joke video at first, but it was pretty interesting. You earned a sub :D
As a fascinating exercise, consider the following: Suppose you help n people receive their own hats by randomly distributing all of them. Then, those with incorrect hats return them to you to randomly distribute again among those still missing hats. What is the expected number of iterations it will take for everyone to have their own hat? (I more or less have a proof for this, but i also have homework due today 😢)
Funny, i didn't know subfactorials were a thing but I'm learning about recursion in programming and this is helping me get more perspective on a specific problem I've been stuck on for two days. Essentially i was trying to figure out a formula to assess all permutations of something and worked out the subfactorial without knowing it had a name, i just noticed it was similar to a factorial but adding.
this was satisfying bc when saw the approximation before it felt so random yet underwhelming (like what, you just multiply the factorial by an essentially constant factor?) but this explains neatly where it came from
Man. I thought I was doing really well with understanding this and then I got to the beginning of the explanation at 6:30. You explain things very well but sadly there is a minimum bar of familiarity with the subject required that I just don't have. C'est la vie!
I didn't realize there was music until I saw these comments. I was listening at a lower volume, and your voice was quite clear, but there was this annoying beeping sound like a reversing truck every now and then. Turned out to be the high notes of the music track starting at 3:50, from which only they were audible to me.
As is obvious from their accurate, lifelike portraits, the men in this diagram are misidentified. They are actually Man 47, Man 312, and Man 14,703. 1. 2. and 3 all died in the pneumonia epidemic of 1919.
At about the 18:55 mark, when reindexing to start at i=0 instead of i=2, wouldn't this change the ending index to (n-2) rather than n? Once we're considering the limit as n goes to infinity, this change no longer matters, so the 1/e result is unchanged.
Hmmm Derangements are actually exactly what I need for my experimentations on creating sudokus. Is there also a way to easily figure out what those derangements are instead of just their amount?
The "key game" parties for tolerant couples were popular in the US in the 70s. The men put their car keys in a pot and each woman drew a key from the pot and drove home with this man to have a little fun together. Different use case, but the math is the same 😊
But they were men without hats And they were dancing And they were friends The men without hats never claimed they didn't "own" hats. They just didn't have any, currently, as is the case with our dancers
1:32 oh god there was a puzzle like this in proffessor layton, you had to figure out how likely it was that 2 people got their hat but one person didnt. I guessed EVERY number from 1-100% only to realise it was 0%
Seems like there's an easier way to count using a recursive definition: let Ai be the set which has *exactly one* fixed point at index i. This means that index i is a fixed point, and the rest is a derangement: !(n-1). Likewise for Bij (the set which has exactly two fixed points). So the number of derangements is going to have the form !n = n! - ((n choose 1) ⋅ !(n-1) + (n choose 2) ⋅ !(n-2) + ... + (n choose n-1) ⋅ !(n-(n-1)) + 1) Then I'm sure we can do some re-arranging.
In case no one else has mentioned it, really weird harmonics in the background music. It cuts through like an alarm signal on small speakers. Good content though!
4:29 I thought I was hearing noises at this point cause my brain was so bored, but it's actually Mario music that's extremely quiet in the background lol
The chance of any person getting their own hat is 1/n. Thus the chance of every one not getting their hat is (1-1/n)^n. This has the well known limit of 1/e as n tends to infinity.
do we have a mathertmatical sign for "total electrical resistance of parallel resistors", which is: inverse of all resistances, summed up, and the sum is then inverted again: (resistor1^-1+resistor2^-1...)^-1 This also occurs (identical function) in "exponential smooth-step function" for "smoothing (more than 2) meta-balls or more complex signed-distance-fields" (commutatively), the simplest one one that is commutative and that allows for independent "sharpness" factors.
That's H(n)/n, where H(n) is the harmonic mean of n numbers. This operation also appears when finding the equivalent capacitance of capacitors in series.
If you have a permutation of n elements, where element n is not a fixed point, you can form a permutation with n-1 elements by removing element n and 'sewing' the permutation back together: where f(a) = n and f(n) = b, then sew it by setting f(a) = b. If the original permutation was a derangement, then either the new permutation is a derangement or a = b. If a = b, then the set of n-2 elements where both n and a (=b) are removed is a derangement. Thus, !n = (n-1) × (!(n-1) + !(n-2)). This is contrasted with the equation n! = n × (n-1)!
A better analogy would be people celebrating Christmas and picking a ballot with the name of a participant from a pot where nobody is supposed to pick the one with their own name.
