Brain Teasers | Brooklyn Nine-Nine | Comedy Bites
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- Опубліковано 25 гру 2019
- How dare you detective Diaz? I’m your SUPERIOR officer!
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It's obvious that Amy sees Holt as a father figure, but the part I like is that she says "our dads" meaning she sees Rosa as her sister.
That warms my heart!
Amy not ami
I'm teaching father the math
does that mean she and jake are siblings?
Hahahagross
"Kevin's right."
"You're fired."
i acctually think holt is right
@@joaoedu1965 no it's true Kevin's right the odds are in your favour to switch
@@doctorwitch.5548 Oh yeah i just heard it
And then Rosa little “ahaha”
@@doctorwitch.5548 Except they aren't. The "third door" is a red herring. It literally doesn't matter when you're talking about probability of the remaining doors.
Monty makes it *sound* like a sweet deal, and the theatrics of opening a door with nothing behind it makes it *sound* like you might have a "better" chance if you switch, but the reality is it was 50/50 from the start. Monty was always going to open an empty door, meaning you already either picked the door with the car, or the door without it. 1 out of 2 choices = 50%.
Changing your answer doesn't affect the probability at all, it was always going to be 50%.
Scully is actually a very valuable member of the nine nine! They just need to know to use his talents! He speaks french, he knows morse code, and that time he and Hitchcock figure out Holt and Kevin had a pie, also! Due his medical history he must had a lot of medical knowledge as well.
Russian? I thought he speaks French.
Yeah but they can't use Scully because he doesn't want to do any work. He says so in some episode and they were quite the detectives in their day.
@@MellowWater Sure they can! just don´t ask him to leave his chair! and bring him a pizza (Don´t worry about the cheese, he doesn´t) let him use his home chair at work, buy him lunch like Jake did once, that time Scully was even glad to move around and work a case!
he speaks french, not russian.
he doesnt speak russian he speaks french tho
Amy shaking in her chair when Holt is yelling BONE gets me everytime.
Omg great observation
I never saw that before now. I was watching how Holt was reacting to the word “BONE” still funny.😂
I’ve always just watched Holt. Rewinding…
Someone pointed that out to me for the first time about a year ago. It made a great scene even better.
HOW DARE YOU DETECTIVE DIAZ????? BBBBBBBOOOOONNNNEEEEEE??!?!!?!?!?!?
Broadway Bound my absolute FAVORITE SCENE!!!! Even more than the backstreet boys!!!!🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣💀💀💀💀💀💀
🤣🤣💯
@@derickspawn6706 thank you, thank you! *bows* 😁
"Because you understand the maths now"
"Nope"
"Because you guys..."
"YEP!"
Gets me everytime! Lol
@@chemicalcastration "See what happens when your dads have sex"
But the biggest brain teaser will always be how Skully and Hitchcock went from what they were in the 80s to what they are today.
Chicken wings 🤒
Cocaine
Wing slutz
Even funnier is that they're actually good detectives...
@Constant Geek cocaine
"I'm teaching Father the math" might be her best line in the whole series
I don't know how anyone kept a straight face while he was screaming "BOOOOONE!?"
They know better piss off the boss while he's going off especially over personal issue you shake and nod plus that emotional i felt it BONE
Except Amy 🤣😂
BOOOoooooOOONNnnEEEEEEEEEeeEe
I don't know how they didn't breaked.
practise makes perfect
"Congratulations. We did it, and we did it together because we're a team."
"should we do it again next week sir?"
"No. I would hate that."
"u use the seesaw to press down on their necks until fatty confesses" LMAO LOLOLOLOL
“No, I found another key you dolt.”
Capt’n has no chill
Mac SteezyWayz he calls terry a bitch in front of him during a chess match with gina
What was the 4th key?
Bruh. But Hitchcock was extremely funny😂😂😂😂
"sorry we're late, turns out the front door was a push not a pull" ima be using this whenever I turn up late to anything from now on
Well obviously you disassemble the seesaw, take the planks and turn it into a box wide enough to step in. You fill the box halfway with water and have everyone step in individually. Then you mark the water level inside the box by how it rises. Everyone's water level should be equal except for the odd one out, whose level would be lower or higher. Something something mass, density, volume.
And technically, you only use the seesaw once. To make the box. Then you just use the box.
Or something like that. My last brain cell gave up ages ago.
🤯
That wouldnt work actually. The problem is some of the water would cling to the people and it wouldnt be exact. Theres also the issue of volume vs weight. Even if two things have the same volume and make the water displace to the same exact height, they may weigh differently. Its a matter of how dense they are. Someone with more muscle mass may be more dense but the same weight as a bigger person with less muscle mass.
Heres your solution:
Divide them into 3 groups of 4 people.
Put any two groups on each side of the see-saw. (First Use)
Condition 1
If the see-saw balances, we are sure that the oddly wieghted one is in the other group of 4.
