Should You Switch? NO!

Not a real offer, for educational purposes only. I don't have two envelopes filled with money. I don't have any envelopes OR any money.
In The Two Envelopes Paradox -- also called the Exchange Paradox -- you know what the right answer is almost immediately. Until you don’t. And then you do. And then you’re not sure.
The problem isn’t so much the problem itself; it’s figuring out why the setup is wrong. Like several of Zeno’s Paradoxes, we’re lured into thinking about the scenario in a way that leads us down the wrong path… and by the time we realize it, we’re so deep in convincing (or troubling) math that we’ve lost sight of the real issue.
Just like with the Monty Hall Problem, even top academics have trouble elucidating clear, meaningful reasoning for why switching in the Two Envelopes Paradox is or isn’t valuable. That’s why Martin Gardner and others struggled with it for years, and why decades after the paradox (and its variants like the necktie and wallet-switching problems) debuted, it was still of academic interest in math journals and popular recreational mathematics publications.
But the Two Envelopes Paradox is an exercise in logic and probability that continues to be valuable, and probably more so than ever, with implications on how we approach math, science, and the world around us.
** SOURCES **
Barry Nalebuff, “The Other Person’s Envelope is Always Greener,” Journal of Economic Perspectives, Winter 1989: pubs.aeaweb.or...
Miles Mathis, “The Two Envelopes Paradox”: milesmathis.com...
Eric Bliss, “A Concise Resolution to the Two Envelopes Paradox”: arxiv.org/pdf/...
** LINKS **
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Facebook: / vsaucetwo
Hosted and Produced by Kevin Lieber
Instagram: / kevlieber
Twitter: / kevinlieber
Podcast: / thecreateunknown
Research And Writing by Matthew Tabor
/ tabortcu
Huge Thanks To Paula Lieber
www.etsy.com/s...
Get Vsauce's favorite science and math toys delivered to your door!
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Select Music By Jake Chudnow: / jakechudnow
#learning #education #vsauce2

[@Vsauce2]

HELLO HUMANS. I see many people mentioning the Monty Hall Paradox. I actually discussed that in my video, "What Is A Paradox?" You can also learn more about Falsidical Paradoxes like Two Envelopes. Here is that video: kzbin.info/www/bejne/oXvdhK19l8uWZs0

[@Aura96968]

K

[@isaiahthorne7707]

Ohhhh right

[@NAT-ld1sh]

Cool

[@xXRuce007Xx]

This is very interesting! I can't help but think that it is somehow connected to the double slit theory. Can you possibly tie these two together?

[@-RAV3N]

Which was X or 1/2 X then? Why didn't you open one to show if our guesses were right? I will be frustrated by this FOREVER. Trust me I will never forget and probably lose sleep over this. Send help...

[@MrHatoi]

I've been switching envelopes for 20 years now, multiplying my expected value by 5/4 each time. By now, my expected value is several trillion times x. I will keep switching, until one day, I will open the envelope, and I will expect to be the richest man in the world. At that point, it will all be worth it. Empires will fall at my feet as I use my wealth to rule the world. But for now, I keep switching...

[@y2kblackout]

You've wasted your life.

[@empirecrumbles615]

Y2K Blackout bruh

[@twopeople254]

Oof man

[@mr.catfish155]

Patience my friend. Continue the path of the switch, and one day your patience will be rewarded.

[@Zeroneii3]

if x is $1 you will become the richest man by opening rn

[@KaliTakumi]

Kevin is playing the good cop bad cop routine with maths

[@BillboMC]

Kali Takumi that’s actually so true XD

[@enderpup9289]

Is that from Brooklyn 99?

[@KaliTakumi]

@@enderpup9289 Not that I know of. I don't really watch the show

[@CpCubeR]

Right?

[@enderpup9289]

Kali Takumi Kevin( police captain) was arguing with his boyfriend about a similar problem: the 3 game show doors. They argued about switching the door or not

[@HunterTracks]

Me: So, Kevin, should I switch? Kevin: Yesn't.

[@galaxycube1016]

Yeno

[@gangayang]

Non't

[@haverjamarosi680]

it depends on what you play ... if you got offered 2 envelopes than no, cos its 50-50% But if you got something and you can switch to something twice the price or half, then you should ... Cos you play quadripple or nothing. and its worth it.

[@wesgrubbs8437]

Nep.

[@TillURide420]

Nes

[@HeyLookItsThatGuy]

This how multiple choice tests feel

[@dewhiterabbit1337]

Ah

[@juliansingh5182]

Unless you friggin study...

[@dewhiterabbit1337]

@@juliansingh5182 Ahh,

[@theluxiemexpress3656]

Ohh ...what’s studying?

[@juliansingh5182]

@@theluxiemexpress3656 its a sacred and rare ritual that grants you knowledge, i use it to cheat all my tests and exams

[@firedropcutie]

Me: *opens envelope 1* Envelope 1: *contains 1 penny* Me: Ok let me switch Envelope 2: *contains a penny cut in half* Me: oh...

[@nathanrcoe1132]

some currencies have (or had) a half penny

[@Yours--Truly]

But then 0 isn't a half of .01 dollars. It was a counterfeit penny in the first place

[@thatoneguy4584]

But even if it is a fake penny, it is still better than nothing!

[@sylvia7961]

@@thatoneguy4584 actually that would be fun to explain to the cops if they found out

[@fos1451]

@@thatoneguy4584 if you don't know it's a fake pennies and you use it to buy something, you can get in a real trouble

[@tpir0426]

Kevin is probably the only guy that could make a 2 hour video on the answer to 2+2 and keep me interested the whole time

[@squeaksquawk4255]

I want this to happen so much.

[@devoarco]

there's actually a whole book with a thousand pages of mathematical Notation to prove 1+1=2. It's called principia mathematica (not from newton but from russell)

[@chocholic8174]

So if there was 'V' in one of them, what about the other one? 'W' or '\'?

