Can You Solve The Marriage Logic Problem? 80% Fail
Vložit
- čas přidán 11. 06. 2016
- Surveys have found 80 percent get this logic problem wrong. Alice is looking at Bob, and Bob is looking at Charlie. Alice is married; Charlie is not. Is a married person looking at an unmarried person? Answer choices are A) Yes, B) No, and C) Cannot be determined. The video presents the correct answer.
The video also includes a second problem. A drawer has 10 black socks and 10 white socks. With your eyes closed, you pull 3 socks. Are you holding a matching pair of socks? A) Yes B)No C)Cannot be determined.
Sources
Sean Carroll blog post (2009)
blogs.discovermagazine.com/cos...
Guardian puzzler (2016)
www.theguardian.com/science/2...
SingingBanana video (James Grime)
• Married Problem with s...
If you like my videos, you can support me at Patreon: / mindyourdecisions
Connect on social media. I update each site when I have a new video or blog post, so you can follow me on whichever method is most convenient for you.
My Blog: mindyourdecisions.com/blog/
Twitter: / preshtalwalkar
Facebook: / 168446714965
Google+: plus.google.com/1083366085665...
Pinterest: / preshtalwalkar
Tumblr: / preshtalwalkar
Instagram: / preshtalwalkar
Patreon: / mindyourdecisions
Newsletter (sent about 2 times a year): eepurl.com/KvS0r
My Books
"The Joy of Game Theory" shows how you can use math to out-think your competition. (rated 4/5 stars on 23 reviews) www.amazon.com/gp/product/150...
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 5/5 stars on 1 review) www.amazon.com/gp/product/152...
"Math Puzzles Volume 1" features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.5/5 stars on 11 reviews. www.amazon.com/gp/product/151...
"Math Puzzles Volume 2" is a sequel book with more great problems. www.amazon.com/gp/product/151...
"Math Puzzles Volume 3" is the third in the series. www.amazon.com/gp/product/151...
"40 Paradoxes in Logic, Probability, and Game Theory" contains thought-provoking and counter-intuitive results. (rated 4.9/5 stars on 7 reviews) www.amazon.com/gp/product/151...
"The Best Mental Math Tricks" teaches how you can look like a math genius by solving problems in your head (rated 4.7/5 stars on 3 reviews) www.amazon.com/gp/product/150...
"Multiply Numbers By Drawing Lines" This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 5/5 stars on 1 review) www.amazon.com/gp/product/150... - Věda a technologie
Bob may be a dog, answer is C.
@@StefV153 he is unmarried but not persona
You are right.
@@StefV153 Since when have dogs been persons?
Actually this IS thinking outside the box
😂
Bob could be the name of Schrodinger's cat, in which case he could be both married and unmarried at the same time...
But since Bob is seen by Alice, photons must be interacting with Bob, meaning the wavefunction will colapse on either married, or unmarried, and the superposition is no longer obtained. So no....
@@nilsastrup8907 Very nice
Wrong subject man!
@@nilsastrup8907 better still, if bob is a schrodinger's married person it is still 100% that a married person is looking at an unmarried person.
Ie the superpositional states are such that both states have a married person looking at an unmarried person
Not how that works but okay
Everyone knows Bob is married and Alice is giving him a stare down for looking at Charlie.
LOL
Distracted boyfriend meme
Exactly!
That’s what I though 🤣 that Alice and Bob were married and Alice was staring at Bob because he was looking at Charlie lol
lmao
Plot twist: Bob is quantum super positioned such that when we examine Alice, Bob is married, but when we examine Charlie, Bob is unmarried, this making the answer C.
The correct answer
But either way there's still a married person looking at an unmarried person, so the answer is A
This would make the answer B
@@Dusk-MTG Good point
Does not necessarily have to involve quantum stuff. How about country (or state) boarders. Each of them standing in a differnt country (or state) while watching each other. Depending on the countries law a marriage between two people might exists but not by law of the other country.
Even if i drew all socks out of my dresser i would not have a single matching pair...
maybe not lose one sock of all pairs you have;')
Dylon Lachenal That is what i was implying, you ruined the joke.
My brothers eat socks, if my mom buys us like 6 new pairs of socks, they'll be gone without a trace within a week
I thought you meant that you wouldn't have a _single_ matching pair if you drew all the socks from your drawer.
lol me 2
80% fail the test because most of em don't care enough to put any brainpower to them.
This is the reason. Anyone would be able to solve this in varying times.
I Can't Play -
Yes indeed. I also fall in that 80% category because I just didn't care to apply brains a little. As such logic has never been my interest, I'm more inclined to Geometry. But this is definitely easy.
Most underrated comment
Haha I thought about it for 5 minutes and still got it wrong
@DaFoRk_Zenwarrior 16 Yeah I wasn't thinking about logic or anything I kinda just thought, eh no way of knowing
I got so confused on the sock problem because my brain thought a matching pair was 1 white and 1 black sock lol
Haha, me too! I was like: What if you draw 3 socks of the same color?
@@CubeforFun hello fellow cuber!