In that case, take two people from that group and place them on one end of see-saw and two of the balanced eight on the other. (Second Use)
Condition 1.1
If the see saw balances, remove all but one from the seesaw and put one of the remaining two opposite them. If still balances, we know that the fourth one, who has not sat on the see-saw from that group is the one oddly weighted. (Third Use)
Condition 1.2
If the see saw is not balanced, remove one from each end. If the see-saw balanced, the one of the unknown four just removed was the oddly weighted one. Otherwise the one who stayed is the oddly weighted one.(Third Use)
Condition 2
If the two groups of 4 don't balance remember which side was lighter, have three get off one end and the remaining person swap places with one of the other four. Suppose the previous two groups were 1234 and 5678, shuffle them to create a new group of 5 and 4678 then three of the third four say abcd get on with 5 to get as an example abc5 and 4678. (Second Use)
Condition 2.1.1
If the position of seesaw does not change and as an example say 5678 and then 4678 are heavier, we know that either 6 or 7 or 8 is oddly weighted. Now put 7 on one end and 8 on the other. If one is heavier they are the odd one otherwise it is 6. (Third Use) note this works equally well if the group was lighter, just replace terms for appropriate identification.
Condition 2.1.2
If the seesaw reverses, ether 4 or 5 is the oddly weighted one. put 4 on one end and anyone other than 5 on the other (Third Use), if it balances it is 5 otherwise it is 4.
Condition 2.1.3
If the seesaw balances we know that either 1 or 2 or 3 is oddly weighted. Say as example 1234 were lighter. Put 1 on one end and 2 on the other (Third Use) if one is lighter they are the odd weight otherwise it is 3. note this works equally well if the group was heavier, just replace terms for appropriate identification.
Done - easy peasy
It is easier than everyone makes it. A seesaw is binary. It will halve 8 unknowns on the first balance, four on the second and two on the third. Set it up so deduction eliminates everything else and your gold. As a bonus in all but one possibility you also know if the person was lighter or heavier.
(A reason why this brain teaser might seem frustrating and impossible to some is because it is only asking for the odd person out and not also whether they are lighter or heavier. It is impossible to know both for sure in only three steps.)
Edit: In 11/12 cases you know whether the person is lighter or heavier as the seasaw dictates it. The only case where you don’t is 1.1.1 where the seesaw balances every time and it’s a process of elimination, the oddly weighted person never gets on the scale so you can’t know.
JoJo Lerner if the object is floating, water displacement is a factor of weight. Just make the boys float
@@jojolerner113 I thought the same but youre parcially wrong
EXCUSE ME?
DUDE HAS NO ONE NOTICED THAT THE TWO PRODUCES OF THIS SHOW ARE
1. MARSHALL BOONE
2. NORM HISCOCK
FIRE MARSHALL BOONE, NORM SCULLY, AND HITCHCOCK!
Yeah, they named some of the characters after producers. if you watch the first episode, terry introduces holt to Scully, Hitchcock, and daniels. Daniels is also named after a producer. dunno why they didn't continue daniels though? Nice spot with Marshall Boone though.
Why?
I KNOW RIGHT I NEVER SAW ANYONE WHO REALIZED IT AS WELL
You forgot Micheal Scully
sorry i've binged the show twice and idk what Marshall Boone is in reference to?
7:14 is one of my absolute favorite moments of the show
Any figuring it out and saying Kevin is right and Holt’s long pause and says “Your fired” and she got all scared has me in tears
For y'all wondering who's right in the monte hall problem (i think that's how it's called) 7:03 Kevin is right. Although yes it does sound like a 50-50 shot, there's something called variable change. It makes more sense if you have 100 doors. There are 99 ducks and one car behind all the doors. Say you select one door, the host opens 98 other doors revealing 98 ducks, which leaves just one duck and one car. AT first, you had a 1% chance of getting it right. Sure there exists the possibility of you actually guessing right, but 1% is pretty low, so switching would most likely increase your odds cuz the host left your initially chosen door (which is most likely wrong) and the other door (which is most likely right cuz the host knows where the car is). Hope this makes sense if yall need help I can explain it better, or the web can explain it better
is you search it online you can find a simulator that you can play
Hugo Garro statistically that makes a lot of sense actually but I don’t think there’s a wrong way to think about it. Like you said if he opens all 98 other doors accept for the door you picked and 1 other door it’s just a 50/50 guess as to witch door you want to end up with. It just has extra steps, at least in my head
@@thepriorstone4064 But it isn't 50/50. If you pick 1 door out of a hundred, there is a 1% chance you picked the correct door, and a 99% chance you picked the wrong one. If the host removes 98 wrong doors from his group, there is still only a 1% chance you have the correct door and a 99% chance that he does. The percentage chance that the right door is in each group doesn't change, it just condenses onto the remaining door in the group.
Hugo Garro
Y'all are overthinking it,
You guys just need to bone.
That theory is taught in basic statistics. Kevin is right.
Man...Scully had a sad childhood. He was left in i think France and his Dad was POW in Korea :(( but if none if that happened, Scully would be completely useless HAHA
What would win in a fight: every “joke” ever on The Big Bang Theory, or Captain Holt saying one word very loudly
We know the answer, don't need the question!
Ooh I’d love to see Captain Holt and Sheldon go at it with their sassy quotes, that would be hilarious😂😂.
BONE?
BONEEE
BONE!!??
9:06 to 9:23,
Amy's expressions brings about so many childhood memories.
That turned dark quite quickly :D
The fact that Santiago actually had the "official" solution to the seesaw problem still kills me.