[@matheuscordeiroKH]

maybe the other could be 'X'

[@Swampert00260]

It's a V made of two lines, so it'd either be a W or a V made of one line

[@roshanprabu]

It is /

[@keelekeele]

The other envelope was labeled. X is the roman number for 10 while V is 5. As the title stated, you should NOT switch from x. ;)

[@DEFCON1GAMRS]

keelekeele i dont think this is intentional but feels like a good time to get high

[@elainastar1179]

kevin just wasted 13 minutes of my time to tell me that a 50% chance really was a 50% chance

[@thegreatkingofevilganondor1500]

Paul Bean But did you gain half brain cells or double?! Who knows?!

[@chrisjfox8715]

insanity wolf 200%

[@researchbothsidesequally4481]

Is this video a complete waste of 13 minutes? Yes!

[@sai-chan1460]

You just saved me 7 minutes of my life. Thank you my saviour :D

[@whatifyoucouldntchangename7091]

@@researchbothsidesequally4481 the best way to waste 13 mins tho

[@phildcrow]

"...so I can clearly not choose the wine in front of you!"

[@thomasbailey921]

"You fell victim to one of the classic blunders! The most famous of which is 'never go into a land war in Asia,' but second, and only slightly lesser known, is this: 'never go up against a Sicilian when DEATH is on the line!'"

[@tarzaan2603]

Jahahahahhahahahaha

[@amosvideojournals9589]

I sure do love how arbitrarily assigned variables can cause issues.

[@kangalio]

Yeah I feel like that was the real issue all along.

[@alexzander7629]

Caleb Smith Well actually it’s not arbitrary. The 2x and 1/2x is kinda, but the idea holds in all cases because there is some real number you can multiply the amount in Envelope 1 to get the amount in Envelope 2, and this a number you can divide (or multiply the inverse of) the amount in E2 to get the amount in E1. The idea is that by taking into account this mathematical relationship between the two amounts, you can say that it is always better to switch despite “common sense” as it were telling you it’s just a coin flip. And he explains at the end why thinking with that ratio in mind creates the false paradox.

[@amosvideojournals9589]

@@alexzander7629 The part that makes it arbitrary is that either variable could be x. There is no real reasoning to calling envelope one x, when envelope 2 could easily be x as well. He demonstrates this in the video by showing that player one and player two both believe that their envelope is x. Since we don't know which is which in this scenario, we should be assigning two variables here. Edit: Removed a bit near the end. I missed the read more in your comment at first.

[@gidelix]

Just what I wanted to say. I’d go n = 2k or x and y or something. I’m positive doing the maths in that case will net you a system of equations which can be solved for his x/3 and 2x/3. I’m too lazy though so go figure it out yourself. (No offence, just a quote from my maths professor)

[@jacobrzewnicki1832]

Caleb Smith exactly, why assume the envelope you pick is only x but the other can be 2x or x

[@blueburekas4977]

kevin: theres a 100% reason you should switch! me: no kevin: yes! me: yes? kevin: NO!!

[@janzacharias3680]

Doublebamboozle

[@Cam370n_Sm3t]

BlueBurekas 🤣

[@LK1]

OoF

[@nates9536]

Kevin is the USB stick of humans

[@TheBluePhoenix008]

r/storiesaboutkevin

[@Norbal.]

Me: Hears the word "right?" Also me .01 seconds later: Oh no Kevin: WRONG!

[@twopeople254]

Me: he's about to say it

[@TuxMan]

Facts

[@christinabarfoot1389]

And I throw my popcorn every time

[@thatssomethingthathappened9823]

So predictable.

[@thatssomethingthathappened9823]

He does it every time

[@stealthemoon8899]

"If you're confused right now, GOOD!" - Vsauce2 2019

[@DRKongo-vi8ok]

wrong

[@senzu._bean]

I literally read this as he said it🤣🤣🤣 No joke

[@danltg]

You mean ripoff vsauce

[@stealthemoon8899]

@@danltg oh, you're right -Vsauce2 2019

[@Ronnie7X]

jock willink: Thought switching the envelope changes anything? GOOD

[@lukasberner7729]

*plot twist: they contained 0$ the whole time*

[@krum6334]

Not a plot twist. The truth

[@thomassynths]

Or they both contain debt.

[@Wolfiyeethegranddukecerberus17]

Well, one would still contain twice as much as the other.

[@thomassynths]

@@Wolfiyeethegranddukecerberus17 Yeah, but the result is negative, so you would always refuse to switch.

[@syrialak101]

@@thomassynthsHe literally proved in the video that there would be no advantage or disadvantage to switching.

[@filiplykkegaardkastrup1876]

I feel like the pitch for Vsauce2 is a man who went down the rabbit hole and is still falling

[@shinkamui]

and the rabbit that dug the whole is regular vsauce

[@Bigfoot_With_Internet_Access]

Everybody gangsta till Kevin is here with 2 envelopes

[@abed3111]

I, too was a gangsta but then Kevin came with 2 envelopes

[@David-mf3bn]

But I am Gangsta

[@TAussieG]

I wasnt gangsta but became even less gangsta

[@ihatemosquitos1037]

ur everywhere

[@outside8312]

Yes No Maybe I don't know Can you repeat the question

[@Clodagh400]

Your not the boss of me now

[@UODZU-P]

@@Clodagh400 what about my not the boss of you now?

[@Clodagh400]

It's the song for maclom and middle

[@fromthegamethrone]

@@Clodagh400 my not the boss of me now?

[@benjaminevans9205]

Hotel travago

[@goegrog27]

Where X is the smaller amount, the envelope you pick has either X or 2X averaging out to 1.5X (Same as the one the didn’t pick). The flaw is setting the amount in envelope A to X when X can be two different values.

[@ewicky]

Exactly. I came here to say the same thing. I'm not even sure why this is a "paradox" when the solution is so obvious.

[@tmorid3]

It's not - bc you either multiple your prize (if the envelope you chose is the smaller one), or lose half of your amount (if it's the larger one). The thing is that you don't know that x is the smaller one, but just the amount in the first envelope.

