Brain Teasers | Brooklyn Nine-Nine | Comedy Bites 

e

That warms my heart!

Amy not ami

I'm teaching father the math

does that mean she and jake are siblings?

Hahahagross

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%.

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.

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

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"

Chicken wings 🤒

Cocaine

Wing slutz

Even funnier is that they're actually good detectives...

@Constant Geek cocaine

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

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😂😂😂😂

🤯

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

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?

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?

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!!??

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.

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?

Ironically, Gina was the most useless out of all of them

@@BlasterzHD super true

*Sweet home Alabama...*

That turned dark quite quickly :D

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.

"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?!???!

he said it so quietly with so much intensity

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

She's a judge. That's not a brain teaser.

Hooooow dare you De-tect-ive Di-az...I am your SUPERIOR OFFICER

Spideybops BOOOONE??!!

It's Dianne Weist

@@LuisCastro-jw5pi my bad haha

@@LuisCastro-jw5pi *wiest

Because Jake would solve them all in 3 seconds.

The key was in a can of peaches, somehow

Skully ate peaches and found one of them.

Jade Nguyen i felt bad because my first thought was to try six vs six just like Amy

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 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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-tE2dZLDJSjA.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.

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...

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.

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

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.

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.

That's the Nickelodeon TMNT new catchphrase, right?

@@juanmanuelpenaloza9264 yup

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

Which episode

@@sk-124 Not sure sorry. Sorry for the very late reply

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

Clearly, the answer is nine pancakes.

Two?

My god its like none of you watched the clip.

No, it's not 50%, because the information you gained is not only that the revealed door does not have the car, but also that the host was more plausible to reveal that door if the car were in the other that kept closed than if it were in which you selected. This occurs because as he must always show an incorrect door that is not which you picked, then when yours is incorrect he is restricted to reveal specifically the only other incorrect one, but when yours is the winner, he is free to reveal any of the other two, we never know in advance which he will take, because both would be incorrect. For example, if you start selecting door #1 and he shows a goat in #2, we can be sure that if the prize were in #3 he would have revealed #2, as he wouldn't have had another option. But if the correct were #1 (yours), it exists the possibility that he would have preferred to reveal #3 in that case and not #2. That's the reason why it is more likely that he decided to open #2 because the winner is the switching door (#3) than because the winner is yours (#1), and the same happens with all the other combinations. It usually helps to think about the extreme scenarios. Notice that in the worst case from the perspective of your door, the host could have the preference of always revealing the highest numbered option of those that are available for him. That would mean that if the correct were in fact yours (#1) he would have opened #3 and not #2. But as he opened #2 this time, it would mean that the car is 100% likely to be in #3, not in yours.