Same here.
Yeah exactly 😂
@@nourganouni cfop?
When you presented the first problem, I was stumped! But when you presented the 2nd problem the answer was obvious, which consequently made the answer to the first problem obvious! What an EXCELLENT teaching method. I want to see if I can employ this in my own teaching.
if people picked out the answer randomly, more people would get it right
lol true
yep
Ya lol
+Penny Lane lol
Yes but only 10% more
Actually, Bob was wearing the socks, but wouldn't let his wife know if they were matching, so she divorced him, making him unmarried.
Top marks for best answer.
Are you holding a matching pair of socks?
Me: I don't know you didn't tell me to open my eyes
But still you would be holding a matching pair.
@@joshuapere997 I'm holding three socks there's a 33% I'm holding all black or all white ones
@@themartian6722 ...and? Even if u had all white or all black you’d still have a pair, you’d just also have an extra that’s the same color
@@samwood917 oof i struggled there for a bit lmao
@@samwood917 Not true, just because 2 socks are the same color does not mean they are the same size, or style or made from the same material. The only information in the setup is they are the same color without knowing these other attributes of the socks you can not know if they are a pair. The whole point of the original problem is using what you know from the setup but you also need to be aware of what you do not know.
Note that if Alice is looking at Bob, and Bob is looking at Charlie, and Charlie is looking at Dave, and Dave is looking at Eve, and Eve is... all the way to Zach.
And Alice is married and Zach is not, and all the people in between are undetermined, the answer is still Yes.
What about Aamanda? Who comes after zack. Then Bbrian, ccarl etc.
@@NetheriteMiner bro how can you forgot me. i was after amanda xd
@@experience8509 ???
Interesting. Seemed like something that can be proved by induction. Here's my attempt:
Let there be n+2 people, such that person 0 is married and looking at person 1, person 1 is looking at person 2, ..., and person n is looking at person n+1, who is unmarried. We claim that for a chain of n people between the married and unmarried person, we have a married person looking at an unmarried person, where n is a natural number. The case for n=1 is proved by the video. This is our base case (n=0 also works, trivially, if you include 0 in the natural numbers).
Let us assume that this holds for some k people.
Then, for a chain of k+1 people, consider person k+1. If person k+1 is unmarried, then by our assumption, we have a married person looking at an unmarried person. If person k+1 is married, then as they are looking at person k+2 - who we know is unmarried - we have a married person looking at an unmarried person.
Thus, by mathematical induction, this holds true for all natural numbers n.
@@ninesquared81 that sounds like a Presh Talwaker answer. Ignore the actual question and go off on some irrelevant tangent
extremely easy, who is target audience now?
Cannot be determined...
Omg
lol that was great
@Eolicus you fucking genius, you made my night man, kudos.
IKR it's just click bait
was the second prpblem just to make people feel better for missing the first one?
i got the first one but not the second one.. i feel like an idiot now :(
Maybe if you get the second question right but not the first one you might reconsider your first answer.
don't worry about it. There are lots of moments that I miss easy problems too. it's part of being human.
The second problem doesn't say only 10 black and 10 white. There also might be 20 green socks in the drawer.
When I first saw the problem I got it wrong but that question is much easier
"While you're thinking about this problem, I'm gonna give you more to think about" lmao
I got them both right... only one who thought it was easy?
No it was easy indeed 😅
Its ez
ys so easy
Yeah I was a bit disappointed
No, it was easy if one works the possibilities. If you did not it's just guessing, but it's easy. The second problem should say "only" but I assumed that it was only. If we want to do pedantic not all white socks are pairs, but that is too much detail.
Bob could be a horse... the problem doesn't state that Bob is a person, so the answer should be C
Or a widower.
In which the case is married
There are married and unmarried widowers, and widowers are persons
Well of bob was a horse the girl would be looking at an unmarried "thing"
for all we know, he could be the last tem in temmie village 😂😂
Basically Alice loves Bob secretly while Bob is gay. ┐(´д｀)┌
LOL
+galihvevo ranggalih uou can still name a girl charlie. also most males are not gay. duh!
Not most.. But there are! cat Johnson
HA
Lol Chiharu Yazawa
Unfortunately, I’ve seen this before. But it highlights that we often make guesses before analyzing what information we already know.
Thanks for getting my brain moving, now and many times in the past.
lesson learned. 80% of people including me don't bother wasting our last braincel to a riddle instead of watching Miraculous Ladybug eventhough were grandparents already.
I clicked on this video just after watching gang of secrets 😂😂
@@pampanag2626 ahhh i still can't believe mari told Alya she's lb! 😫
@@queensupit Clearly, they weren't joking when they said each episode will drop a bomb.
@@pampanag2626 ikrr!! 😭
@@pampanag2626 And what if later Adrien would tell Natalie that he's cat noir and Natalie has to keep the secret from Gabriel cause she didn't want Adrien to get hurttt!!!! JAKASNKALAAHVSKAMSUSHSMDJAS I don't think I can take it anymore 😩
I didn't get the socks problem at first, because I thought it meant that with your eyes closed you literally draw a sock... like, a drawing, a painting XD
I can't believe I'm that stupid lol, sorry I'll just leave the internet and never come back
lol you're not the only one
Lmao I'm dead 😭
Meeee I was like wtf do you mean, man?