[@Ooopsi]

The issue is that the amount in each envelope, even if we don't know it as the player, is fixed before the choice of switching is even offered. If the lower value is 50 dollars, we know the other one is 100 dollars and there's no benefit for switching as both are equally likely. The expected value leading to 5/4X nonsense assumes that the value of the first envelope is fixed, and the value of the second is not decided yet. It assumes that if the first envelope has 50, the other could have 25 or 100, in which case it makes sense to switch. Unfortunately, it doesn't work that way with predecided values because either you start with the lower value X and double to 2X, or you start with the higher value 2X and drop down to X. There's no X/2 here, the X/2 is simply 2X/2=X.

[@nickm3694]

Michael: "true statement. Or is it?" Kevin: "true statement, right? WRONG! Actually RIGHT!"

[@bastiliedtk4390]

Nicholas McGinty Wrong

[@Jordan.m.martinez]

We miss micheal

[@simon6071]

Man to woman in bed: Is it true that when a woman in bed with a man she loves means yes when she says no? Woman: No.

[@christoperjohnteodosio2688]

¤😆😕

[@decyted6847]

@@Jordan.m.martinez where is he?

[@Splox5]

So "eeny, meeny, miny, moe" _doesn't_ exponentially improve my odds of picking the correct envelope? Darn.

[@SebastianHasler]

TL;DR: The expected value must not depend on a random variable.

[@luck3949]

Thanks. I didn't understand this paradox before reading your comment.

[@hetgenie]

Still: you have nothing to begin with; in a 50-50 scenario you can double your gain, I'd take that risk anytime. Worst case scenario my profit is halved, but I still have more than I began with. People play the lottery with way slimmer chances to win anything at all.

[@AverytheCubanAmerican]

Nintendo: Do you switch? Vsauce: No

[@anthonybusch]

especially with the joy-con colors on his face

[@ocomentador7444]

*wrong*

[@kanakousakis]

@@ocomentador7444 right?

[@somerandomguy___]

Vsause1 : or DO you?

[@Ultrasonix3]

WRONG

[@marissasmith2827]

Should you switch? Vsause: "Well yes, but actually no."

[@Vivi-yw1eu]

but actually yes

[@krajsyboys]

@@Vivi-yw1eu but actually no

[@mwperk02]

@@krajsyboys But actually yes.

[@bang1054]

@Mathew Perkins But actually No

[@alanosoelpoderoso1912]

Bang 10 But actually, as the expected valué states, Yes.

[@ArcticFoxWaffles]

*When you need to meet the 10 minute criteria:* "Soooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."

[@ivyvetniss3148]

Lol

[@bb-xw1np]

well it's a 12 minute long video soooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

[@Woldhardestgamer]

@@bb-xw1np i scrolled really hard after expanding you comment and i'm quite underwhelmed...

[@bb-xw1np]

@@Woldhardestgamer "o" is the best letter you just have bad taste in letters

[@irenaveksler1935]

Trash Duck is it tho?

[@gmeztubenation]

Anyone else think “ohh no another Monty hall” when this started

[@kruton93]

and then it got much, much, much more complicated.

[@oni8337]

true

[@nuansakautsar5687]

yep

[@lilzachx4629]

Yup

[@nirorit]

@@kruton93 monty hall is just as complicated

[@icefang111]

God the final bit saved me, the whole time I thought I was losing my mind "why are you adding in X when neither contains X by the rules you're outlining or did I just miss something????" the X is a lie

[@niapor]

@@nathan5160 from what VSAUCE2 says, ( if it is only one bill always ) one envelope has to be X and the other Y will be = X/2 or 2X as the other envelope ( from what I understand ) can be more than X or less than X but never equal , the only constants are that if X is 1, than Y is 2 as an example, but as 2 does not have double than both 2 and 1 are out, 5 does not have half thus is out, since 5 is out 10 is out, which means 20 is out since 10 is not available and there is no 40 bill, 50 is out because there is no 25, thus eliminating 100 as now 50 is not available PLUS there is no 200 bill, thus we can never have this premise now if we go with premise that X is "either" X/2 or 2X compared to the other envelope ( but NOT BOTH necessarily ) than we have as possible sets only 4 ( and we can know what the other is if we can see the chosen envelope before possible switch with 6 possible bills, with only the 10 dollar bill being one that is a true 50-50 choice ) 1-2 5-10 10-20 50-100

[@akkok5059]

It's simple. Choose one, it's 50-50 lol don't overthink it

[@icefang111]

@@akkok5059 Exactly!

[@Koby777]

The method for calculating Expected Value, as demonstrated at 4:18 is wrong. Expected Value also takes into consideration the cost of your decision. In each case, we lose the value of x from either decision, so it looks something like, 0.5(2x-x) + 0.5(0.5x-x) = 0.5x - 0.5x = 0 therefore no expected value gain.

[@boghag]

@@Koby777 0.5(2x-x) + 0.5(0.5x-x) = 0.5x + 0.5(-0.5x) = 0.5x - 0.25x = 0.25x (expected gain, this is equivalent to 1.25x expected value, which is 5/4x as shown in the video). Your math was wrong, sorry

[@bwpbwp9613]

The 5/4 X result comes from faulty math. X is used to mean 2 different numbers which is not the proper use of a variable. It need to be set to either the higher value or the lower value and shouldn’t change within the same equation

[@toastman3955]

That’s explained in the video

[@gopalsinghrathore1346]

@@toastman3955 but isnt that how probability works. I coudnt understand why behind it

[@toastman3955]

@@gopalsinghrathore1346 In the video, he explains exactly why the beginning math is really false. He plays off the premise that it is true until the end where he brings up the exact problem that the other envelope cannot be both 2X and 1/2X, but the correct way to write it would be an overall loss or gain of X/3 which balances out to make switching not optimal or not optimal.

[@juliannaf.3074]

me: so you should switch? kevin: **angry space baby noises**

[@jumbocattle8597]

Switching to Geico could save you 15%

[@e2gaming337]

or more

[@guy3nder529]

indeed kevin: no!

[@Mandyyyyyy258]

Or more on car insurance

[@edgarmiranda1822]

wrong!!!!!!!!! or does it????

[@niapor]

I always do the right choice, because I always stay at the Holiday Inn Express

[@toddbiesel4288]

Q: Should I switch or not? A: Yes.

