Holt's Impossible Brain Teaser | Brooklyn Nine-Nine

Holt challenges Amy, Terry, Gina and Rosa with a brain teaser in exchange for Beyonce tickets.
Season 2, Episode 18 'When Jake's father (Bradley Whitford) comes to town, Jake is excited to see him, but Charles is wary of his intentions.
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ABOUT BROOKLYN NINE-NINE
"Brooklyn Nine-Nine" follows the exploits of hilarious Det. Jake Peralta (Andy Samberg) and his stoically ever-professional captain, Raymond Holt (Andre Braugher), along with their diverse, lovable colleagues as they police the NYPD's 99th Precinct. In this final season of the series, Jake and the squad must try to balance their personal lives and their professional lives over the course of a very difficult year.
Rounding out the ensemble is Sgt. Amy Santiago (Melissa Fumera), who has a passion for organization and a weakness for dork dancing; Lt. Terry Jeffords (Terry Crews), a mountain of a man whose most powerful muscle is his sweetheart; Jake's best friend and human puppy dog, Det. Charles Boyle (Joe Lo Truglio); and the incredibly secretive, tough-as-nails Rosa Diaz (Stephanie Beatriz). Also part of the Nine-Nine are veteran officers Det. Scully (Joel McKinnon Miller) and Det. Hitchcock (Dirk Blocker), whose only skill as police officers is their ability to make a passable pot of coffee.
Created by Emmy Award-winning writer/producers Dan Goor and Michael Schur, "Brooklyn Nine-Nine" won the 2014 Golden Globe Awards for Best Television Series - Comedy or Musical and Best Performance by an Actor in a Television Series - Comedy or Musical (Samberg). Emmy winner Braugher earned four Emmy nominations for Outstanding Supporting Actor in a Comedy Series for his role as Capt. Holt.
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[@sreeharivenat8002]

"I've already consumed the required calories for this day period. " That's Robot Holt at it's best 🔥🤖

[@ma133ha_99]

I love how Gina is thinking so realistically about this riddle but *This brain teaser actually got me thinking, What is the answer* ?😂🤨

[@reeceytaylor]

Number the islanders 1 to 12 First weigh 1, 2, 3, 4 vs 5, 6, 7, 8 on the seesaw. If they are equal then… Weigh 9, 10, 11 vs 1, 2, 3. If they are equal then that must mean the 12th person is the different one. Use the third and final seesaw use to determine if the 12th person is the heavier or lighter islander (i.e. 1 v 12) If you found that the second group weigh was unequal, then weigh 9 vs 10. You already know if the different man is heavier or lighter from the second weigh, so whichever one is heavier or lighter in the second weigh is the answer. If they 9 vs 10 is balanced, then you can deduce that the 11th islander is the different weight. From another comment

[@ma133ha_99]

@@reeceytaylor oh wow thanks!

[@aravindvirinchi9917]

I thought of a simpler solution. Weigh 6 on one side, 6 on the other. Obviously one side will be heavier. Take that side, split them into 3 each, weigh them against each other once again. Again, get the heavier group out, select two random people and weigh them. If one among them is the heavier person, you have your answer. If both are equally heavy, you still have your answer. There's only 3 people in the group.

[@footbelike4537]

@@aravindvirinchi9917 you can't just take the heavier side because you don't know if the person is heavier or lighter. That's the point

[@SerGlushko]

@@reeceytaylor what if the first comparison was unequal? You have no idea which half your islander is in and 2 more comparisons can't guarantee you'll find him

[@MikeMJPMUNCH]

I love that Hitchcock and Scully have to write which side of the pizza is theirs

[@umbreensikandar6189]

"don't just tease my brain captain ,really go to town on it" *one of the most iconic lines of amy santiago*

[@CaptainBluePirate]