LOLLL
You better come back to correct your 'cime'
the real answer to the second problem is no, you are holding a pencil
to draw the socks
With your eyes closed, you magically draw 3 perfect socks with a pencil. Did you trace a sock? A. Yes. B. Naa. See. Cannot be determined.
+Sean Martin (Czechoslovakia) it socks. get it?
You get an A+
IKR
For the third one, this was my thought process:
If you draw three of the same color socks, then you won't have a pair.
*Looks at the problem again*
*Questions thought processes and writes it down*
Dude literally same omfg I’m so ugh
I was also sure that in order to get a pair you need a one black and one white, so 3 identical is not a pair :P Maybe I'm not that good at socks.
@@jarlfenrir i thought the exact same thing lmao
3 of the same colour indicates you have 2 of the same colour which makes a pair.
@@diyaagrawal145 At least in math (0, 0) is not equals to (0, 0, 1),
0 representing a white and 1 repr. a black sock.
So if you consider all socks you draw and hold in your hand as one unit you would never draw a pair, because a pair has only two elements while you always draw 3. Although i doubt they would accept that as answer.
There's a famous Brazilian poem about this called "Quadrilha" ("quadrilha" is a tradicional type of square dancing).
João amava Teresa que amava Raimundo
que amava Maria que amava Joaquim que amava Lili
que não amava ninguém.
João foi pra os Estados Unidos, Teresa para o convento,
Raimundo morreu de desastre, Maria ficou para tia,
Joaquim suicidou-se e Lili casou com J. Pinto Fernandes
que não tinha entrado na história.
João loved Teresa who loved Raimundo who loved Maria who loved Joaquim who loved Lili who didn't love anyone. João went to America, Teresa to the convent, Raimundo died of disaster, Maria became a spinster, Joaquim committed suicide and Lili married J.Pinto Fernandes who had not participated in the story.
I identify as bi-marital. I am neither married nor unmarried
That would be a-marital. Bi-marital would mean you are both married and unmarried. Regardless, I am triggered.
Thats logical impossibility. Thats one of the most important rules of logic: something cannot be true (1) and untrue (0) at same time.
Lepa TheWarrior Sure it can google Schrödinger's Bob Cat.
Would you be engaged then?
+Lepa TheWarrior is that what the patriarchy told you?
Got the first one wrong, but the second right - shame on me
Same here!
same 2nd is fairly easy
same
You do know he's wrong, right?
+Nebula Mirage he's right
the question only asked if A person is looking at AN unmarried person
we don't know whether Bob is married but if he was then a person would be looking at an unmarried person
if Bob wasn't then Alice would be looking at Bob who is unmarried
the question didn't ask for anything more specific than what it states
it only needs any one person to be married looking at any unmarried person
This was an absolutely excellent math problem. I had to write it out, step by step, to understand the logic, but you're absolutely right. Man, I am humbled.
The first problem, I just thought that since Alice is looking at Bob, she’s married to Bob, so the answer must be A. Lol
lol same
What if Bob is married to Charlie?
@@maksim_tak *_it's given that Charlie is unmarried_*
The answer is A though?
Alice and Bob are always sharing secret messages.
without watching:
the answer is yes.
if bob is married, Bob looks at Charlie = True
if bob is not married, Alice looks at Bob = True
second problem : yes, you choose at least two socks of the same colour, since there are only two colours
Proud of u
And if Bob isn't human...?
+MatheusLCF then Bob is unmarried so its still true
+Michael Gerwitz "Is a married PERSON looking at an unmarried PERSON?"
The second problem is short on information. It did not specify "matching color" pair of socks, only "matching socks". It did not specify available types of socks, meaning you could draw a white ankle-high sock and a white knee-high sock, but still not have a matching pair.
Excuses excuses lol
I thought the same thing. Yes, you are guaranteed two socks of the same color, but not necessarily matching.
This reasoning is true, but it is also the "I didn't get the answer right but I'm going to blame it on this possibility so I sound smart" reasoning.
+ILikeWafflz I'm with him on this. The question is too easy that you somehow second guess yourself, and then you find other factors to the problem that would make you change your answer. If you went with your first answer without even questioning if that was technically right, then that just makes you one of the majority of people who doesn't even think twice when answering this kind of question.
I thought it C because the problem never said that you opened your eyes to check. (I'm not good a puzzles, or math. I'll find another channel)
I really found it highly fruitful... And it also proves that it is always better to go deeper into a question rather than just jumping to a conclusion... A really mind-boggling question and equally amazing solution... Really great..
Well if Alice is married to Bob and Charlie's a girl I bet they are all gonna be unmarried soon..
If Charlie is a man it is even worse.
@@azziademir lmfao
@Ademir Azzi Jr. Why is it worse? Go on. Tell me.