Well, that was fun!
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i agree, that was fun
That was very interesting
😢😢😢àaaaaaaaaaaaàaaaaaaaaa😢aàaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaà😢😢😢aà😢😢😢a😢😢@@kilgorezer
You say subfactorial. I say the Secret Santa function.
Frick this easyer formula ⬇️
!n=(n-1)^n-2
I feel like "It's 1/e, isn't it?" is the "he is right behind me, isn't he?" of maths
Great analogy!
Euler is always there. Sometimes he's downstairs.
Was not expecting to see a 2021 Abu Dhabi reference here 💀
It's always an expression to do with pi, e, or ln of something it feels like in these higher level math videos
@@blasphemer_amonit's not a reference to "2021 Abu Dhabi", it is a reference to the thousands of badly written movie scripts with tiresome clichés
IT'S ALWAYS e LEAVE ME ALONE EULER
I love that even in your furious use of caps lock you still have the respect for mathematics to properly write e as lower case
My rage doesn't make me disregard how symbols work. I have standards.
This smaller case e makes it seem like you say e calmly, and shout the rest
Euler will chase you for your life
I love this reply lmao 🤣
“A dark room with several men who aren’t wearing any hats.”
Oh! The nightmares I’ve had about that!!
😂😂
Nightmares? I had fantasizes about that
That's Inception level 1
The deeper nightmare is none of them getting their hats back.
This is a branch of Applied Mathematics I never knew I needed.
Oh, _hats_ not wearing hats, right. That's what they aren't wearing.
It might not be the lack of hats in such a situation that's hazardous, but the lack of clothing well south of the hat region. Or if not hazardous, perhaps adventurous.
e ...xactly!
I have a feeling the men without hats are happy to be partying in a dark room. After all, they can dance if they want to, and they can leave their friends behind.
True, but one has to worry about the hats going wherever they want to, a place the men may never find
And, if they don't dance, well they're no friends of mine
It will be definitely a SAFETY DANCE! 😉🤗😄 ua-cam.com/video/1p_BvaHsgGg/v-deo.html
I'm glad these men are so safety-conscious while dancing.
"A derangement" is such a hilarious term for something in math. I love it.
I would bet whoever came up with subfactorials was just goofing around.
0:38 this is getting very spicy
several men without hats
Spicy indeed, in a world of hatless men, where do we find God?
@@WrathofMaththe God is a man without a hat
Did they leave their friends behind?
@@tweer64 If their friends don't dance, well they're ... no friends of mine.
*"There's no party like a Diddy party"* 😏
as a programmer, seeing "!n" just makes me think "logical not n" which evaluates as either 0 or 1 depending if its non-zero lmao
bitwise not also works
I'd be more inclined to think bitwise not as I usually write logical not as ¬ but yeah this would be confusing 😭
@@cosmnik472 For that ~ is used like this ~n
or strict n in haskell
Those who have programmed in some home computer BASICs will be thinking "pling n".
I'm a game developer, and a strikingly similar scenario - and result - came up awhile back when doing a deep dive on some item drop rate adjustments.
Imagine you have a monster that drops an item when defeated at a rate of 1 in 100 times and then you defeat 100 of that monster. What's the chance you've gotten at least 1 of that item?
Due to the "at least 1", this is easier to count the inverted result of "how many times did you fail to get the item" and repeat 100 times. So: (1-1/100)^100.
And then invert that result: 1 - (1-1/100)^100. Giving a result of approximately ~63.4% chance of getting at least once.
Generalizing this as n instead of 100, and then letting an n approach infinity, we get the result: Lim n->inf [1 - (1 - 1/n)^n] = 1 - (1 / e)
I remember mentally calculating critical chance for multiple critical chances way back in the Warcraft 3 days and my rough math always seemed to end up near either 33% or 66%
Geometric distribution is a bitch.
Brings a whole new meaning to "Statements dreamed up by the utterly Deranged"
(from the "stop doing math" meme)
I love the goku voiced version
A dark room with many men in it... Sounds like a Berlin nightclub.
And then e showed up.
@@empathogen75 only if it was one of Those clubs, with hatless men.
It does indeed sound like a Berlin nightclub, but the calculation was not gender specific, and it could therefore also be applied to any dark room where people remove their hats on entering.
I feel like there's a Flight of the Conchords song about this kind of situation ...
@@coyetsyes, obviously. Did you miss the joke?
