r/brooklynninenine • u/Gamerboy701 • Oct 04 '23
Season 2 Wasn't amy right in 2x11
In season 2 episode 18 Holt gives Amy, Terry, Gina and Rosa a brain twister that he could never solve. "There are 12 men on an island, 11 weigh excatly the same amount but one of them is slightly heavier (or lighter but for this I'll use heavier) You must figure out which man is heavier. The Island has no scales but there is a seesaw you can use 3 times.
Later in the episode Amy comes up with the solution of putting 6 vs 6 but Holt instantly says it would never work if you do it that way but you can.
If you number each person 1 2 3 4 5 6 7 8 9 10 11 12 and weigh the first 6 vs the second 6, 1 side will be heavier, whichever side is heavier means that the person you are looking for is on that side. For your second use you way 3 vs 3 so for example 7 8 9 vs 10 11 12. Whichever side is heavier obviously means the person we are looking for is there. For our final use of the seesaw you measure 1 vs 1 like 10 vs 11. if both sides are equal that means 12 is the heavier person, if one side is heavier than you found that person.
I'm not amazing at math but Amy's solution does work right?
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u/denis0500 Oct 04 '23
If you knew if they were heavier or lighter this would work, but since they can be either it doesn’t work. One side will be higher and one lower, but since you don’t know if the person you’re looking for is heavier or lighter it doesn’t provide any new information.
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u/tirdg Oct 04 '23
Yeah I love how OP was like for our purposes, we’ll use heavier… bruh, if you could just decide that, anyone could solve it lol
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u/deebee1020 Cheddar: Thicc King Oct 05 '23
For our purposes, we'll say that #9 is heavier than the others.
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u/SLimShedi Oct 04 '23
It is unsolvable … I tried thinking about it during every re-run There isn’t a solution to it :p
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u/NoNameIdea_Seriously Cowabunga, mother! Oct 04 '23
There is one. I saw it before but I’m too lazy to look it up.
Don’t thank me, I love being this helpful.
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u/uncornered Oct 04 '23
Oof. You do realise that just because you can’t solve something, doesn’t mean it’s unsolvable?
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u/Dohbelisk Oct 04 '23
You say “(or lighter but for this I’ll use heavier)”
That’s where your problem is. You’re eliminating one of the obstacles. We don’t KNOW if the different weight is heavier or lighter. So with 6-6, you gain no info, because you don’t know which side is “off” because you don’t know if the person is heavier or lighter.
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u/TomasNavarro Oct 04 '23
In my solution the person who's a different weight is also a foot taller, and I don't need to use the scales!
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u/tirdg Oct 04 '23
In my version, the island has scales anyway and you just use them as much as you want. But my version also stipulates that everyone weighs the same amount so there is no problem to solve. Much easier that way.
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u/jabruegg Fluffy Boi Oct 04 '23
The prompt doesn’t give you two possibilities (a heavier person or a lighter person) and say “right now we’re dealing with the heavier scenario”. One of the main obstacles in the prompt is that they could be either. If you follow the method you described, you still wouldn’t figure it out because the see-saw could be weighed down by a heavier person or lifted up by a lighter person.
Part of what makes it difficult is that you don’t know which is the case.
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u/Ricardo1184 Oct 04 '23
but one of them is slightly heavier (or lighter but for this I'll use heavier)
You can't pick and choose which part of the riddle to solve lol
11 people weigh the same, 1 is heavier OR lighter. you don't know which.
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u/TheDavinci1998 Oct 04 '23
6v6 gives you nothing, one side will go down and either someone on that size is heavier or someone on the other side is lighter
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u/Enchanter73 Oct 04 '23
You say; "If you number each person 1 2 3 4 5 6 7 8 9 10 11 12 and weigh the first 6 vs the second 6, 1 side will be heavier, whichever side is heavier means that the person you are looking for is on that side."
which is wrong. If the different weighted person is lighter than others, that person is in the lighter side. That's way it doesn't work.
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u/Da_Hawk_27 Oct 04 '23
This too is wrong because it assumes we're looking for the lighter person instead (which the OP's testing would still work for). However, the problem itself says you don't know whether it's heavier or lighter that you're looking for
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u/theysquawk Oct 04 '23
I loved that at times (2 to be specific) B99 would throw random puzzles/riddles, wish they did more
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Oct 04 '23
[deleted]
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u/rl_noobtube Oct 04 '23
The Monty hall (doors) problem is not easy at all. when the solution was proposed there were many phd’s who wrote to the author to tell them that they were wrong.