@@azziademir yeah, why is it worse?
@@azziademir * Better
You're not holding a pair at all because you're holding three socks not two :p
+Imitationsfilms I'm just being a goof. I understood the question
"one socket"? What we're now bringing electrical appliances into the mix?
+Harrison Harris
Great... now we're all toast!
+Joshua Williams
was that supposed to be a pun because it was electrifying
Ellis Brown
Yeah, but honestly, it was shocking.
what if bob is in a quantum state of both married and unmarried?
then the answer would be both yes and no at the same time; aka undetemined
No becouse whetever he is married or not the answer is yes. So if Bob is in a quantum state it would still be yes, simply in two various ways.
Well as soon as Alice observes Bob he will be in one of the two states leaving the answer at yes.
noodles6669
someone has to observe Alice though, since they're part of a closed system
what if bob was an asshole who cares.
Ah, that explains why he collapsed when I got the answer. (I assumed it was the rum punch.)
"While you're thinking about it I'll present another problem"
*whoa there, my mind can only take in so much*
Actually I don't like percentage "facts" on video (like 80% can't solve this) because it often doesn't come from real statistic data.
I agree with this
Int the first question, it needs to specify that Bob is human. In the second question, it needs to specify that there are no socks that are a color other than black or white, and that all white socks and all black socks match. Otherwise, the answer cannot be determined in either of these questions. It's funny that the maker of this video got both of the questions wrong even though they were the one explaining them 😂
What a fucking jackass, use common sense and stop trying to be smartass. You're supposed to assume Bob is a person dumbass.
Cesar Tolosa We are taught in school not to make any assumptions in logic problems. My apologies for being more educated than you.
+Neil Oh yeah, I forgot a CZcams question determines who has more education. ARE YOU STUPID? You probably got an IQ of error since you're so stupid.
+Neil you're not making yourself seem smarter, you're just being a fucking idiot. You're the type of person to stop someone mid sentence just because their grammar was slightly incorrect. Nobody likes that type of person.
With all of the identifications these days we do not know what "Bob" is. For example, my name is Bob and I am a cat. Therefore, I can neither be married nor unmarried as humans cannot understand interfeline relationships.
The sock problem is flawed. There are many more variations between socks than just color. The other problem is much better, because a person is either married or they are not.
Yes and also we don't know if there are only 20 socks in the drawer. There could be 10 black socks 10 white socks, 7 red socks and 16 blue socks for all we know.
Good point. They didn't nail down all the possibilities. With the other question, you had three distinct individuals with enough information to form a definitive answer, even if there were other people involved.
you are just overthinking the problem. these questions are simple logical questions and are not designed for getting into cia or Mib
+Ajit Singh Maybe you're underthinking this question.
I get u but I think we know what they meant
What I've understood here is that there are 3 people and 3 socks, so at least one of them is wearing a single sock (if not all of them).
The first one did confuse me at first, but the sock problem was pretty easy to comprehend😊. Tysm for making these fun videos that challenge ourselves
Yes, because if Bob is unmarried, then Alice is looking at an unmarried person. If Bob is married, Bob would be looking at an unmarried person. I'm so proud of myself
The problem never specified that all three of them are human. Bob could be another species altogether, in which case Alice isn't even looking at a person at all.
I assumed unique left and right socks for the second problem. Guess that wasn't true. Hahaha.
there's a difference between the Socks Problem and the Gloves Problem. socks count as being simmetrical, so can go on either foot
This is even most clever than the original answer!
Socks are usually but not always symmetrical, so during an important test that would raise a doubt and put smartest people in danger :)
I misinterpreted 'draw', thought "i can't reasonably imagine what socks i would illustrate if i were to illustrate socks" and got mad at the question =/
That's creative thinking though. :P
A question I have is: Why can't the guy draw 3 black/ white socks. It's randomised, so why not? So it can't be determined.
Me thinking about the logic and answering "Yes" in the 2 problems...
* I feel 🌟genius🌟 * Lol
I had seen the sock challenge previously, but not the Alice, Bob, Charlie problem. Thanks for explaining!
this video made me feel stupid.
good job...
+Fuzzybucket True XD still trying to figure out why this was in my feed though...
+Fuzzybucket are you dumb? If there is 2 colors and you have 3 socks how the fuck can you not have a pair? There would have to be 3 colors for it to be C😒
No if you didn't know the answer's you don't feel dumb you just are😒
there's a comment on here somewhere where they say "since they used the term "draw" my smartass answer was no because I can't draw for shit and I picked no socks" I couldn't stop laughing but it was kinda true at the same time so we all see this problem in different ways some right some wrong so no answer was right or wrong
Actually the second question is wrong, there may be more than just black and white socks
This is just a simple brain teaser. Stop adding unnecessary factors.
in a hypothetical you are suposed to assume the conditions given are the only conditions so asuming another color sock is there breeks the hypothetical creating a new hypothetical
Hot Dogger Hehe, I shall not be wrong at any cost!
It clearly says that 10 are black and 10 are white. Stop adding random crap.