[@mantis3574]

I showed my cat this. She's now working at the bank

[@therobotics1rthegreat157]

I think I did something wrong. Showed it to my hamster an it is now running a multi level marketing scheme and wants me to invest.

[@RobotAnimals]

Instructions Unclear: My dog is now the boss.

[@David-mf3bn]

@@therobotics1rthegreat157 good one

[@bobwebber8521]

When I first came across this I thought "How could this drongo have any subscribers"? But I watched on and was surprised to see he was actually quite entertaining and the subject can be looked at so many ways. Well done you. Perhaps if more teachers had this style their students could absorb lessons better.

[@Vsauce2]

Did you subscribe to the drongo?

[@mariafe7050]

Well, did you?

[@aartigodia3422]

@@Vsauce2 yesn't

[@RafaelMunizYT]

@@Vsauce2 do the maths and find out by yourself

[@seazein]

No one: Sprint commercials: Should You Switch? YES!

[@JamSparing]

I am watching this in the middle of the night and I am like, transcending several planes of mental gymnastics. Help.

[@Dark_Slayer3000]

This was but a warmup. Get ready to get high on math

[@zt3853]

*snorts geometry*

[@OMGclueless]

If it helps: if you lose x/2 by switching, then x must have been the bigger number. If you gain x by switching, then x must have been the smaller number. So you can't just add up the gains and losses in terms of x because the value of x is not independent of whether your number is smaller or larger.

[@blutwirbel3180]

@@OMGclueless I don't think that helped

[@finedig1854]

@@blutwirbel3180 especially since the guy sent it 3 months after op was confused

[@GaryoXGaming]

Vsauce: Theres just no value in switching Verizon: WELL SCREW YOU

[@user-qh5jk1mn5i]

Alpha Mit3 *breathes out of nose slightly

[@jamesmoriarty9433]

AT&T 😤👌🔥💯

[@GaryoXGaming]

Marshall Kimber haha yes yes

[@GaryoXGaming]

The Donovan At&t is better🤔

[@jamesmoriarty9433]

@@GaryoXGaming Yee

[@kncle]

I've seen so many videos that say that "you should always switch because math" and ends there. The conclustion at the end makes this video so worth watching

[@filipklominek3527]

question: od you switch? Kevin: Well no, but actually yes but actually no.

[@qladinado]

kevin: 50% is not 50% me: no it is kevin: YES! me: Yes? Kevin: NO! me: 50% = 50% kevin: no, actually 50% = 50% me: wat kevin: 25%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@kittycypher]

XD

[@levithe2nd]

@@kittycypher XD 2XD 1/2XD 5/4XD

[@kittycypher]

3.14159XD is equal to infinite XD arranged in the form of Pi. 🥧 :)

[@artluka727]

Kevin: Right? me: ye... Kevin: WRONG, it's 50%

[@gangayang]

Kevin: 50%=50% Everyone: yes Kevin: _NO THAT'S WRONG!!!!_

[@zeronothinghere9334]

I think I understand it. You need to factor in that there is only one total sum, not two. So, the problem with 2X vs 1/2 X is that in one, you have X and the other 2X, in the other one, you have 2X, and the other one X. 3X (the total) however has a different value in each, depending on what value your envelope has. So let's assume you have 10$. In one case, X = 10$, the other envelope has 20$, total is 30$. In the other case, X = 5$, the other envelope has 5$, total is 15$. Note that the total changed. But the total doesn't actually change, and the calculations need to reflect that. They do not reflect that, instead they view it from the perspective of having one envelope.

[@rfvtgbzhn]

Yes, that's the case in the first example. In the second example the total isn't fixed because the coin toss happens after you chose the envelope. This also explains why it makes sense for both players to switch in the second example.

[@SilverLionGR]

"""This is what you probably already knew until we got math involved !!! !!!""" WHAT A FANTASTIC thing to say to anyone involved in Math for making a living !!!

[@Rikri]

*Math or logic. Without math we couldn't debunk this paradox that would have worked with some logic anyways.

[@MinerDiner]

Kevin: But don't worry, it gets weirder! Me: Thank goodness I was worried for a second that it wasn't going to

[@jimbroomy8207]

I think the error is assuming that 2x and x/2 are equally likely. If you know that's true then you should definitely switch. For example, if you know that the higher value was chosen from a uniform distribution and that x is at most half of the maximum value, then switching is correct. If there's no maximum value then it's not a uniform distribution (and higher numbers will typically be less likely, immediately solving the issue). If you don't know how the numbers were selected, then it's not a well-defined problem. Just because you selected one at random doesn't mean that (x/2, x) and (x, 2x) are equally likely to be the case.

[@TranslatorCarminum]

Wait a minute. I've watched this multiple times, and I think the explanation at 8:48 - 9:51 is just a roundabout way of restating the argument first made at 6:12 - 6:22. Since (2X)/3 = 2(X/3), if we assign Y = X/3, then one envelope contains Y and the other contains 2Y, and we're right back where we started, with the crucial variable merely re-labeled. The expected value of switching is then calculated as (1/2)(2Y - Y) + (1/2)(Y - 2Y) = 0. The denominator of 3 is completely moot. It doesn't seem to add anything useful to the calculation. All you have to do is substitute X/3 for Y in that last equation, and the result will be practically identical to what Kevin wrote on his dry-erase table. Also, while this isn't quite a recapitulation of the Monty Hall problem, all it would take to become little more than just that is a third envelope, maybe with a potential prize of 3X. It would be slightly different, since the two losing envelopes wouldn't be of equal value (like the two goats in Monty Hall), but both would still be significantly less than the winning one. So while an application of expected value might work out a bit differently, your binary chances of choosing the single winning envelope (the one with the largest prize) would be subject to the same paradox.