What you mean?
INcorrect, normally you would be right(6 on each side, then 3, then one, and if they weigh the same then it’s the third guy), but the key note is the guy could be heavier OR lighter, meaning that you have a 50/50 chance of screwing up on the first measure since you don’t know what to look for, the side going up or the side going down.
@@walkermcmullin3106 Thank you, I really wanted to know what I was missing :D
Nope, she's wrong, the 1st move is 4 vs. 4
@@JMacSD but there's 10 man in total.. which 2 to leave out?
re: Monty Hall problem, you have a 33% chance of selecting the correct door initially, or a 66% chance of selecting the incorrect door. When provided an option to switch, the only time you would be correct in not switching is if you correctly guessed the 33% door initially (the statistically less likely option of the two) whereas if you switch you are 'selecting' the 66% chance retrospectively
8:23 I just love the way how both Jake and Amy see Holt as a dad
*Sweet home Alabama...*
Child: mom's right.
Mom: indeed.
Dad: you're grounded.
BOOOONNNEEEE!?!
holy shit 1.7K likes for a two syllable word?
WHAT DID YOU J U S T S A Y!?
"How dare you detective Diaz I am your superior officer!"
I don't actually understand why he yells that. It doesn't really make any sense.
Affan Khokhar He's offended and defensive because it's crass but also 100% right.
Nic Q if he finds it so crass then why would he yell it aloud in his workplace?
“Should we do it again next week sir?”
“No, I would hate that” Captain Fucking Holt😂
Ironically, Gina was the most useless out of all of them
@@BlasterzHD super true
I love when Captain Holt screams "BONE!?" Amy looks so cute rocking herself like she's boutta cry,and idk why thats so cute to me.
6:07 Did Holt just say "Buyakasha!"? Ali G would be proud.
he said it so quietly with so much intensity
Gina always asking the right questions
There are 12 people, 11 weigh the same, 1 weighs slightly heavier or lighter than the rest. We can only use the seesaw 3 times. Identify the odd one out.
My answer is kind of messy and there's probably a cleaner way to solve this but this is how I did it.
Lets label our people A, B, C, D, E, F, G, H, W, X, Y, Z.
First measure: 4 vs 4
ABCD vs EFGH
If they weigh the same, the odd one out is W, X, Y, or Z. Let's label this Situation 1.
If they do not weigh the same, the odd one out is A-H. This is Situation 2.
(It does not matter which side is heavier or lighter because we do not know if the odd one out that's causing this imbalance is doing so due to it being lighter than the rest or heavier than the rest.)
For the sake of the argument, lets say that ABCD are lighter than EFGH.
ABCD = light group (L)
EFGH = heavy group (H)
WXYZ = safe (S)
Situation 1 Second measure: 2 vs 2
WX vs AB
Keep in mind in this scenario A-H are safe. No need to measure WX specifically with AB.
If WX weigh the same as AB, the odd one out is either Y or Z. If the seesaw shifts, the odd one out is either W or X. Let's say the seesaw shifted.
Situation 1 Third measure: 1 vs 1
W vs A
If W weighs the same as A, X is the odd one out. If the seesaw shifted, W is the odd one out.
In the previous step, if it was discovered Y or Z is the odd one out, do the same thing for that pair (YZ) as we did for this pair (WX).
Situation 2 Second measure: 4 vs 4
ABEF vs CGWX
LLHH vs LHSS
Here is where things get messy.
Most of the people that were weighed the first time around is here again, except for D and H. So, if this measure is balanced, that means the odd one out is either D or H. If this is the case, measure D against Z (or any other person in the safe group). If D is lighter than Z it is the odd one out. If the seesaw is balanced, H is the odd one out.
If ABEF turns out to be lighter than CGWX this tells us that the odd one out is either A, B, or G. This means A or B is lighter than the rest, or G is heaver than the rest. (A, B, and G are singled out)
If ABEF turns out to be heavier than CGWX this tells us that the odd one out is either E, F, or C. This means E or F is heavier than the rest or C is lighter than the rest. (E, F, and C are singled out)
Let's say ABEF is heavier than CGWX.
Situation 2 Third measure: 1 vs 1
E vs F
If the seesaw is balanced, C is the odd one out. If the seesaw shifts, the heavier one is the odd one out.
In the previous step if ABEF turned out to be lighter than CGWX, we would measure A vs B. If the seesaw is balanced, G is the odd one out. If the seesaw shifted, the lighter one is the odd one out.
Daniel Na you are a genius
I went with starting with 5 on 5. I think it works.
@@vinoandretti I think it does, here's my logic anyway:
If you do 5 vs 5 on the first try 2 people are left out - but that doesn't matter to us as, if the scale's balance shifts, then we know that they weren't important as they must be one of the 11 who weigh the same else if the balance doesn't shift that means that one of our two people left out are the culprit we are looking for which would mean we'd only have to weight them against each other to find out our culprit which would leave us with a remainder of 1 go on the seesaw. For this case we will assume the scale tips in favour of one of the group of 5, which removes 1 from our 3 scale tries. We are left with 2 tries left.