+Dulcie.Panda you still have a pair
I got this wrong, but I did learn something valuable. I answered the wrong question.
I was trying to find out, "is Bob married or single?" but that was not the actual question.
The actual question was, "Is a married person looking at a single person?". Irrespective of whether Bob was single or married, in both cases the answer is yes.
Thank you for the wake-up call!
Second case is even simpler. You overthought it. You're going to have matching socks by the simple virtue that there are not three different types in the drawer.
I thought c at the second question, I tought of left and right socks
ah true
my socks are ambiguous
Tim Rodd
lol forgot those existed
I never thought of it that way. Technically, you are correct.
I didn't thought of left-right, I just thought that same color socks may not pair (a very frustrating truth when I try to much the socks I just washed)
anyone else feel stupid after the first one?
Nope. I felt like I got it right but also learned a different way of thinking as a result. If you also learned from it you aren't feeling stupid, you're feeling smarter.
+Glenn Hurd That's such a positive take on it 😂✌ thanks pal
EpicTrolledYouBro lowkey
nope just you
no it's VERY easy
For the sock question, just because socks are the same color doesn't mean they are matching, so the answer is still C. There's a whole set of other criteria that need to be met for socks to be matching (size, length, style, whether they're toe socks or not, etc). The drawer could have 10 different white socks and 10 different black socks belonging to 20 different people.
If Bob is married then hes looking at Charlie who is unmarried so yes, if Bob is unmarried then Alice is looking at Bob so still yes. This is the easiest question i ever seen what did i miss?
Apparently people are bad at uncertainty, or think they must also know who is married and unmarried
Yes, but it never states whether Bob is a person or not.
some people are lazy to think and just say " ah we dont know about bob so the ans is C "
@@randombanana640 to be fair i sat there for quite a while thinking about it and then still said c, i think it's less laziness and more intelligence. maybe that's just me being autistic though
I like the example James Grime gave for this to show how this method applies in real mathematics.
Question: Can an irrational number (= a number, that can't be represented as a fraction) to the power of an irrational number be a rational number?
Answer:
We're taking a look at the irrational number sqrt(2).
sqrt(2)^sqrt(2) = x
We don't know, whether x is rational or irrational.
x^sqrt(2) = ( sqrt(2)^sqrt(2) )^sqrt(2) = sqrt(2) ^ (sqrt(2)*sqrt(2)) = sqrt(2)^2 = 2
So if x is rational, then sqrt(2)^sqrt(2) = x already proves our conjecture.
If x is irrational, the second line: x^sqrt(2) = 2 proves that our conjecture is true.
So we've proven our conjecture without being able to give a single, concrete example of it.
(PS: x is irrational)
You made a very uninteresting video interesting, and this in just one comment! Good job!
***** If you like that kind of stuff: What about transcendental numbers like e and pi? (Only on the real number line, e^(pi*i) = -1 is too easy ;) )
Questions like this may well come up in a first year Analysis course at the start of the course, it reminds me to some extent of problems you could be asked and the skills used in solving the question in the video are needed for Maths proofs in the course.
easy e^(ln2)=2 both e and ln2 are irrational.
May I put a question to the mathos here (I'm pretty hopeless at maths)? Does the A4 paper-folding kite problem need algebra, or can it be shown by a classical straight edge and compasses method? A4 paper (all A series paper) has a width to length ratio of 1 to sqrt(2). Two folds give a kite where two adjacent sides each have length sqrt(2), and the other two have lengths such that the perimeter is a rational integer (4?). The algebraic solution is simple, but I prefer appeals to symmetry and such!
So he just assumes that bob is married and therefor he is married..? That question makes no since, lack of information makes it cannot be determined. The question states the married person is looking at bob and bob is looking at an unmarried person, the question asks if a married person is looking at an unmarried person but it doesn't give the information of whether or not bob is married.
Regardless of wether Bob is married or not the equation holds true. Did you not watch the video? This was clearly explained.
Yes, there are two scenarios
A- bob is NOT married. Alice, married, is looking at bob. A married person is looking at an unmarried person.
B- bob IS married. Bob is looking at Charlie, who isn't married. A married person is looking at an unmarried person
Bob can only be either of two variables, married or unmarried. If we assume Bob is married then Alice is a married person looking a married person so the answer is not yes YET. Since Bob is married and he is looking at Steve (the question states Steve is NOT MARRIED) then the questions answer is yes.
If we assume Bob is unmarried then Alice is looking a bob so the questions answer is now yes.
So for both options of Bob being either married or unmarried Alice/Bob (one of them) will be looking at an unmarried person and so therefore there is a 100% chance that the answer is yes to the question and so we have enough information.
What a stupid idiot, watch the video and its so easy to solve.
+T U C K F R U M P - Must be one of the 80%.
Apply the same principle to a problem involving N people. Prove that a married person is always looking at an unmarried person.
So long as you can confirm that at least one person is married and at least one person is unmarried you can prove this to be true for all values for N greater than 1.