This is Weird : I was just doing some economic modelling and this result popped out. This explains the entropy of a market, as it estimates states.
That sounds very interesting, will have to research
You might find it useful that derrangements can be represented by an integral: int_{0}^{/infty} (e^(-x))(x-1)^ndx
@@joshuaiosevich3727 That's true. But I was finding the ratio by construction - ie a fractal result. This explained why the result converged on this.
@@joshuaiosevich3727 linear algeba ruins everything
@@815TypeSirius I'm afraid I'm too dense to get what you're saying.
this is literally how i process every combinatorics problem hoping all the terms cancel out when they dont
Finding this guy in a math video is a fever dream
1/e also shows up in another famous math problem, which I'll poorly paraphrase: When dating, what % of the total pool should you check out before committing to one? The answer is 37% of the pool, 1/e.
With how big dating pools are now due to modern transportation and dating apps, I still have a significant amount of work to do 🤣
It also marks the point where the graph y = x^x stops decreasing and starts to increase
that's quite a deranged equation.
You could say that
11:00 just a note: I think here either the sum should be marked with i < j (not i =/= j), or, if written in this way, there should be 1/2 in front of it. You don't want to repeat elements: A1 intersection A2 and then A2 intersection A1. But the result is correct, the non-repeating sum over i < j is equal to n choose 2.
Came here to say this. The n choose 2 fixes this, since it ignores permutations of the choosing, which is why even though the left hand side is wrong, the right hand side is correct. This also means the higher order indices in the summation later on for the inclusion exclusion formula need to be written as i < j < k ... for all free indices from the set, or put another way 1 =< i < j < k ... =< n.
Your pace and depth are perfect. I would not attempt those formulas on my own, but you made perfect sense if them. Thank you so very much!
In my head I thought "its definitely something like e, or 1/e", and imagine my surprise when I saw the result! Although, not much of a surprise, whenever probability is involved, e will show up.
Yeah, for the mathematically initiated, it may be far less surprising. but it’s so amusing to take a silly word problem like that regarding hats, and the answer is 1/e 😂
1:04 say that again
Absolute cinema
Hat Tuah
bro💀
5:38 "it will blow your socks off"
What a relief that it won't blow my _hat_ off 😌
The hats are long gone at this point 😂
I wasn't wearing socks and my toes blew up.
Sorry 😞
a small price for science
@@WrathofMath np, that solved having to trim toenails.
Unfortunate
Your fault for having toes
i was scared for just a second when you started drawing that hat rack
I was wondering how he would manage to hang the hats on it.
Well paced, engaging, and playful. Euler would be proud.
I appreciate the gentle transition to combinatorics via hats and Zeus's wrath :))
I remember finishing our subfactorials a year ago and I loved them so much. I made a scratch project to plug a number in to give the sub factorial of the number.
Super fun!
@ Hell yeah!
I’m excited to integrate a “deranged“ math lesson into my sons home-study curriculum!
poor son😂
Oh cool. I love taking an integral of the subfactorial function.
i'm excited to show this to my math teacher
The night theme of Hateno village made this an emotional hat story for me...
Takes me back to sophomore year of college
This is why nobody likes mathematicians. They mix everybodies hats up when noone is looking.
Great explanation, you show how it’s calculated, how it’s relevant, and the end result is actually simple to calculate.
Thank you!
That was well worth the wait. Thanks for such a fun explanation.
I think that if we asked 10000 people the question (for some arbitrary number of hats, like 20) to give a percentage from 0-100 that no-one gets their hat back there would be quite a peak at 37%.
Yeah, there would likely be one at 73% too.
Omfg it's really everywhere
Why would there be a peak at 73% as well?
@@Kapomafioso Second most popular "random" number from 0 to 100.
6:39 WHAT YOU'RE LETTING AI?!?!?!
AI's taking everyone's jobs, not even the set of all permutations where i is fixed is safe....
gotta do what you gotta do
I will never write a "3" or a "2" as legibly as this man did @ 2:30. 😢
I was cooking with those
I did once, and then never again. 😢
However; not equal looked indistinguishable from +/- @11:30 :)
0:42 "there's no party like a Diddy party" 💀
Fixed points in this sense are cycles of length 1. The obvious generalisation is to permutations with other minimum (and indeed maximum) cycle lengths. These would be practical things to know.
e and pi are the most interesting numbers like “oh you have a weird value for this problem which no field of mathematics even comes close to?” Plug in e or 1/e or e^2 or the eth root of e or e^pi just keep plugging variations of e and pi and it’ll probably work and if it doesn’t even Euler can’t help you
So, if you need to calculate a subfactorial for some reason and you wish to save a lot of time, just divide the factorial by the number e then round to the nearest integer. It works every time.