Whichever group of 5 tipped the scale will be measured against each other this time, once again we have to leave out someone as to not tip the scales in favour of one of the groups. Just like before if it tips we've cut the number of people down once more (to just 2), else we've found our guy. This subtracts 1 from our 2 tries - we only have 1 try left!
We're down to 2 people, we have one go left on the seesaw and it's all we are going to need to solve our mystery if it still remains elusive to us. The two are pit against each other on the two seats for the seesaw, and we finally see the one which weighs more/less.
What this entails is that realistically we have used up all 3 tries on the seesaw but have come out with an answer of who weighs more/less. Although there is a chance we find that answer with 1 try or even only 2 tries - though that alternative is less likely and would cut this explanation short without a full explanation as to if this works without luck on our side.
4A=4B
2A=2C or 2A>2C
1A=1C or 1A>2C
1C is the answer
4A>4B
2A=2B or 2A>2B
1A1B>2C
1A is the answer
4A>4B
2A=2B or 2A>2B
1A1B
Can't you simply split twelve men into three groups of four. Put any two groups on opposite ends of the seesaw, and you can see which group is heavier from having the odd man. If they are equal then the odd man is in the third group. Split whichever group has the odd man into two groups of two. Set them on seesaw. Then split whichever of those groups has the odd man into one man against the other on the seesaw.
"Captain Dad is just my boss"
Me: Ahahahahaha, that must be the best line ever
"BONE!?"
Me: Wait...
0:26
😂😂😂Damn..hitchcock and Scully marked their territories 😂🍕🍕🍕
What’s funnier is that he had to leave the door frame and then go back a half hour later for that rant
I love the little details like when Kevin and Holt having that math problem argument, Amy reacts worried when Raymond says “and now it’s a scene.” They were in character in every moment. Wonderful acting.
You’re fired.
What?
*ha ha*
I see a lot people trying to overthink this puzzle but you guys just need to bone.
"WhAt'D yOu SaY?"
Robert Callaghan
*I said you need to bone.*
BONE?!
@@excitedcat9517 HhHhHow, DARE you, Detective Diaz, I am *yOUR-SUPERIOR-OFFICER!*
BOOOOONNNEEE?!???!
You can see how traumatic amy looks when holt got angry at rosa
Holt, Gina, Hitchcock, and Scully in the Escape Room is still my favourite segment of the show :) xD
The one word to never say in front of Cpt. Raymond Holt:
*BONE*
Hooooow dare you De-tect-ive Di-az...I am your SUPERIOR OFFICER
Spideybops BOOOONE??!!
Incorrect..
*Damn*
..and disturbing.
Its a very diaz answer it be weird if she wasn't answering something extreme/violent just makes it better she's a cop
Holt, this is Diaz we are talking sbout, you know that right
Apparently Jake isn’t the only one who sees Holt as a dad
The biggest brain teaser is how Captain Holt calls his mother YOUR HONOUR😂
She's a judge. That's not a brain teaser.
1:20 - Actually, touching Rosa IS how I want to die.
The trick to the Monty Hall problem that makes it not 50/50 is that the host can't pick A) the door you chose or B) the car. If you didn't pick the car first time (2/3 chance), that rules out 2 doors, leaving only the car so switching wins 2/3 of the time. If the door was merely opened at random, you wouldn't be able to deduce any extra information, and it would indeed be a 50/50 chance.
*Funky cats and their feisty stats*
I thought I knew the answer to the seesaw one since I first saw the episode. Turns out I was wrong as soon as I heard lighter OR HEAVIER.
9:15 the best moment in this scene
4:26 Hitchcock kills me everytime lol
The last time i was this early .. i was diagnosed with dianne weist infection....... likeeeee yeasssssst
It's Dianne Weist
@@LuisCastro-jw5pi my bad haha
@@LuisCastro-jw5pi *wiest
"YO, HOW MUCH YOU GUYS WEIGH?"
done.
The easiest way to understand 'monty hall' for me:
you have 3 doors: wrong door nr1 (w1), wrong door nr2 (w2), and the correct door withe the car (c)
if you don't switch you have three possibilities with equal probability (1/3):
- picked w1 => you lose
- picked w2 => you lose
- picked c => you win
if you switch (changing to the only other door) you have three possibilities, same prob. (1/3):
- picked w1 => host opens w2, change to c => you win
- picked w2 => host opens w1, change to c => you win
- picked c => host opens w1 or w2, change to w2,w1 respectively => you lose
so not changing wins in 1/3 cases, while changing wins 2/3
dont just tease my brain captin...really go to town on it 😂😆😆 best thing ever
3:54 Capt. Holt gets me everytime! 😂💖
IDK how they pulled all of these clips and not a one had Jake!
Because Jake would solve them all in 3 seconds.
Don't just tease my brain. Really go to town on it....
They only found 3 keys though: Knight, TV, Phone.
The key was in a can of peaches, somehow
Skully ate peaches and found one of them.
I imagine there were just many other clues to help with the coordinates and morse code during the escape room, just Scully and Holt had other experiences.
I love that during the BONE scene, Amy's rocking back and forth at 5 and 25 minutes later 😂
See saw answer: divide the 12 into groups of 4. Put two groups onto it. If even, then the other group has the heavy man. Take the heavy group and split it into two groups of two and put them on. Then the heavier group into one and one and onto the see saw. Bam
The issue here is we don't know whether the man is heavier or lighter. So if one group is heavier than the other, you cant be confident which ones has the man.