In cases where there is a mix of married and unmarried people, looking at the next one in line, the proof is obvious, so we only need consider the edge cases.
If the first person is married, and all other people in the line are not, then a married person is looking at an unmarried person.
If every person in the line is married except the last, then the second to last person in the line (married) is looking at the last, unmarried person.
@@resdamalos yeah but if every person in the line is UNmarried except the last, then it doesn't work
Here's my attempt at a proof (by induction):
Let there be n+2 people, such that person 0 is married and looking at person 1, person 1 is looking at person 2, ..., and person n is looking at person n+1, who is unmarried. We claim that for a chain of n people between the married and unmarried person, we have a married person looking at an unmarried person, where n is a natural number. The case for n=1 is proved by the video. This is our base case (n=0 also works, trivially, if you include 0 in the natural numbers).
Let us assume that this holds for some k people.
Then, for a chain of k+1 people, consider person k+1. If person k+1 is unmarried, then by our assumption, we have a married person looking at an unmarried person. If person k+1 is married, then as they are looking at person k+2 - who we know is unmarried - we have a married person looking at an unmarried person.
Thus, by mathematical induction, this holds true for all natural numbers n.
Who else thought that the questions were easy and didn't take the time to think?
My answer is YES. No matter what Bob's status is, there always be one married person looking at an unmarried person. If Bob is not married, YES. Alice (who is married) is looking at unmarried person (Bob). If Bob is married, YES. Bob is looking at unmarried person (Charlie). PS: I wrote this comment before I watched the whole video
This doesn't make sense.
+Petit Orenji you definitely need the brain nutrition supplement.
+Petit Orenji it makes perfect sense. If Bob isn't married then Alice who is married is looking at him, meaning in that situation the answer is yes, a married person is looking at an unmarried person. If Bob IS married and he's looking at Charlie who is unmarried then the answer would also be yes, a married person is looking at an unmarried person. In both of the 2 possible situations the answer is yes which means the answer can be determined as yes :)
youre actually right, the guy in the video is wrong
Too many ifs, therefore cannot be determined 🤷♂️
Stop easy problems!
you want a hard problem?
okay, here it goes:
What causes the phenomena known as dark mass and dark energy?
lelwat
***** This channel is about math so we want hard math problems
this is not related to math puzzles
mathworld.wolfram.com/UnsolvedProblems.html here, I hear there is money to if you solve them ;3;
The answer in the sock problem makes an assumption that the problem does not: that there are no socks that aren't black or white in the drawer. If, for example, there are ten white socks, ten black socks, and also two blue socks, there are ten white socks and ten black socks, but I could pull three socks and get {black,white,red}, and not have a matching pair.
I got the sock question without even thinking about it. Even though they're the same basic principle, for whatever reason the married question through me for a loop because the second I knew we didn't know the married status my brain automatically decided there wasn't enough information.
What if you grab 3 black Socks but they're different brands? (Ex. 1 black Nike, 1 adidas, and one unbranded Walmart sock.) if they're different brands they're not matching making them not a pair. Plus, if you actually care about matching socks you ball them up so you just grab them At once, not one by one. Making the person grabbing the socks stupid anyway.
U r so right
I guess if you make assumptions instead of following the logic of the question, then yeah.
+ShootLikeOrlando no, because you are holding 2 pairs of matching socks
if you grab 3 white or 3 black socks then you will have a pair
since this is a hypothetical we assume at this point that collor is the only variable that needs to match to make a pair do to this being the only characteristic we know
Wow. I'm surprised I actually got both right. I knew the sock one right away but the marriage one took slightly longer.
It seems so difficult at first. But if we just break down the question or explain it briefly to ourself,we can easily get it. Thanks 😊.
I failed the Sock problem due to my pre-supposition that Sock pairs normally consist of Left and Right socks to make a pair. My failure for adding unasked for information into the question. However 'pair of socks' is actually undefined in the question - so is it correct to use the colloquial meaning?
Taking a guess before finishing the video.
A[M] -> B[?] -> C[UM]
Originally I was think C) can't be determined since we don't know Bob's status...but that's likely what 80% of people thought. Looked again and realized there are only 2 possiblities.
A[M] -> B[M] -> C[UM]
Therefore married Bob is looking at unmarried Charlie
A[M] -> B[UM] -> C[UM]
Therefore married Alice is looking at unmarried Bob
Solution:
A)Yes
C[UM]. Enough said.
Harrison Harris Glad someone noticed.
+Harrison Harris lol
+CrackaBox B[UM] More needed to be said.
"A[M]" not yet i still need sleep
For the marriage one: Yes because no matter if bob is married or not it is yes. If not, Alice is looking at bob. And if yes then bob is looking at Charlie. Both scenarios of a married looking at unmarried. For the sock one, every scenario results in a matching pair, whether you grab all white, all black, or a combo, you’ll at least have 2 of one color
Bob could be a dog, answer C
technically it doesn't specify that all black socks are the same or all white socks are the same, so if one wanted to get nitpicky then the answer to the second question would be c. we dont know if the 2 black or white socks are identical, we can only infer this information not actually know.