We weigh ABCD against EFGH. There are three possibilities here: the seesaw could be level, ABCD is heavier, or EFGH is heavier. Suppose it is level. Thus, the odd coin must be among IJKL and therefore, all the rest are normal. Pick two among this third set of 4, say I and J and weigh them against each other. This is our second measurement. Again, there are three possibilities but let’s think of it simply as two possibilities: they weight the same or they don’t. • In the case they weigh the same, then either K or L is the odd one out. For our third measurement, weigh K against B. B is a normal coin so if the seesaw is level, then K is normal and L is odd. In this case, we don’t know if L is heavier or lighter. If the seesaw is not level, then K is odd and we can determine whether it is heavier or lighter. • In the case that I and J do not weigh the same, then one of them is odd. Say, I is lighter. Now for the third measurement, weigh it against a standard coin, say, B. We use the reasoning as the prior case to determine whether it is I or J that is odd. If I and B match, then J is the odd coin and it is heavier (because we now know I is a standard coin and J was heavier than it). If I and B do not match, then it must be that I is the odd coin and it is lighter. Now we get to the trickier bit. Suppose that after our first measurement from above, ABCD is lighter than EFGH. For our second measurement, we remove ABC, move FGH to join D and have IJK join E. That is, we’ll now be measuring DFGH against EIJK. Here is an easier way to remember this. We originally had: ABCD and EFGH and IJKL. We cyclically permute ABC and EFG and IJK via (1 3 2). Let’s make a few remarks. First, IJK are all normal coins. Secondly, we want to remember that after our first measurement, D was in the lighter group ABCD. For our second measurement, D is now placed with FGH which were in the heavier group EFGH. Again, we look at the three possibilities: • Suppose the seesaw is now balanced; so all the coins are the normal. This means that when we removed ABC, we removed the odd coin. Not only this, ABCD was lighter compared to EFGH. Therefore, not only is the odd coin among ABC, we know it is lighter. We make our third measurement: weigh A and B against each other; if they are balanced, then C is the odd coin out and it is lighter than all the rest. If they are not balanced, the lighter one is easily found. • Suppose the seesaw is configured such that DFGH is heavier than EIJK. We know from the first measurement that FGH belonged to the heavier of the two past groups: ABCD and EFGH. Well, coin E cannot be lighter since it was in the heavy group and it cannot be heavier now that we see DFGH is heavier in our second measurement. The heaviness cannot be attributed to coin D because it was originally on the lighter side after the first measurement. Therefore, the odd coin is among FGH and we know that it is heavier. We can use similar reasoning with a third measurement as above (with the case of ABC) to single out the odd coin. • Lastly, consider the configuration in which DFGH is lighter than EIJK. Coin E was originally on the heavier side and is again on the heavier side after making our new arrangement. Similarly, D was on the lighter side before and is now again on the lighter side. On the other hand, FGH were originally in the heavier group and are now on the lighter group. We may conclude that FGH are normal coins. Combined with the fact that IJK are also normal coins, we conclude that either D is lighter or E is heavier. Our third measurement will be used to determine which is the odd coin. We weigh D against A. If they are balanced, then E is the odd coin out and is heavier. If they are not balanced, then it must be that D is lighter than A and thus we’ve found that the odd coin is D. Of course, D cannot be heavier than A by prior reasoning.

[@Twirly_Sticks]

Wow I think this is it

[@ahboosamuelfung4542]

its lighter or heavier not just lighter

[@debmitra4384]

1:40 and 1:57 classic Hitchcock

[@Dinesh.201]

One of most promising lines for Amy's character building 🤣

[@sharathkumar26]

"You're going down "😀

[@ajayowen]

Anyone just binge watch these & just cannot help but LOVE B99

[@orianneo2041]

This can be solved if they specifically say if the different person is heavier or lighter Divide 3 groups of 4 each and weigh them if they are equal take the third group and if they aren't take the ones that has been specified (like lighter or heavier) then split into 2 each as a group and in next iteration you'll get the answer. 3 tries.....

[@TheJinjitsu]

it is actually possible as stated by holt just the second weighing is difficult :)

[@notthinkumar1015]

Ur hired Welcome to the nine nine

[@charisantoniou5640]

or you can split them in 2 groups of 6. After you see which group is lighter/heavier you split that group into 2 groups of 3. Then you put 2 people on the see saw. If one of them is lighter/heavier you found him if not he is the other guy

[@charisantoniou5640]

before you put 2 people on the see saw you weigh the 2 groups of 3 to find which one is different

[@whateverihateyouwtf]

you dont need to know in advance if the oddball is heavier or lighter. doesnt matter.