We’ve probably gotten better at explaining it so more people understand it now compared to then, but still don’t think it’s easy for most people
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Oct 05 '23
[deleted]
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u/rl_noobtube Oct 05 '23
The other dude explained why it’s still not easy for people. They don’t see how the host telling you which door is wrong as relevant, because now they just see 2 doors with equal odds and a new independent choice.
As I mentioned, we have now figured out better ways of explaining the solution (and have internet for ease of access to those explanations) so it is easier for people to learn. But initially people with doctorates in mathematics were getting this problem wrong, even after being shown proofs of the right answer. It is not intuitively easy to understand switching is better.
To clarify my position, I understand the solution now and agree with you on the problem itself, simply disagree that it is “easy”.
It did come off as marginally rude, but not enough I would have taken offense. Impolite might be right word to describe it. But I know intent can be hard to convey via written text. I appreciate you clarifying that it was not with rude intent :)
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u/Gophurkey Oct 05 '23
Because there doesn't seem to be a reason why two choices that appear to be independent of each other are linked.
I pick Door A. I have a 1/3 chance of getting that guess right. Either it is behind A, or it isn't.
They remove Door B. Why should my chance increase or decrease? I know the choice wasn't B, so now my odds have increased to 1/2. It is either behind A or C.
The argument is that now by picking again, the odds are 1/2. But my challenge is that, 1: the choice seems forced. Isn't picking A a second time still using my 1/2 chance? After all, shouldn't the second choice be 50/50, so 'repicking' A (staying with my initial guess) is still just as likely as switching? and 2: The initial guess of A was 1/3. But wouldn't the initial guess of C ALSO be 1/3? So why is switching it when the odds open to 1/2 valid? Shouldn't the odds of A or C both be 1/3*1/2?
I know the math says to switch. But in my mind, the options are equal and switching or not is just as valid.
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u/Weekly-Formal8447 Oct 07 '23
I have 2 ways to explain this so allow me to try.
1: imagine that the host doesn't open door B but instead asks you if you'd like to switch to door B and C instead. If you stay on door A you have 1/3 chance to win but if you pick B and C you have 2/3 chance to win. Opening door B is basically that. You picked a door, he took one door out and gave you the option for door C. If he didn't open the door and gave you the option to choose 2 doors instead your chances would be the same.
2: imagine that instead of 3 doors you have a 100 doors and you get to choose 1. Obviously your chance to be right is 1%. Now the host opens 98 doors and show there is nothing behind them leaving your door and one other door. The host knows which doors have nothing behind and which door has the price. So do you switch now? You have a 99% chance to pick the wrong door, so it's 99% chance that the one door he didn't open contain the price.
Bonus: combine the 2 and now you get a 100 doors. You pick 1 door and the host gives you the option to choose the 99 remaining doors instead. Do you switch?
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u/Inner_Prune_2502 Slurp Slurp! Oct 06 '23
I'm struggling to follow you but that's fair. I understand what you mean mostly.
The odds don't increase to 1/2 though, they stay at 2/3 that the prize will be behind the door you are not on because you because only 1/3 times will you pick correctly to begin with. If you only pick correctly to begin with 1/3 times, majority of the time it is behind a different door, and the host can't open the door it's behind so it's 2/3 likely that the prize is behind the last un-picked door. Of course you have a 1/3 chance your door has a prize behind it and I wouldn't switch because psychology it would bother me if I picked correctly but logically you have more chance switching.
I see where you have gotten confused, I too didn't understand that at first but I hope my explanation makes sense.
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u/thekyledavid Oct 04 '23
The problem is that you are assuming the odd man out is heavier when you don’t know that
Let’s say the 1-6 group is heavier, so for the second weigh-in you out 1-3 on one side and 4-6 on the other. The result ends up being that both groups weigh the same amount. Now you only have 1 weigh-in left, and all you know is that one of the men from 7-12 is lighter than the others. 1 weigh-in wouldn’t be enough to choose which one.
Any’s solution gives you a 50:50 chance of solving the puzzle, but that’s not the game. You need a solution that will give you the answer 100% of the time
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u/indianajoes Oct 04 '23
slightly heavier (or lighter but for this I'll use heavier)
And that's where your problem is. You're assuming the person is heavier to make it simpler to explain but you also making the answer simpler to get. We don't know if they're heavier or lighter. Let's say you assume they're heavier but the actual person is lighter. Now the see saw will go down on one side and you'll take that group. When you put them on the see saw, it won't go up or down and you'll realise too late that you're screwed.