Plus what if the first group is lighter or heavier
@@RogerYu612 thats true, but you dont need to find out which one is lighter or heavier just if they are lighter or heavier. So you can go through the same process. 4-4 if even then do one of those groups with the last one. If it goes up then theyre lighter, down heavier.
@@RogerYu612 if they arent even in the first group you can still use the see saw another two times, so you can weigh them all in groups of four and find out if theyre lighter or heavier.
@@RogerYu612 for ex. If we use groups A, B, and C and they all consist of 4 ppl. AA then A is lighter. A=B, BA then C is heavier.
The Monty Hall problem was the most mind blowing thing I came across before I finished high school. I understood the logic but intuition (though faulty) said otherwise. I was convinced when I wrote a problem to simulate the problem.
I have watched Holt freak out over the Monty Hall problem so much, but I still need an extended version of that clip.
,"BOOONE" ,"diaz what happens in my bedroom is none of your business" ,"BOOOOOONE"
Hitchcock wanted to be in this so bad 😂
The bone clip will endure
The solution is really simple you divide the 12 population into three groups randomly i.e 4 people in each group and then select two group and weigh them
case 1 the see-saw is balanced and the heavier person lies in the other group
case 2 the see-saw is unbalanced and the one side which goes down has the heavier person
you take the group which you have identified to be the heavier and then divide them 2-2 and weigh them the side which is heavier will go down and then weigh the 2 people remaining
and thats how you identify it with using see-saw just three times
The riddle says the person is heavier OR lighter.
@@mrsperkin7198 When it's lighter OR heavier:
Divide them into three groups (A, B, and C). Weigh A and B on the seesaw.
Case 1: The seesaw is balanced, one of the four men from C is the odd one out.
1. Take three of the C men and weigh them against three of either group A or B.
2. If it balances, the fourth guy from group C who hasn't been weighed yet is the odd one out.
3. If it doesn't balance, then rename the three from group C: they are now either H or L depending on if their end of the seesaw was the heavier or lighter side.
4. Balance two Hs or two Ls against each other. If they balance, the third guy is the odd one out. If they don't balance, then if they are Hs, the heavier one is the odd one, and if they are Ls, it's the lighter one.
Case 2: The seesaw is NOT balanced, so one of the eight men from groups A or B is the odd one out.
1. Everyone on the lighter side is renamed L and everyone on the heavy side is renamed H. The leftover group is still group C.
2. Make three Hs get off the seesaw and replace them with three Ls, and then on the other side, have three Cs and one L.
3. If the heavier side is STILL heavier, then either the single L on one side, or the single H on the other, is the odd one. Pick one of them (say L) and weigh them against one guy from group C. If L and C balance, it's H. If L and C don't balance, it's L.
4. If shuffling them around made the previously-heavier side LIGHTER, then one of the three Ls you moved to that side is the odd one out. Pick two of them and weigh them against each other. If they balance, the third L is the odd; if they don't, the lighter L is.
5. If shuffling them around made them balance, then one of the three Hs you got rid of is the odd one out. Pick two of them and weigh them against each other. If they balance, the third H is the odd; if they don't, the heavier H is.
Cribbed heavily from this TED-ed video: ua-cam.com/video/tE2dZLDJSjA/v-deo.html
@@mrsperkin7198 it doesn't matter apply the same logic the lighter one will go up go on selecting that groups
@@mauktiktiwari4589 but if you dont know if it's heavier or lighter and the seesaw is unbalanced which group do you choose?
Your answer only works if on the first weigh the seesaw is balanced
Divide into 2 groups (i know holt said you couldnt but im gonna assume it was comedic thing). Weight them. 1 side will be heavier, split that group into 2 again and weigh it. If it is balanced, then the man is in the lighter group, and is lighter. If it is unbalanced then the man is heavier and lies in the original group. Only requires 2 uses of the seesaw.
Here's the solution to the see-saw problem. The solution starts by weighing 4v4.
(1). If unbalanced (let's say to the left side), then the person is in the 8 people that were weighed. Weigh 3v3 by removing 3 from the left side, moving two from the right side to the left side, and then adding one from the original 4 (control group) to the right side.
- (1.1). If the test in (1) is unbalanced to the left side, the person is either the one that remained on the left side or the two that remained on the right side. Remove the one on the left side, and move one from the right side to the left side.
- - (1.1.1). If unbalanced to the left side, the person is the one that remained on the right side.
- - (1.1.2). If unbalanced to the right side, the person is the one that was moved from the right side to the left side.
- - (1.1.3). If balanced, the person is the one that was removed.
- (1.2). If the test in (1) is unbalanced to the right side, the person is one of the two that were moved from the right side to the left side. Weigh one against one of the original 4 (control group).
- - (1.2.4). If balanced, the person is the one that was not weighed.
- - (1.2.5). If unbalanced, the person is the one that was weighed.
- (1.3). If the test in (1) is balanced, the person is in the group of 3 that was removed from the left side. Weigh two of the three by adding them to the left side they were originally on, then move one of them to the other side. Leave one out.
- - (1.3.6). If unbalanced to the same side, the person is the one that was on the original side.