[@sleepyboi5804]

Kevin and Holt probably used the Beyoncé tickets 😂

[@Calico9288]

You start by measuring 4 on each side. If the scales aren't even you can immediately discard the four you haven't used. This is where you start thinking about the weight itself, cause you now have 4 that are potentially heavier and 4 that are potentially lighter. Since all except one weigh the same, if the outlier is heavier then it must be on the down side of the seesaw on this weighing and if it's lighter it must be on the upside. The 2nd try is three on each side, this is the complicated part. Take two potentially heavier (doesn't really matter if you remove two heavy or two light) ones and put them to side. Then on each side place one personal that is potentially heavier and two people that are potentially lighter. If the scales on the 2nd try aren't balanced you can remove the the two potentially lighter ones that end up on the down side this time and the one potentially heavier one that ends up on the up side. So now we have two potentially lighter and one potentially heavier left. So now for the third attempt, we're down to our final three options. Take the two that are potentially lighter and measure them one on one. If the scales aren't even then whichever side is up is the outlier which we know now to be lighter than the rest of options. If the scales are even then we know that our final potentially heavier option is the answer. If the scales are even on the first weighing then it's fairly straightforward, as you can eliminate the 8 on the see saw, leaving you with 4 left. For attempt two, you take two of your remaining possibilities and weigh them one on one. If attempt two is unequal, then you know it's one out of these two options, if they are equal then you know it's one of the two you didn't weigh. For attempt three you take one of your remaining two possibilities and measure it with another person that you've already established from your earlier weighs, is equal to the rest. If they're uneven you know the one potential option you chose for this test is the outlier, if they're even then you know it's the one you didn't choose. If the seesaw isn't even on the first attempt but is even on the second attempt then you know the outlier is one of the two potentially heavier ones you set aside and for your third attempt you can simply measure them one on one to see which one is heavier.

[@owntmpsn]

Everyone knows the answer is to waterboard the islanders until they tell who weighs differently

[@love_from_india]

Number the islanders 1 to 12 First weigh 1, 2, 3, 4 vs 5, 6, 7, 8 on the seesaw. If they are equal then… Weigh 9, 10, 11 vs 1, 2, 3. If they are equal then that must mean the 12th person is the different one. Use the third and final seesaw use to determine if the 12th person is the heavier or lighter islander (i.e. 1 v 12) If you found that the second group weigh was unequal, then weigh 9 vs 10. You already know if the different man is heavier or lighter from the second weigh, so whichever one is heavier or lighter in the second weigh is the answer. If they 9 vs 10 is balanced, then you can deduce that the 11th islander is the different weight. This is definitely not my calculation but what Scully and Hitchcock would have searched on the internet to get a free pizza from Holt.

[@Alanhallow]

This will never work You just wasted your and my time

[@love_from_india]

@@Alanhallow That's literally what Scully and Hitchcock do! Waste everyone's time including their own!

[@Lewis_Bell]

What if the first weigh of 1, 2, 3, 4 vs 5, 6, 7, 8 is unequal?

[@love_from_india]

@@Lewis_Bell Then I will cry in the corner of the precinct

[@Lewis_Bell]

@@love_from_india Lol same this puzzle breaks my mind

[@fardan50]

One of my favorites puzzles

[@issahchaconjuarez4255]

Just weigh 3 against 3 first, that's already 6 of the 12. If it's stays leveled, then the odd one out is on the other half, so then just weigh 3 against 3 again, see which side is heavier or lighter, take two out of the three on that side, weigh them, if the odd one is one of the two it'll be obvious who, and if it stays leveled then you know the one you didn't weigh is the one. In case the seesaw doesn't stay leveled on the first 3 against 3 (thus, the odd one is on one of either sides) then you just skip one step and take the three out of the side that lighter or heavier and carry out the last step. Though the 4 against 4 solution is also valid, this is a problem based mostly also on probability considering you have a certain amount of chances for the seesaw to have the odd person, but either way, just getting all the ones that aren't the odd person by any mean would work perfectly fine. I think the 6 vs 6 wouldn't work solely because that's way too complicated, too many steps, also how would 6 grown men fit on one single side of a seesaw? Or maybe I'm wrong but idk, that's just how I percived it

[@garydemonic]

Solution: and before anyone says this wouldn't work, it all depends on what you consider is the act of using the teeter totter. the assumption is that using a teeter totter is the action of going up and down. However, "Use" is a loose term. Using the teeter totter could just mean the act of using the device as a whole, so getting on and off of the teeter totter is all part of using the device. if this is the case, as you start to place groups of two men on the either side of the teeter totter, you'll eventually get to were you find the one guy that is heavier (or lighter) while loading the Teeter totter up to be "Used". once you know that one of the two men that changed weather the teeter totter is up on one side or level, then you just have to swap out one guy, and it will tell you who the Heavier, or Lighter person is in only 3 Uses.