[@miyuiismells]

This man wrote an entire freakin essay

[@raihanalam9]

Words

[@vedanshbudhia8148]

Wtf

[@knewbod]

The point of the explanation is to re-frame the problem. The paradox occures because one is mxing two different universes and is using X wrong. The reason for X 1/3 and X 2/3 is to make it clear one must think of the total sum of the envelopes as constant. Once you do that you stay in the same universe and the paradox goes away. Saying one envelope is Y and the other 2Y, is just saying the entire universe has a value of 3Y (and then doing the math for both switched from there). This works but it's confusing, since we seem to be using Y the same as the original X, when we are not. In the origonal example we are mixing the universe with a total value of 3X and the universe with a total value of 1 1/2 X. In the new one we're not. Changing X to be the total value is an attempt to break out of that thinking.

[@unliving_ball_of_gas]

Wow......so long

[@pietergeerkens6324]

This once again reminds me of an old adage by Mark Twain: "There are Liars; Damned Liars; and Statisticians."

[@insertclevernamehere2506]

It's a math problem, not a statistical one, though I'd side with Twain anyway :)

[@nmotschidontwannagivemyrea8932]

@@insertclevernamehere2506 Statistics is a form of math. Your comment is like if I said to NASA, "Getting to Mars is a science and engineering problem, not an astronomical one."

[@OldBooTofu]

plot twist: one of the envelope contains a reverse card

[@martinmilchov4065]

So the other one contains a double reverse card... Or is it a half reverse card?!?

[@zt3853]

Мартин Милчов 0-o

[@irenaveksler1935]

Мартин Милчов LOL

[@collinsnow2203]

So does that mean the envelope chooses you to open up instead lmfao

[@lusamine7925]

It actually contains v sauce 2... So does the other contain v sauce 1, or v sauce 4???

[@carbongaming9612]

Wait so at the end the one Envelope had "V" so that must be the smaller one. The other contains "W"?

[@Epic_Pingu]

maybe the other one contains "\" or "/" though. hmmmmm...

[@carbongaming9612]

But the other envelope contains 'double' the amount. / And \, as far as I can tell one isn't twice the amount of the other

[@overloader7900]

\ and / are halfs of V

[@carbongaming9612]

Ah I see I'm thinking wrong!

[@DaP84]

The other one could be X, two V's mirrored vertically. :P Or "diamonds"

[@galaxycube1016]

There was a bit of a flaw. Rather than one being "X" and the other being "2X or 1/2X", it should be both of them being "2X or X"

[@thundersheild926]

Exactly. The mislabelling of x is what leads to thus apparent paradox. If you swap out x for what it actually represents in the two probabilities, you get (1/2)*(100)+(1/2)*(50)=75, which makes perfect sense. Still, it's fun to try and spot the error.

[@ireallydontknow3299]

Yeah, that's what the solution at the end addresses. Regardless of how much is in one of the envelopes, their total doesn't change. Saying yours has X and the other 1/2 * X neglects that fact, as it makes the total for the "I have the smaller one" equal to 3X, while "I have the larger one" becomes 3/2 * X, which is absurd.

[@nasdfigol]

True, since 1/2X × 2 = X, not 2X

[@HeHe-du4le]

I didn’t think of it that way, interesting. Thanks!

[@TheGrapeApe22]

Yeah but that makes a possibility that both values are the same...

[@lilnut9423]

Kevin- Yes? Me- yes! Kevin- NO! Me- Maybe? Kevin- PARADOX

[@JesperoTV]

Hotel- trivago

[@jrrwtch]

xXjesperoXx WoW just wow

[@Hallowed_Ground]

This one is dumb. Seems like it's just copying everyone else's joke.

[@kit_nine]

@Juan Cortez Muro TAKE THIS *REPORT* TO THE FACE

[@jujment699]

This is my favorite V sauce channel. I like the other channels, but when i look at my youtube and see that Kevin uploaded a video i get particularaly excited. Does anyone else agree? Kevin if you read this we want more uploads from you!!

[@gamingworld8355]

Me

[@RaivoltG]

You're hilarious when you say one way is the best way and then you change your mind, like the beginning of this video! RIGHT! WRONG! Your showmanship is very unique and entertaining! Your explanations and use of examples are much easier to understand than other people's explanations. You've got a great show, well thought out, very informative and has the perfect amount of clever humor! Thank you!!!

[@brettsutherland8390]

This video made me burst out laughing so many times... All while learning!

[@feetfinderguy7044]

- "Hey! do you watch Vsauce?" - yes - "do you understand it?" - no

[@Daniel_WR_Hart]

- WRONG!

[@alexbitzan8747]

It’s not THAT difficult to understand. It just takes a little thought

[@JohnDoe-kc5kv]

yes no yes no

[@IkissyJesus]

Sliding the envelopes away and erasing the numbers at 1:14 was slick as hoopla.

[@_prince_4703]

Kevin: Right? Me: Uhhhh... ye.. Yes Keving: WRONG!

[@Литотерапия101камень]

Kevin: Right? Me: Uhhhh... ye.. Yes Kevin: WRONG And I have a scientific math explanation, right? Me: Ya, tell me that Kevin: *WRONG!*

[@tylerhayes9808]

Youre counting it as "losing" something if you get the lower number. Dont forget youre always gaining x.

[@HJHawley7677]

Tyler Hayes that’s not the problem here.

[@LLPTV]

You're only gaining X if the other letter is 2X, otherwise the you're gaining 1/2 X, with the other letter containing X. The value of both letters, need to be considered, otherwise you're "losing" imaginary value.

[@WilliamElliottHinson]

Opportunity costs. If you could get 200 but you get 100, you lost 100.

[@elniko88fuegorojo37]

Me: Oh, I think im understanding it... Kevin: WRONG!!

[@bogdan-cristiantimofte5625]

9:56 "Is probably what you knew until we got math involved"

[@lusamine7925]

School in a nutshell

[@yinyang1217]

"Should you switch?" "Yesn't." "Because human brains are randomized and not like computers."

[@averagemilffan]

Bruh I'm just some high school kid but I think human brains are made up of complex circuits. We use these types of circuits in A.I and neural network

[@BatuhanDere]

michael in vsauce: your confused? here is a quick runner up for you kevin in vsauce: 8:17

[@evank3718]

The light angle in this video makes you look like you’re in a presidential poster

[@hyche_-6790]

In math we trust

[@jamesmerkel1932]

NO YOU CAN!!!