- - (1.3.7). If unbalanced to the other side, then the person is the one that was moved to the other side.
- - (1.3.8). If balanced, then the person is the one that wasn't weighed.
(2) If the original 4v4 is balanced, then the person is in the remaining 4 that weren't weighed. Weigh 2 of the 4 against each other.
- (2.1). If the 1v1 in (2) is unbalanced, the person is one of the 2 that were weighed. Weigh one against one of the original 8 (control group).
- - (2.1.9). If balanced, the person is the one that wasn't weighed.
- - (2.1.10). If unbalanced, the person is the one that was weighed.
- (2.2). If the 1v1 in (2) is balanced, the person is in the other 2. Do the same test as in (2.1).
- - (2.2.11). If balanced, the person is the one that wasn't weighed.
- - (2.2.12). If unbalanced, the person is the one that was weighed.
12 different outcomes based on which person is the odd one out.
"Is this how you wanna die, Hitchcock?"
LOLLLLLLLLLLLLLLLLLLLLOLOLOLOL
It's funny because Captain Holt is viewing the decision to switch doors as an independent event rather than conditional one and is therefore overlooking the fact that he should use Bayes' Theorem to solve for the probability that the prize is behind the other door.
The fictional police captain doesn't understand that Monty Hall knows which door hides the car thus him deciding which door to reveal ends equal probability. That's kinda funny, ok.
What's really funny is than when a columnist posted this problem & solution in a newspaper years ago, many readers wrote in declaring she's an idiot, some of them claiming to be math professors.
Teeter Totter of Taunts
Jade Nguyen i felt bad because my first thought was to try six vs six just like Amy
Terry: I'm saving them from this weight-obsessed nightmare island.
Amy: By murdering them?
Terry: I had to mercy kill the donut holes!
September 28, 2020, 6:18pm
No need Santiago it's all good
Yep
Nope
Yep
I will never stop laughing at this 😂😂😂
HOLTS "OH DEAR GOD" KILLED ME HIS ACTOR IS PERFECT
The solution:
Step 1: Weight 4 Vs 4 leaving 4 left behind
case 1: they are equal
case 2: they are not equal
Case 1 continue:
let us denote the last 4 left over as ? and the earlier 8 that are equal in weight as N
Step 2: Weigh ??? VS NNN
3 possible situations:
S1:??? is heavier
S2:??? is lighter
S3:??? is equal to NNN
S1 solution: Denote ??? as HHH since ??? is heavier than NNN
Step 3: weigh any 2 H together like so: H VS H
the heavier one is the one that weighs more than the others, if equal the one left is heavier than the rest
S2 solution: as above but denote ??? as LLL and the one that stands out is lighter instead of heavier
S3 solution: Weigh last ? VS N, if ? is heavier, then it is the unique one and is heavier, if it is lighter, then it is the one who is lighter, as all other units are normal weight
Case 1 solved.
Case 2 continue: they are not equal
Denote the 4 who are heavier as H and 4 who are lighter as L and the 4 left as N
you now have HHHH LLLL NNNN
Step 2 Weigh HHL VS HNL note there are 2 L left and 1 H left
if they are equal, then weigh the last 2 Ls left over: L VS L to determine if one is lighter
if they are equal again the last H is the one who is different weight and heavier than the rest of the island.
if one is lighter then it is the one that stands out, and is lighter than the rest of the island
as such, there are 2 possibilities left in which HHL and HNL are not equal
if HHL is heavier, weigh the 2 H in HHL: H VS H, if they are the same then the L in HNL is lighter than the rest
if they are not the same, then the one who is heavier is the one that stands out, thus heavier than the rest.
if HNL is heavier, then weigh the H of HNL with any normal in H VS N, if H is heavier then it is the one who is heavier than the rest.
if it is equal then the L in HHL is lighter than the rest of the island.
Case 2 solved.
thus, this is one solution for the puzzle. there are other ways to solve it, depending on the second weigh. but this is one of the easier ones.
You made that way more complicated than it needed to be.
Yeah, I did similar
Unknown=U, Maybe high=H, Maybe low=L, Common weight=C.
1st: 3 groups of 4 UUUUvsUUUU
2nd: path (=) UUUvsCCC or path(> or
@@bobafettjr85 I honestly can't think of an easier way to solve this question. The problem is that you need two pieces of information to answer this question. Both if the person is heavier or lighter, and who it is. If you already know the person is heavier/lighter, the question becomes much easier to answer.
@@jackieliu5775 The solution you used is the simplest solution it's just the way you wrote it out was longer than it needed to be and way more confusing than it could have been.
Your not trying to find which of the 12 is different your only trying to find out if the different one weighs more or less.
1. 3vs3 if the scale is balanced use the other 6
2. The new 3vs3 will be unbalanced if the first 6 were balanced.
3. Replace the lighter side with 3 from the other group that you know are balanced.
If the result is balanced the odd man weighed less and if the result is off balance than the odd man is heavier.
Double G correct.
had to thumbs up just to keep other people from having to scroll down to see the answer.
Or
Unknown=U, Maybe high=H, Maybe low=L, Common weight=C.