[@Cici900]

There is a Ted ed video that explains how this exact riddle but with coins.

[@desultorypsycho9729]

Sinple version. Divide each side equally and take the off balance side, then repeat the process leaving a remainder of three men for your last use. Pick any two or the three and they will either be uneven or even, by process of elimination you will have your answer. Now give me my Beyonce tickets Captain Raymond Jacob Holt and my hundred bucks Diaz🙂

[@wordplay1527]

Ccliickkkk.....

[@MoyaBrennan6825]

I wanna know what blush Amy is wearing

[@alexdwyer8591]

This is so easy and amy was right… idky holt said that would never work

[@jasonmatos5462]

does anyone know the answer to the riddle.

[@whateverihateyouwtf]

Piers Anthony told me how to solve this 30 years ago. Anyone remember? weigh any 3 vs any other 3 if same you know the odd one is in the final group. weigh one of them against another. if theyre the same, you know the odd one is the 3rd one. weigh it against one of the others to see if its heavier or lighter. if the 3vs3 wasnt the same, take either 3 away and put the final 3 on. this will show you which group has the odd one and ifnits heavier or lighter. now just weigh any one from that group against another. if the same, you know the 3rd one is odd. if they arent the same, you already know if the odd is heavier or lighter, so you have your answer.

[@whateverihateyouwtf]

dont tell me Terry wasnt a Piers Anthony fan.... he should have known this

[@mitsu.B]

Came here after the ted ed Riddle lol

[@Juliahehe2124]

I know the awnser

[@TheBillJohnsonVarietyHour]

sudocruise

[@lowlevell0ser25]

why didn't he just look it up

[@paulg903]

It's impossible to work out, I googled this years ago, even mathematicians couldn't solve it.

[@rickjones3340]

Dude, it absolutely is possible, it's just a bit complex. Of course mathematicians have solved it. You can Google the solution.

[@paulg903]

@@rickjones3340 tell me the solution then

[@rishith8595]

Can anyone confirm this woul work? So first we weigh 6v6 qnd take the set of heavier 6 men and split them in half and weigh them agian 3v3 and then we would find set of 3 being heavier, Now again we take the heavier set of 3 men and pick out two and weigh them , if the they weigh the same the other one is the heaviest or if one is heavier than the other we agian cqn conclude we found the heaviest

[@Twirly_Sticks]

Holt said he can be heavier or lighter

[@adityatiwari482]

Many people have given std solution but heres slightly different approach i thaught which works too Solution is use only 6 people ie 3 vs 3if see saw tips one side then unequal one is in that group else it is in rest of the 6 then out of the people with unequal weight pick random 4 and weigh 2 vs 2 if if it dips on one side then we are down to two people else if it is equal on both side then also we are left with 2 people of 6 with still 1 use to go

[@Alanhallow]

It doesn't work

[@adityatiwari482]

@@Alanhallow why??

[@Alanhallow]

@@adityatiwari482 should is really have to explain We don't know the person is heavier or lighter According to your logic on the third third weigh what if it goes equal we still left with 2 people Also the question not to find the odd person, but , to find the odd person along with figuring out that he is heavier or lighter

[@adityatiwari482]

Agreed with your heavier lighter but none has solution fr that part And about 3rd part we only used it twice and if it weighs equal you have 2 people left and 1 use whats wrong with that??

[@Alanhallow]

@@adityatiwari482 Here is how to do it The seasaw positions are L - left slanted C -centered R - right slanted Give the islanders 12 number On the 1st weigh 1 2 3 4 against 5 6 7 8 On the 2nd weigh 1 2 5 9 against 3 4 10 11 On the 3rd weigh 3 7 9 10 against 1 4 6 12 This is the possible outcome Heavy | Light 1- LLR | RRL 2- LLC | RRC 3- LRL | RLR 4- LRR | RLL 5- RLC | LRC 6- RCR | LCL 7- RCL | LCR 8- RCC | LCC 9- CLL | CRR 10- CRL | CLR 11- CRC | CLC 12- CCR | CCL You can find the answer with this

[@ZSGreenFalcon]

But why would 6 on each side not work?