[@jebkerman5422]

I will create a great paradox and I will make your logic pay for it MAKE MATH GREAT AGAIN

[@notyepdranel961]

Or in almost any movie poster

[@ICountFrom0]

When math doesn't match reality, that means that you've got a flawed assumption somewhere, da?

[@davidbarroso1960]

The flawed assumption is when he was assigning the variable x in the first half of the video. X isn’t constant in that case because it could be two different values. It’s a mathematical trick. There’s really two variables that he’s assigning as one which leads to the paradox. It’s just bad math. In the last part, he assigns the total sum as x, and x is now a constant because the total sum can only be one value, which gets him the right answer

[@abijo5052]

@@davidbarroso1960 yeah you can see why Schrödinger could have come up with it, it's very similar to a lot of quantum mechanical questions. He let X be a variable that can take on one of two states, but then treated it like a regular variable, which doesn't work.

[@ryanodd4149]

@@davidbarroso1960 Wait what if you were to open one of the envelopes? Then X is assigned, and if I'm correct, it is always worthwhile to switch?

[@Nukestarmaster]

@@ryanodd4149 Nope, opening the envelope still doesn't change anything.

[@unliving_ball_of_gas]

@@Nukestarmaster Wait,waa.. Doesn't that means that I will know how much there actually is inside so I could decide to switch or not?

[@DevilMaster]

It's much easier to analyze all possible cases. So, let's assume the larger amount is in envelope 1: - You choose 1, you don't switch -> you win - You choose 1, you switch -> you lose - You choose 2, you don't switch -> you lose - You choose 2, you switch -> you win So, do whatever you want. No special behavior will increase the probability of winning. This, by the way, is exactly what does NOT happen with the Monty Hall problem, where there are 3 doors, a car is behind one, a goat is behind each of the others, and in every case, you're shown one of the other doors, which always contains a goat. In the Monty Hall problem, there is a high probability to benefit from switching, and it can be easily proven. If we assume the car is behind door 1: - You choose 1, you're shown a random door, you switch -> you lose - You choose 2, you're shown 3, you switch -> you win - You choose 3, you're shown 2, you switch -> you win - You choose 1, you're show a random door, you don't switch -> you win - You choose 2, you're shown 3, you don't switch -> you lose - You choose 3, you're shown 2, you don't switch -> you lose

[@uniworkhorse]

Friends: Don't over think things Me:

[@vasilzahariev5741]

It's the same as "Deal or no deal", in the end, when there are only 2 boxes left, the banker almost always offers to switch boxes (at least in the show that I watched). A lot of people who switched got screwed over, but then a lot of people who didn't switch were also screwed over.

[@Edyledyl]

This is actually a different problem despite the fact that it seems similar. In Deal or No Deal you gain information as boxes are opened, and this information actually increases the expected value of switching at the end. Deal or No Deal is an extension of the Monty Hall problem, and there actually is a benefit to switching because of the information we gain as boxes are opened. For the two envelope paradox though, there is no increase in expected value from switching. The two envelope paradox is actually related to a quirk in the law of total expectation.

[@weir9996]

@@Edyledyl Deal or No Deal isn't quite the same as Monty Hall as you open cases randomly. Monty knew what was behind each door and acted accordingly. In Deal or No Deal the odds are the same whether or not you switch

[@duck1sgood]

@@weir9996 wait but you see the boxes that were open so you definitely can make out the new probability of the switching being worth it right? I never watch the show so Im just guessing

[@richjhart]

@@duck1sgood Once there are only two boxes left, it's 50/50. You do have all the information about the open boxes - the only information you have about the remaining 2 boxes is that they definitely have different values in them. So it is exactly the same as the two envelope problem.

[@iamadooddood4331]

@@Edyledyl The given situation however refers to when there are only 2 boxes left, by which it reduces to the 2 envelope paradox.

[@zappyeats2579]

I never seen the 2 envelopes before but the whole time I was thinking the 2x was wrong math and you needed to use total sums. Thank you now I feel smart for the day.

[@mikeabrahams679]

The problem with the math in the switch option is that x is arbitrarily assigned to your envelope. You could just as easily assign the value 2x or 1/2 x to that and x to the other envelope.

[@ella8070.]

At 6:27 you are changing your value of x. When one envelope is X and the other is 2X that means X is easily to the lower value. When one envelope is X and the other is 1/2X or 2X than x is equal to the value in the first envelope. To clarify in situation one x is always lower but in situation two x could be lower or higher

[@BEN-ys6gu]

This problem scares the shit out of me. Thanks for reducing my confidence in any kind of logical reasoning that I had to 0. If the paradox is not obvious for this extremly simple scenario, I can only imagine it hidden in a complex scenario

[@gamingzzfornuthinzzz6067]

Kevin would be a great math teacher

[@nasdfigol]

Agreed

[@Sanguinaryyy]

I'm fckin high and I am gonna watch this, o'boy!

[@jonasrogier]

✧ Sanguinary same😆😆

[@guy3nder529]

oh yes thats always fun

[@san4os94]

weird flex but ok

[@Calliopeassociatio]

you're not alone, bud

[@sados7884]

You wanna talk about it bud?

[@iem123]

Kevin is the king of solving paradoxes I didn’t know exists.

[@quakquak6141]

the expected value only makes sense if we try the same experiment a lot of times, but we need to keep X constant (only with infinite tries probability truly converges) and yeah, if you can see the content of your envelope (let's say it's 5$) and there's a 50% chance switching will get you 10$ and a 50% chance switching can give you 2.50$ you only get the expected value if you play the game a lot of times, if you play it 100 times, always starting from 5$ and always switching you'll get 625$ on average instead of getting 500$, this only happens cause we fixed X, we decided it to be 5$ which makes the second envelope's possibilities restricted, either it contains 10$ or 2.50$, but if X is not fixed we need to take that into account when calculating the expected value, and the result is undecidable cause X is any random amount

[@iainmacleod7169]

This is a great comment, thank you. Am I right in saying we can't try the same experiment an infinite number of times such that probability converges if we have no fixed value for X?