1st: 3 groups of 4 UUUUvsUUUU
2nd: path (=) UUUvsCCC or path(> or
Well, thats one way to see it, in which case you are correct. But if you try to find exactly who is imbalanced, then the answer is really complex
That's simply not true lol
"You must figure out which" can be taken as both "Are they heavier or lighter" or "Who is heavier /lighter"
Your answer is either incomplete or wrong, depending on what is asked...
1:00 If I did this right, I'm pretty sure I have the answer. First you would divide the 12 people into 3 groups of 4 (11 weigh the same + 1 diff = 12/3 times to use a seesaw = 4 people in each group). Then you take two of the three groups and have them both get on the seesaw to see if they weigh the same. Using this will lead you to either having a balanced seesaw or an unbalanced one. From this point on, it becomes simple process of elimination. So, say you have an unbalanced seesaw at the start, perfect, now you would take one of the first groups (A) and balance it with the one you didn't (C). If they are balanced, then you now know that the person is in group B. Now, you have 4 people with 1 more try to figure out if the person is lighter or heavier than everyone else. This part is easy and just relies on how seesaws work. If you take the now two groups of 2 and position one on the side closer to the fulcrum you will be able to see if one group is heavier or lighter (If the group closer to the fulcrum [D] is able to continue using the seesaw, then they are heavier than the other group [E]. However, if group D is unable to continue using the seesaw, then they are lighter than group D). If you have a balanced seesaw when you start with groups A and B, then you just continue to measure groups D and E. Am I right?
7:15 Another comment explains this better, but Kevin is right. Imagine you have a 6-sided die, you have to roll a 2 or a 5 with your eyes shut (1/3 chance or 33.3% odds). After you roll, your friend removes 3 and 6, then asks if you want to roll again or stay with you roll (Your 6-sided die becomes a 4-sided die, you still have to roll a 2 or a 4). Your odds don't increase if you stay with your original roll, they stay the same (You still have a 33.3% chance of being right if you remain with the first roll). However, if you decide to roll the now 4-sided die in an attempt to roll a 2 or 4, your odds become 50% (1/2 chance). Therefore, your chances increase if you switch doors ergo, Kevin is right.
Amy was right.
6vs6, you know which group its in
3vs3, you know which group its in
1vs1, either the see-saw is still unbalanced, or its the 1 thats not on it.
@@dyent 6v6 doesn't do anything. One dude is EITHER havier or lighter.
Nice solution for the first one , but it would not be accepted. The problem has many different versions , some of wich are about coins and a regual scale ( with 2 plates) so you can't use the real properties of a seesaw. But the first step is correct , you do divide them in 3 groups of 4
1:44 When i try to spend the day studying.
The way the entire office just stopped and went silent when Jake accidentally called him "Dad", compared to absolutely no one noticing Santiago's antics in that episode. 🤦🏿♀️🤣
"There are 12 men on an island, 11 weigh exactly the same amount, but one of them is slightly lighter or heavier, you must figure out which. The island has no scales but there is a Seesaw. The exciting catch, you can only use it 3 times." ================ ========Read on to figure out the solution!----------->
You only have to figure out if the person is lighter or heavier, not who the lighter or heavier person is (Mistake #1 by most people). So, you divide the 12 people into 3 groups of 4. We'll call them groups A B and C. We weigh each group against the other 2.
A vs B
A vs C
B vs C
The (lighter or heavier) person that's apart of its group has now for sure been apart of 2 of the 3 times we used the Seesaw.
During those 2 times, 1 of the groups(A B or C) will be apart of both instances the seesaw went up or down.
If the group that was present both times the Seesaw moved, was raised both times, the person is lighter, if it was lowered both times, the person is heavier. Solved.
1.
I think you can get closer to figuring out who it is. You first do 3 vs 3. If they are equal then you eliminate 6, if they are unequal you also eliminate 6 (and you know which side was heavier/lighter than the other). *leaving 6 remaining*
2.
Then you take 3 of the confirmed (eliminated) same weighs and compare them to 3 of the unknowns. If they are equal then you eliminate 3 people. **Note that If in the previous test they were unequal then you now know if they are heavier or lighter**
If they are unequal then you now know if the person is heavier or lighter (so for your criteria you are done) and you also eliminate 3. *leaving only 3 remaining*
3.
Best case scenario you now have 3 possibilities and you already know if the odd person is heavier or lighter. You compare 2 of the unknowns and now you know who is heavier/lighter and if they are heavier or lighter.
Worst case scenario (round one and two were both equal which is a 1/4 probability) you now have 3 possibilities and you don't know if the odd person is heavier or lighter. You can now compare them to 3 confirmed same weights and you will figure out if they are heavier or lighter.
You can just put 6 on either side, whichever side is heaver you split into a group of 3, put 1 on either end and either one will be heavier or they will be the same and the other person must be heavier. Only need to use it twice, I mean you could always use it a 3rd time to prove it but if the riddle is right then it's pointless.
How come nobody has brought up that the lady basically didn't even allow one of the questions over the radio? The "how do we get out of here" question. Either she's obligated to answer them or the questions become meaningless (clearly there should have been additional rules stipulated on the note.)
Santiago shaking in her chair when Holt screams BONE! Unforgettable.