[@leonjustus6961]

Because we do not know if the wanted person is heavier or lighter. Putting 6 on each side would not tell us on which side the odd one out is.

[@joevictor53]

Okay, you do that and one side goes up, one goes down. What do you do next?

[@leonjustus6961]

@@joevictor53 this is a nice mathematical problem which I cannot explain as well as some websites you can find when googleing this riddle

[@joevictor53]

@@leonjustus6961 I'm not asking you to explain it. I'm asking the person above what they would do next after that step

[@charisantoniou5640]

@@joevictor53 after you weigh 6 on 6 you split the lighter or heavier group into 3 and 3. Then you weigh the 2 groups. You find the one that is heavier / lighter . Lastly you you weigh 1 on 1. If they're equal then the heavier/lighter person is the third one. If not then you can see it

[@Sora-md1ye]

Let's just say that the odd one out is slightly heavier than the rest. You could either split the islanders into 2 groups of 6 or 3 of 4. After weighing the 2 groups of 6, you check for the heavier group and split them in half. Weigh them again to check for the heavier trio. The third weigh in will be for two of the last three islanders. If they weigh the same, your target is the one not on the seesaw. The 3 groups of 4 route is similar. Groups 1 and 2 are weighed while group 3 stands to the side. If they weigh the same group 3 has the target islander. Split group 3 into 2 pairs and weigh them. The heavier pair will be split for the last weigh in and you have your target.

[@Twirly_Sticks]

This only works because you assume that he's heavier but you don't know that. With the way holt told that riddle, you have no way to know if you should be weighing the heavier or lighter group

[@tylerklasell7936]

To determine the odd man out among the twelve men in three weighings using a see-saw, you can follow these steps: First Weighing: Divide the twelve men into three groups of four (A, B, and C). Weigh group A against group B. If they balance, the different person is in group C. If one side is heavier, the different person is on the heavier side. If one side is lighter, the different person is on the lighter side. Second Weighing: Take the group (A, B, or C) that has the different person and divide it into three pairs (1, 2, 3). Weigh pair 1 against pair 2. If they balance, the different person is in pair 3. If one side is heavier, the different person is on the heavier side. If one side is lighter, the different person is on the lighter side. Third Weighing: Take the pair that has the different person and weigh one person against the other. If they balance, the different person is the one not weighed. If one side is heavier, the different person is the heavier one. If one side is lighter, the different person is the lighter one. This strategy allows you to identify the odd man out in just three weighings using the see-saw.

[@Twirly_Sticks]

If group a is heavier than group b how do you know whether that means someone in group a is heavier or someone in group b is lighter?

[@mask9900]

Hhhh

[@barneybramley-maye1247]

First you weigh 6v6 the one group which is heavier you split into two groups of 3 you weigh those groups and take the heavier group then you take two from that group if neither is heavier the one you didn't weigh is the heavier one

[@rickjones3340]

You cannot solve this by starting with 6v6 (as Holt notes in the clip.) You assumed the odd man out is heavier, but that is not what the puzzle says.

[@Rose_Butterfly98]

Just watched this episode. I find it amusing that detectives took so long to solve it. Kinda too unrealistic when someone as dumb as me solved by the time Rosa said Marcus loves Beyonce lol. Well, I figured the way I would find out but to actually do it I'd also need to write a list of how the seesaw would tilt depending on who it was and if they were heavier or lighter. Wait, is it easier if you're dumb? Maybe that's why they had to take Jake out, he would've solved it.

[@okimapullup2050]

Early

[@molie7712]

Wow I'm first!

[@molie7712]

@@manharkashyap_006 Thank you ! It's my first time ever ! 😍

[@deepanshnarang1364]

I solved it😳😳😳😳😳😳😳😳😳😳 fr U have 3 chances so, Divide those 12 men in 3 set of 4people 1st turn- take one set and measure ot with one other, if the see- saw weighs equal then theres no impostor in those 2sets, 2nd turn- exclude both these sets, take the third set, take a single men out of them, and compare with another, if the see- saw shows same then these 2 arent impostors 3rd turn- now 2 left, take one from the previous 2 non-impostors, and compare with one of the 2 left, is it is same then the last 4th one is the impostor or if it is different then that 3rd person is impostor i.e. with different weight...

[@manavudgirkar3345]

what Amy was going for was right.. right?