Thanks for two easy ones. Really made me feel better. Most of your problems demoralize me. LOL.
There’s an issue with the second problem.. it never says they are the same style so you could have 10 different styles of socks in white and black. A white ankle sock would not match a knee-high or even another ankle sock with different branding
the clock in the background
HOW THE FUCK
Clock?
what clock? where?
Shikhanshu Agarwal It's faint as shit
+astphaire Really?
All depends on what City and State you live in....
How?
ye how???
Bruh someone put this in a cringe compilation
*cringe*
yes cringe
It took me a few seconds, but I eventually got the first one right. The second one was extremely easy, but since I'm a programmer who deals with writing lots of sorting functions, it's natural to me.
1. Yes, if Bob is unmarried then Alice (married) is looking at an unmarried and if Bob is married, then Bob(married) is looking at Charlie(unmarried)
2. Yes, if you draw 2 socks then the only way there isn't a pair is if it is 1 of each, drawing another ensures you have a pair
The answer for the first one is No. Charlie could be gay and be married but it isn't recognized as a normal "marriage". JK, but in this day and age, it could happen.
i love your ass! 😂😂
Yeah dude... Sometimes I feel stupid when I'm really smart. The answers of these particular questions are not certain. I could tell what's going on, but the tester is being a dick. When the answer is not satisfying, I am considered as "fail".
When you do logic problems like this you have to compare the information given. Same with things like math problems in school, overthinking it doesn't solve the problem.
+Glenn Hurd Agree. But sometimes in life, we find some people trying to assess our intelligence with unfair unexpected test. When we answer it pragmatically, the tester actually intended an outside the box answer, and vice versa. This is nauseating :/
+Nendra Affandi Exactly, I don't remember the title of a book I read about success. But something that I remember is that there was data that showed a high IQ person did poorly on a test of imagination (like you have a rock a sticks: what can you do with it). People with lower IQ tests proved to be better (they gave more ideas than those who simply answered the question). I'm pretty sure the author of the book was Malcolm Gladwell.
you said pair, not color. just because they're the same color doesn't mean they're a pair.
that's dumb
exactly not enough info they didn't say if they matched or even if the socks had to match if their exactly the same or what so not enough info for anyone to solve the problem
If the information is not provided in a logic puzzle, it's because the information does not exist. That's just the nature of logic puzzles. The puzzle did not specify different kinds of socks exist in the drawer. Therefore, they don't.
***** It didn't make a claim either way. If you wish to prove that different socks do exist, then that burden of proof is on you. And, since the only evidence about the situation is the information given, you're going to fail. Therefore, there are no other socks.
"A drawer has 10 black socks and 10 white socks."~ It is as simple as that. Clearly they are just plain socks since there are no extra details. Don't create information that isn't there.
"Are you holding a matching pair of socks?"~ You don't need to be Einstein to figure out that two black socks are "a matching pair of socks" and that two white socks are also "a matching pair of socks. Again, don't create extra details, unless you have reasonable deduction for it.
first time i feel good answering yes in "did you figure it out?"
"Is a married person looking at an unmarried person?"
Honestly, it says "a married person" so it could be technically ur parents looking at someone
no one says there are only two types of a marriage status. you can be divorced, thats not married, nor unmarried. what if he is engaged? or his wife is dead? first correct answer is: C
dodekaedius I think the assumption is that you're either married or not married. There's really no other option there. Divorced, engaged, widowed, and whatever else is all in the not married section.
A divorced person is still not currently married = unmarried. The problem assumes (in my opinion correctly) that by definition, a person is either married or unmarried
Unmarried and divorced are very different things.
+Ralin Unmarried should just mean the state of not currently being married
timmifutz I get that that is the most sensible meaning of it, but that's not what it means. Divorced, Widower, Widow, these are all separate categories of marital status.
The one about the socks didn't say the drawer only contains the 20 socks listed. If it also contains red socks, then perhaps you get one of each color when you draw 3.
The second question did not specify if there were coloured socks in the drawer with the black and white ones. For that reason, the answer cannot be determined. If the drawer has ten black, ten white, and ten yellow socks, then the initial statement would still be true.
The second question only specifies colors, but not any other characteristic that will determine if they will make a matched pair, (such as size, fabric, or style) so, while you will definitely have two socks of the same color after drawing three socks, there is not enough information to say that those two socks will actually be a matched pair.
Alice is looking at Bob, Bob is looking at Charlie, and so on, in a chain of unknown length, until Yngwê is looking at Zhukov. Alice is married, Zhukov is unmarried. Is a married person looking at an unmarried one?
The answer would still be yes.
If everyone besides Zhukov is married, the Yngwê would be looking at unmarried Zhukov.
If everyone is unmarried except for Alice, Alice would be looking at unmarried Bob.
If the case is different than the above, you would just have to pick a married person and an unmarried person, and the problem would repeat itself.
So even with an infinite line, the answer would be YES.
You went for Yngwé over Yasmin, Yuri or Yvonne? And Zhukov over Zoë?