[@f.p.5410]

Nope, doesn't work that way. As soon as you start doing your game with X constant, you've changed game. It's not the same game anymore and its results don't tell you anything about the original game. Also, you absolutely can play this game (both the right and the wrong interpretation) infinite times even when X is a random amount and, as long as the hypothesis of the law of large numbers are satisfied, the mean will converge. The result is never undecidable but it depends on "what kind of random" X is.

[@iainmacleod7169]

@@f.p.5410 Can you elaborate on that last sentence?

[@quakquak6141]

@@f.p.5410 the game is different, I agree, and this is where the math makes it look like a paradox, cause we are discussing about 2 different games at the same time without realizing it

[@quakquak6141]

@@iainmacleod7169 I'm not sure, I think if we don't have a fixed value of X we must consider all possible Xs, I'm not sure, but it would be a really weird infinite sum, it might converge to something, Idk

[@thebashfulturtle9987]

I feel like the faulty math comes from when you assume that the other envelope either contains twice X or half of X. If you actually gave the values of the envelopes values, unbenounced to you, you'd see how that math doesn't work out. If x is the lower value, it doesn't double the value of the other envelope because the value of both envelopes was predetermined. It's a little confusing but basically I can try to sumerize. The value of the other envelope is predetermined so it could be 2x or 1/2x, but the probability thereof isn't actually 50% it's 100% one or the other, even if you don't know, even if you guess right or wrong.

[@withtheworks]

That x/3 solution was so satisfying

[@EnzoHetJoch]

"But that make no sense!" -Kevin 2019

[@Daniel-gk1dp]

Enz Be shut up with ur quotes

[@EnzoHetJoch]

Appelspeler op kankeren

[@KoaFidCZ]

I feel like half of these guessings here are "but that makes no sense!" :D

[@penguinchess]

Plot twist: both envelopes contain $0 and Kevin just wasted 12 minutes of your life

[@ultimaxkom8728]

Plot Twist: The envelopes don't exist, you won't get anything no matter what.

[@jayantkolapkar2926]

Kevin looks like Jim Moriarty from BBC's Sherlock Holmes.

[@bradleybobbs]

This is a great apparent paradox! It took me a while to figure it out. But once I figured it out, it's really very easy to understand. I don't know why Martin Gardner would think that it was difficult to communicate. All you need to do is to see, in the expected value equation as shown, that the two Xs used are not the same X. Assuming that the Xs are the same is the error that leads to a paradox. The 1st X applies when it is the lower amount, while the 2nd X applies when it is the higher amount, and so should actually be 2X. Hence, the correct expected value is found to be 3X/2, not 5X/4. The expected value is found to be 3X/2 whether you switch or not. Thus, the apparent paradox is seen to not be a paradox.

[@isleohagger5455]

If you open the envelope and it's $20, then x = 20. Now you have an option to switch. You're telling me it's not a 50/50 chance anymore of getting $40 or $10?

[@bradleybobbs]

@@isleohagger5455 No, I'm not telling you that. But now I see that I still haven't figured it out. It still looks like I have a 50/50 chance of getting $20 more or $10 less. Oh no, this paradox is blowing my mind again!

[@GF-gl9xo]

The algebra in this “paradox” is all wrong. To understand it you simply need to put in some values. Use the lower value as 5$ and higher as 10$ If the first envelope is “x(5$)” the second envelope can be “2x(10$)” but can never be “0.5x(2.5$)” because that would make a third value in a problem with only 2 values. The only time the second envelope can contain 0.5x is when the first envelope changes it’s value to the higher of the 2. In this case being 10$. So you cannot keep the value of the first envelope a constant “x”. Making this algebraic equation to get 5/4x totally wrong, because the second envelope can never be both 2x or 0.5x where “x” is kept a constant value. You’re welcome

[@josiahferrell5022]

Thank you. The whole time I was thinking to myself "wait a second, that algebra smells odd"

[@Karavusk]

Yeah like "x" changes definitions based on larger or smaller assumptions. In these two cases we are not talking about equal "x", they are different. That is more like comparing y and x.

[@MetroidPrime35]

I was thinking this the entire time. But the real kicker is that the value in each doesn't matter, it doesn't change the probability of getting one or the other, and it is a probability equation, which is simply one out of two or 1 / 2

[@gieljanw]

There is also a difference between the paradox and Nalebuff's variation. When the contents of both envelopes are both unknown the expected gain of switching is 0. But let's say you get to open one and it contains 100 dollars. You now have the opportunity to open the other envelope at the cost of 100 dollars. Seeing how the remaining envelope contains either 50 or 200 dollars, the expected gain of switching is 25 dollars. Giving you the 1.25 from the start of this video.

[@AlistairCrooks44]

What you're saying is true, but only if we know the total value of the envelopes. If we don't, then the paradox holds. I think the problem in the paradox is more to do with the assumption of fixing the value of one envelope and changing the value of the other. It's not fair to do this, because the envelopes are dependent on each other.

[@kidkecleon]

The 5/4 isnt the value from switching, its the average return for playing the game. Not very confusing

[@Corey_Bee]

No, 5/4 is just bad math.

[@ertertz9408]

@@Corey_Bee I love how smug he sounds while posting (mostly) wrong stuff on KZbin.

[@VaporMagyk]

its only 5/4 because you assume that the other will either be double or half. at the beginning he said you had 2 envelopes with unknown values so you dont know if one is double or half it could literally be like +/- 0.01 and that would change the 5/4 expected outcome. he is literally changing the scenario to choose what he wants to do

[@honeybadger_sc9816]

Your average return for playing is the mean of the 2 numbers in the envelopes. 3x/2

[@Broockle]

I can't even follow the line of reasoning for that. It's all still a 50%chance anyway.

[@haloiscool1235]

The problem is that both envelopes are x. It’s wrong to think of one as x and the other 2x or 1/2x. Cause you don’t know if your x is either x, 1/2x, or 2x. You have two choices keep or trade. Keep your x so your assuming you have x or 2x. Or trade assuming you have x or 1/2x. Both are sound logic cause it’s impossible to know which one is x. Edit- Btw both are x cause both are separated equations. Try and solve a

[@Spanner1971B]

The way I like to understand this one is : You either have X (and would switch to 2X), or you have 2X (and would switch to X). X is defined as the smaller value. It's a slightly more readable version of x/3 and 2x/3 .