Seems like a useful way to calculate the secret santa arrangements where no-one gets its own gift.
Thanks for the video, I wasn't sure if it was a joke video at first, but it was pretty interesting.
You earned a sub :D
What I love about mathematicians is that they'll mention Greek gods like it's 1200 BC.
When the Wrath of Zeus meets the Wrath of Math.
"Our fight will be legendary!"
If you gather enough men without hats, they can do the Safety Dance
As a fascinating exercise, consider the following:
Suppose you help n people receive their own hats by randomly distributing all of them. Then, those with incorrect hats return them to you to randomly distribute again among those still missing hats. What is the expected number of iterations it will take for everyone to have their own hat?
(I more or less have a proof for this, but i also have homework due today 😢)
I bet the proof wouldn't fit in the margin, either.
is the expectation not just 1/n! ?
Love how this is done sooo much easier with like a for loop and if statement in programming
I kept waiting for the hilarity, and was left with the impression that you math types are easily amused!
Beautiful.... Thanks you! (Dr Mike Ecker, retired PSU math professor)
Thanks for the presentation! This is a clssical gem of discrete mathematics.
This was really fun, and the notation wasn’t too difficult. Thanks!
Beautiful explanation. Thank you.
A wonderful and beautiful result, thanks.
Thank you for watching!
1/e is so hilarious, I'm amazed there isn't a sitcom about it.
Same!
Funny, i didn't know subfactorials were a thing but I'm learning about recursion in programming and this is helping me get more perspective on a specific problem I've been stuck on for two days. Essentially i was trying to figure out a formula to assess all permutations of something and worked out the subfactorial without knowing it had a name, i just noticed it was similar to a factorial but adding.
I didn't get the formula on my own, just that the subfactorial is involved lol
this was satisfying bc when saw the approximation before it felt so random yet underwhelming (like what, you just multiply the factorial by an essentially constant factor?)
but this explains neatly where it came from
4:23 I hear your Easter Egg of putting the Select File Theme from SM64, as Pannenkoek does! Very clever!
Man. I thought I was doing really well with understanding this and then I got to the beginning of the explanation at 6:30. You explain things very well but sadly there is a minimum bar of familiarity with the subject required that I just don't have. C'est la vie!
Enjoyed the SM64 music 😊
Oh wow that changes a lot
Hot example, honestly. This is also the most lucid presentation of inclusion-exclusion I have ever seen. Well done.
Thanks a lot!
1:03 there's hawk tuah🗣️🗣️
HAWK TU-*fucking spits*
Bro
You are NOT funny
you should uncomment yourcomment now
I fw with you vro, keep going
5:34 Missed opportunity to say it would blow our hats off. Booo!
I will not be tempted by your cheap puns!
Great explanation!
Thank you!
oh hey, alternating harmonic series, love to see it
What a fantastic video, such an enjoyable watch, the mario 64 music was just the cherry on top
Thanks so much! Trying to get the music at the right volume, I think I got it just about right this time.
I didn't realize there was music until I saw these comments.
I was listening at a lower volume, and your voice was quite clear, but there was this annoying beeping sound like a reversing truck every now and then.
Turned out to be the high notes of the music track starting at 3:50, from which only they were audible to me.
2:25 that sounds deranged
Great expanation of the inclusion/exclusion proncople
As is obvious from their accurate, lifelike portraits, the men in this diagram are misidentified. They are actually Man 47, Man 312, and Man 14,703.
1. 2. and 3 all died in the pneumonia epidemic of 1919.
At about the 18:55 mark, when reindexing to start at i=0 instead of i=2, wouldn't this change the ending index to (n-2) rather than n?
Once we're considering the limit as n goes to infinity, this change no longer matters, so the 1/e result is unchanged.
I now wanna question what n!n is, is it n! × n, or n × !n...
parentheses are your friends
Id say n × !n bc you put numbers before variables like 2x
!n!
stuff like this is why im not better at math, that’s very ambiguous
well, clearly, the bodmas order wasn't enough 😢😂
Dude your penmanship is so nice
Thank you!