Yes, and you can prove it by induction:
Let there be n+2 people, such that person 0 is married and looking at person 1, person 1 is looking at person 2, ..., and person n is looking at person n+1, who is unmarried. We claim that for a chain of n people between the married and unmarried person, we have a married person looking at an unmarried person, where n is a natural number. The case for n=1 is proved by the video. This is our base case (n=0 also works, trivially, if you include 0 in the natural numbers).
Let us assume that this holds for some k people.
Then, for a chain of k+1 people, consider person k+1. If person k+1 is unmarried, then by our assumption, we have a married person looking at an unmarried person. If person k+1 is married, then as they are looking at person k+2 - who we know is unmarried - we have a married person looking at an unmarried person.
Thus, by mathematical induction, this holds true for all natural numbers n.
The fact that anyone can fail this test is mind boggling. How did those people made it through school?
Some of us have a different interpretation of "matching". When I get caught wearing two different socks, which I do quite frequently, I tell people they really are a match and I can prove it because I have another match just like that at home. So I have ended up feeling that a "matching" pair are two different socks, ie. one black and one white in this case, hence the answer is C ;)
Alice and Bob are married. Alice is looking at Bob disapprovingly as her husband, Bob is looking at Charlie and Alice thinks her husband would cheat on her with Charlie.
Everyone knows why Alice is looking at Bob , he is married and looking at Charlie . Married to Alice to be exact .
For the second question, my mind initially answered yes, but after some thinking, came to option C,. ( because my brain flipped and said to me that a matching pair means a black and a white)
I got the first one right after some thinking because I'm a genius. I got the second one wrong because I'm not smart. :(
i believe the second could be cannot be determined if were tqlking specifics, because when it said "a drawer has 10 black socks and 10 white socks" it didnt say the drawer ONLY has 10 black and 10 white socks. POSSIBLY there could be different colors. but the answer in the vid works aswell
I just roll my socks up into pairs anyway so why would I need to pull out 3 individual socks?
wtf bob is still undetermined you can't place single status or married status on him if the question doesn't state whether he is married or not. the socks one I do understand.
Either way, a married person would be looking at an unmarried person.
Bob could either be married or unmarried. If either option proves the question correct, then you don't need to know whether he's married or not as the answer would still be yes.
+Undead Nightmare um or bob could be neither. he could live in another demension.
no but if he was married then he would be looking at an unmarried person. and if he wasn't then the first person that's married would be looking at an unmarried Bob. so if he was either married or unmarried, a married person would be looking at and unmarried person. you just have to look at all the choices
+Animares 822 that's the stupidest thing ive ever heard. He is in inception not the matrix/other dimension, i would know cuze i'm bob😢
exactly
I liked that one my first impulse was c but then i turned on my mind :P
Technically it isn't stated that there are no other coloured socks in the drawer, it could have 20 yellow, 30 red 10 black and 10 white in it, the statement that there are 10 black and 10 white inside drawer would still be true.
Question 2 assumes without stating that socks of the same colour are a matching whereas in reality you could socks with varying sizes thus making it undetermined.
The second one I chose b because it says with your eyes close draw then socks, then proceeds to ask if you are holding a pair of matches. I chose no because you weren't holding anything in the first place.
just interpreted it different
+Taiquan Evans i thought of course not, i'm drawing, i'm holding a pencil.
ikr!!!
+auzzty41440 G 😶was that some shade bruh?😂 I thought this was like a trick question because these are tricky questions
You're looking too deep into it, that's the problem. Just see it as is. If you pick 3 socks of two different colors, you will ALWAYS have a pair. In this case, it was black or white. The possibilities are; white, white white; black black black; black white white; black black white; white black white; and black, white, black.
But what if you pull three white socks in a row.
White + White = Pair + extra sock
I thought he meant drawing a sock.... Aka with pencil .
+Random Guy well fucking done dude
Then you have a pair of white socks and an extra white sock...
+Random Guy that could possibly used as Well (I think) just determines how you see the word draw in which case if you can't draw for shit your answer would be no and you didn't pick up any socks so... yeah or you can have different brands of socks in which case you probably graved 3 different socks so not a pair but that's if u assume there's a million different ways you can solve this problem they just used the most logical explanation welp that's depends on what the person thinks and the different ways humans see things so none here is dumb nor smart
The first one i didn't get but the second 1 i got instantly, the second question walks u too the solution while the first just confuses u, so it seems more like a trick your friend would tell you, while the second is like something u would find on an exam.
If bob is married, he is looking at an unmarried but if hes unmarried, alice is looking at an unmarried
Hi, I'm unmarried like it said in the video
You made this comment four years ago, are you married?
@@britishsongbird I forgot I even left this comment
So are you married?
married yet lmao
single pringle charlie for lifeeeeee
For the socks problem, I choose c) because it is not determined if there are more sock types. XD
Maybe I complicated it too much.
what if you pull a black sock, white sock, and a white sock that says it color identifies as a yellow sock?
Finally I figured out both problems after not guessing all the others from previous videos. I feel my brain expanding