[@guy3nder529]

when you understand the correct solution throughout the entirety of the video.

[@guy3nder529]

@Joely7 haha yes! Perfect!

[@devdog7409]

This paradox is simply caused by having X represent two different numbers. If you had $5 and your opponent had $10 then by switching you would gain $5, so representing your original value with X you end up with 2X. When you consider the possibility of having $10 and your opponent having $5. That's half, so you assume you end up with X/2. The problem occurs when you compare 2X to X/2 and it looks like you have an advantage switching, not realizing the X equals different values in each possibility. X equals 5 in the first and 10 in the second. If you compare like this 2(5) to (10)/2 you can see there is no advantage.

[@DeepDishDepression]

The problem lies in that depending on which card you initially choose, the value of x is different, so I feel it should be delineated as y for my explanation. Let's say the cards have values x and 2x. If you pick the 2x and set it as y, then switching loses you an x. If you set the y to be x, then switching gains you an x. The expected value of switching ends up as 0. The problem lies within setting the card have picked's value to x, which changes the path of switching depending on what your x actually is, the 2x or the x. Hopefully this was explained well enough

[@chrisg3030]

I think I have an alternative explanation which doesn't involve another algebraic term, y. There are 3 equally probable possibilities for the contents of the first envelope you choose. They are x, 2x, or x/2. If it contains x, then if you swap you get either 2x or x/2. So you stand to gain x or lose x/2 If it contains 2x, then if you swap you get either x or x/2. So you stand to lose x or lose 3x/2 If it contains x/2, then if you swap you get either x or 2x. So you stand to gain x/2 or gain 3x/2. The total of all 3 possible gains is x + x/2 + 3x/2 = 3x. The total of all 3 possible losses is x/2 + x + 3x/2 = 3x. So no overall advantage or disadvantage in swapping. The popular mistake is to only consider the first possibility, namely that the first envelope contains x. This is due to the way the problem is posed. People think that once they stipulate the first envelope chosen contains x, then they can't have it possibly containing x/2 or 2x.

[@funwithsiblings7310]

@@chrisg3030 I might be wrong but you can't have 2x and x/2 together as the paradox stated that one of the envelopes contain double of the other.... But, 2x is 4 times of x/2 so that changes the situation completely

[@jolotabani]

I would pick the one with less money so Kevin can have more.

[@lusamine7925]

But how do you know which one?

[@spencerkeene6493]

Literally every time he made me go “what that’s not possible” I stopped the video to think about it for a sec and then literally thought up what he was just about to say

[@Breadaronie]

Simple solution: *don’t switch cuz if u get the lower dollar envelope u will feel bad*

[@Leo-ss5ss]

you should do a video on the Bertrand paradox

[@joharding5469]

1. Pick envelope 1 2. Switch to envelope 2 3. Switch back to envelope 1 4. ... 5. Boom! infinite money

[@ostrich_dog]

I thought it was going to be Monty Hall's paradox again

[@Lattamonsteri]

Me too :P Part of me still thinks that.

[@hakesho]

@@Lattamonsteri In a way it is. In Monty Hall the benefit of switching comes from the fact that information is gained after your first choice. If you ran a modified Monty Hall without this added information then there would be no benefit of switching. This is the reverse. No information is gained after your first choice so there is no point in switching, but if we ran a modified game where you do gain information then this could change. For example, if the value of the second envelope was decided after your first choice and was either double or half the value of your envelope, then you should switch every time by exactly the calculation given in this video. Even though that feels like the same game (there are still 2 envelops, one still has twice the value of the other etc.) the change in information introduces asymmetry to the choices.

[@floopmaster4252]

Video idea: Get the game called Farkle, then answer this. When is the best time to stop? When should you keep going? Thanks, I just got done playing with my family and barely lost which made me think if I had done something different if I could’ve won.

[@gizmoteknodekker7060]

Damn I love farkle!

[@manuelgehmeyr5626]

In this experiment we assume that X is uniformly distributed between 0 and infinity and this is not possible. We simply cannot work in a probability setting and terms like expectations are not well defined.

[@Draidis]

THANK YOU! I have been stressing over this for YEARS since I took stats in university. I always said the math was being done wrong but even my prof said that you had a better chance if you switched. Now I finally know I was right!

[@Max-ui5gc]

You trust a random guy from the internet over a maths professor?

[@Draidis]

@@Max-ui5gc lol fair point. In this case it seems to make sense but you're right.

[@trevorgrover5619]

Paradoxes happen when you don't know how to apply math to your problem.

[@user-ox5ir2rd6g]

Was thinking of having a separate unit instead of just saying envelope 1 is “x” (like you set it up at the beginning), only to realize it really just was that simple… I love these puzzles, they make you think a bit but they’re not too difficult to where you get frustrated trying to find the explanation.

[@cmilkau]

3:44 is the mistake. X is not the same value in both cases, but you later mix the two X's in one equation as if they were the same value. A more detailed explanation: say the left envelope contains $100, for example. It doesn't matter, it is just less confusing with a definite amount rather than an abstract X. Now you are saying the other envelope has a 50% chance to contain $50 and a 50% chance to contain $200, which is nonsense. Only one of both can be true, because you either definitely put $150 into the envelopes total or definitely $300 total, not randomly one of these. Now, of course you could say fine, I put $300 there for sure, but I want to refer X to the content of the left envelope no matter whether it contains $100 or $200. But then, X is not a single value, its value depends on the actual location of the $200 (you could call it a "random variable" or just a function). The correct formula then becomes 50% × x1 + 50% × x2, where x1 and x2 are the two possible values of X. You obtain a sum of $150, which is what you would expect. This is the correct answer and doesn't change by switching. The paradox, as always, comes from a misunderstanding. In this case, the misunderstanding is mixing up these two interpretations. The first one gives you a single value of X, the second one allows you to calculate the expected gain, but none can do both.