My favorite part about it is that the sequence appears more slightly strange (the terms don’t all end in 000…)
Hmmm Derangements are actually exactly what I need for my experimentations on creating sudokus. Is there also a way to easily figure out what those derangements are instead of just their amount?
Latin squares
@buddermybacon well latin squares only have n derangements not every possible one though
the file select theme in the background makes this so beautiful
It's a classic!
5:33 Now I just want to know how likely is it that none of the viewers get both their socks back?
I gotta say when i read the title I was expecting the same as factorial except with division instead of multiplication
This video amazed my mom as I watched this since she never heard of subfactorials before
I love how at 2:00, the row for the hats was above the row for the men like they were wearing them lol. I hope that was intentional
0:25 yes, it means 1 factorial times n = n
Nice. You can instead of divising by e multiply by the precomputed 1/e or e^-1
The "key game" parties for tolerant couples were popular in the US in the 70s.
The men put their car keys in a pot and each woman drew a key from the pot and drove home with this man to have a little fun together.
Different use case, but the math is the same 😊
Clearly, they were not doing the Safety Dance.
(Gen X earworm, activate!)
Well, then they're no friends of mine!
But they were men without hats
And they were dancing
And they were friends
The men without hats never claimed they didn't "own" hats. They just didn't have any, currently, as is the case with our dancers
I had exactly the same thing in my head 😂
i was trying to work out a comment like this…well played!
1:32 oh god there was a puzzle like this in proffessor layton, you had to figure out how likely it was that 2 people got their hat but one person didnt. I guessed EVERY number from 1-100% only to realise it was 0%
Slide to the left!
Exclamation point:
!
!
!
!
Now Charlie Brown!
Seems like there's an easier way to count using a recursive definition:
let Ai be the set which has *exactly one* fixed point at index i. This means that index i is a fixed point, and the rest is a derangement: !(n-1).
Likewise for Bij (the set which has exactly two fixed points).
So the number of derangements is going to have the form !n = n! - ((n choose 1) ⋅ !(n-1) + (n choose 2) ⋅ !(n-2) + ... + (n choose n-1) ⋅ !(n-(n-1)) + 1)
Then I'm sure we can do some re-arranging.
12:45 writing (-1)^(j-1) feels illegal
What happens when we move the exclamation mark to the other side? I imagine it gets quieter instead of louder.
Honestly amazed by how you could confidently go deep into a topic in such an entertaining way. Might be slightly jealous...
music credits?
Excellent video!!!
Thank you!
In case no one else has mentioned it, really weird harmonics in the background music. It cuts through like an alarm signal on small speakers. Good content though!
This music is making me think I'm watching a Zullie the Witch video
I was thinking panenkoek
4:29 I thought I was hearing noises at this point cause my brain was so bored, but it's actually Mario music that's extremely quiet in the background lol
the moment he said "the probability will surprise you" I thought "probably 1/e or sth like that"...
The chance of any person getting their own hat is 1/n. Thus the chance of every one not getting their hat is (1-1/n)^n.
This has the well known limit of 1/e as n tends to infinity.
It's here!
"it's the on left now"
do we have a mathertmatical sign for "total electrical resistance of parallel resistors", which is: inverse of all resistances, summed up, and the sum is then inverted again: (resistor1^-1+resistor2^-1...)^-1
This also occurs (identical function) in "exponential smooth-step function" for "smoothing (more than 2) meta-balls or more complex signed-distance-fields" (commutatively), the simplest one one that is commutative and that allows for independent "sharpness" factors.
That's H(n)/n, where H(n) is the harmonic mean of n numbers.
This operation also appears when finding the equivalent capacitance of capacitors in series.
20:05 I was just waiting for it (it was worth it)
Was the file select music intended to be a reference? I really hope it was.
probably a dumb question. but is this extensible to continuous functions in a similar way to the Gamma function?
I don't think anybody tackled this problem
If you have a permutation of n elements, where element n is not a fixed point, you can form a permutation with n-1 elements by removing element n and 'sewing' the permutation back together: where f(a) = n and f(n) = b, then sew it by setting f(a) = b. If the original permutation was a derangement, then either the new permutation is a derangement or a = b. If a = b, then the set of n-2 elements where both n and a (=b) are removed is a derangement. Thus, !n = (n-1) × (!(n-1) + !(n-2)). This is contrasted with the equation n! = n × (n-1)!
that's called an admin command
A better analogy would be people celebrating Christmas and picking a ballot with the name of a participant from a pot where nobody is supposed to pick the one with their own name.