It's an unsolvable problem that's it. They are fairly common in maths, especially with a probability question
Which episode
@@sk-124 Not sure sorry. Sorry for the very late reply
The seesaw problem:
1) weigh 4 vs 4, if it’s stable use the 4 you’ve left. If not use the heavier side.
2) weigh 2 vs 2, use the heavier side next
3) weigh 1 vs 1
Done
That may not work. "1 of them is slightly lighter or heavier" he may be lighter, and choosing the heavier side would not have found him in 3 moves.
6v6. Then 3v3 then 1v1 if it’s level then the person not on it is a different weight.
It wouldn’t work because the man is either slightly heavier or lighter, if you only weight the heavier side you may not find the man because he was actually lighter than the rest...
Like Holt said, that wouldn't work.
I solved this question in 15 minutes but Holtz set up the question in such a way that its impossible to know the outcome. He cannot say lighter or heavier. He should have said one or the other. As it is the riddle is impossible to solve
4:55 The TVs being a dick. Hilarious.
I did this puzzle twice as a kid, once in class and once as an interactive science museum exhibit, I don't remember the exact logic but you do it in 3v3's and it all hinges upon finding the people of equal weight and using them to eliminate the odd one out, rather than trying to find the heavy or light person.
I'm wine drunk, please don't ask me to solve RN
BOOOONNNNNNNEEEEEEEE!!!!!
Honestly, Amy and Kevin were right about the Monty Hall problem. Macro-size it to understand it. If there were 100 doors, you picked 1, and then the host showed you the 98 doors where it *wasn’t*, it’d be obvious that you need to pick the one you hadn’t picked before.
Not at all, since it could still be the one you chose. Since it left 2 doors for you to choose, it'd still be a 50/50 situation
@@Koraki0000 You had a 1 in 100 chance when you started. Then you have a 50/50 chance after they showed you all the wrong doors. You have much better odds if you switch.
@@1217BC theoretically. But it’s still the same door.
Theory doesn’t mean squat.
@@brmbkl It's not theory, it's statistical probability. The chance your first choice was right is 1 in 100. The chance the other door is right is 1 in 2.
Kevin was right tho. Switching doors will grant you a 66% success rate vs a 33% when staying
Based on psychological factors, not statistical ones
@@pickcollins9910 It's not psychological. When you make your first choice, there's a 1/3 chance that you chose the right door, and a 2/3 chance that you chose a wrong door. If you chose a right door, then the eliminated door is irrelevant, but if you chose a wrong door, then the remaining door is correct. So it's a 1/3 chance that you're right if you stay, 2/3 if you switch.
KaiRaine when you make your first choice you have 3 options. You pick 1 out of the 3. Then one door that doesn’t contain the prize is opened and now you can choose between 1 of 2 doors. Yes statistically there is a 33% chance the prize is behind that door when there were 3 viable options, but now there is only 2.
Pick Collins no, the host can’t open the door in front of you but he can open both doors you didn’t choose so if he opens one and it isnt there that means that it can be either in front of you or in the third door but as he cant open the door in front of you it means that the car is most likely behind the third door
Because he did choose to not open that one
Never in my days have I ever thought that Raymond holt would say... "boyakasha"
That's the Nickelodeon TMNT new catchphrase, right?
@@juanmanuelpenaloza9264 yup
8:22
This whole sequence of amy lines has the feel as a episode later with Jake "Officer Dad"
This was an extra credit question on a Chem test one time when I was in High School
Amy has the best facial expressions
I loved when Gina pushed Hitchcock away by his Face and Rosa said is this how you want to die Hitchcock. Such an awesome show
That "BONE!!!!" moment is *e v e r y t h i n g* to me 😂
0:54 Easy
The fact he says "lighter or heavier" I believe makes it impossible. If you know one is definitly heavier for example, then you can do 4v4, then choose the heavier side, split into 2v2 then 1v1 after that. If they were both the same, the person is in the 1st group, and you can do 2v2 then 1v1 also. However because the person could be heavier OR lighter you wouldn't know which side of the see-saw to choose. Unless I'm missing something then this is impossible
It’s possible. I found a solution that works.
Weigh 6vs 6....take heavier side and weigh 3 vs 3 then from the heavier 3 just take any two and weigh them...if the scale is equal the guy who is left out is the heavier....if the scale bends over to one side u got your heavy guy again
The seesaw answer I think is, keep on adding a person on each side one by one untill the balance is shifted.
To clear out any doubts, last two persons on any turn to sit on the seesaw causing the balance shift could be again weighed. The seesaw would only be used twice.
you’re right! In the turn where there is a balance shift, you will need to replace one of the two newly added people. The seesaw will either return to balance or go up on one side. Either way, you should be able to tell which one is abnormal, and if he’s lighter or heavier.
You weigh 4 ppl on each side, if the see saw balances out then then the third group has the fat guy. Otherwise, use the group that weighed down the see saw. Then simply weigh 2 by 2 and 1 by 1 based on which group had the fat guy.
Lol, adding people is a NEW Use of the see saw mate! don't try to be smart... haha
7:00 The Monty Hall Paradox is well known.
You have a statistical advantage is you switch doors.
@8:58 best part.
"Is this how you wanna die Hitchcock."
7:19 Amy's brain is getting teaseee baby