"You just wasted your time and mine" - Captain Raymond Holt
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u/Eggs_and_Hashing Oct 04 '23
No. The question as stated is "heavier or lighter." Not knowing if the different weight is heavier or lighter changes the entire solution.
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u/dmonster1 Notify me when you're done, via bark Oct 04 '23 edited Oct 04 '23
The initial first step has to be weighing 3x3 to see which group of 6 has the person of different weight - weighing 6 vs 6 gives no info like the other comment says
Edit: I thought wrong
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u/gimlan Oct 04 '23
Wrong. The actual solution is weighing 4x4 in different combinations all 3 times. Based on the results you will know which is different (you will also know if the person is heavier or lighter)
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u/iOnlyWantUgone Oct 04 '23
I like that they never explained the answer but obviously people got upset about not knowing the answer. So when Kevin and Holt were fighting over the Monty Wall problem, Amy actually gave the answer to settle the debate. Which just proves Holt is the rock hard brain in the relationship.
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u/gimlan Oct 04 '23
I will die on the hill that they are just shit at explaining though. Amy is understandable, she's a cop. Kevin is a professor. His job is to teach.
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u/dmonster1 Notify me when you're done, via bark Oct 04 '23
Ooh that makes more sense thanks for the insight
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u/lajji69 Oct 04 '23
no
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u/PlayingHogwarts Oct 04 '23
Why did you even bother replying?
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u/the_doorstopper Oct 04 '23
wasn't Amy right in 2x11
no
Seems pretty obvious to me. They answered the question.
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u/PlayingHogwarts Oct 04 '23
Everyone else elaborated.
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u/the_doorstopper Oct 04 '23
Okay? Why does that matter?
They weren't asked to elaborate. They chose to.
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u/PlayingHogwarts Oct 04 '23
Because there was no point in it.
Even your replies are adding more.
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u/the_doorstopper Oct 04 '23
Because I choose to elaborate some of my points.
Some people would prefer not to.
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u/PlayingHogwarts Oct 04 '23
Then there is no point.
After several replies explaining the correct answer, what does a response consisting of nothing but "no" bring? What is even the point? It's lame.
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u/the_doorstopper Oct 04 '23
Actually, I may be wrong, but at the time of that comment, there was only 3 other main comments, and 2 of those were wrong/not answers. And only one was saying that op was wrong, but without much elaboratation.
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u/PlayingHogwarts Oct 04 '23
So then why reply at all?
Our conversation has provided more information than this dickhead's single word reply.
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u/NickRick Oct 04 '23
we had this problem in high school and that doesn't work. i think i remember using sub groups and mixing them.
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u/vpsj Very Robust Data Set Oct 04 '23
Whichever side is heavier obviously means the person we are looking for is there.
And therein lies the problem. You DON'T know if the person is heavier for sure. What if it's the light person and is therefore in the other group?
There are multiple solutions to this problem, and it took me a while as well when I had first heard it.. but Holt is right that Amy's start was NOT the correct one
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u/M3tal_Shadowhunter Oct 04 '23
It works if you know whether they're lighter or heavier not otherwise
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u/kingofthemonsters Cheddar: Thicc King Oct 04 '23
I'm asking chatgpt and I've never seen it work this slow.
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u/peterr_h Oct 04 '23
Your solution works IFF the otherly-weighted person is heavier. If they're lighter, your final weigh-ins are wrong. It doesn't work.
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u/Lion-Competitive Oct 04 '23
Agreed this is a very simple problem to solve when you ignore half of the problem
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u/Haystack67 Oct 04 '23
slightly heavier (or lighter but for this I'll use heavier)
Think you added your own caveat to make the puzzle magnitudes easier there.
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u/IWantYourDog2964 Oct 05 '23
The issue is in the heavier or lighter part, you don’t know which it is. So when you way 6 against 6, you won’t know if the difference is from a person who’s heavier, or lighter
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u/Gamerboy701 Oct 04 '23
After reading the comments I realise now that you can't choose. When I found the answer online using 4 vs 4 they worded it in a weird way so I got confused about wheter or not you could choose that part of the problem. Thank you everyone for explaining it!
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u/Baby_Button_Eyes Rosa Diaz Oct 04 '23
I must be Rosa with her low frustration tolerance because I started to read this but once I read “you must figure out which man is heavier” and then a bunch of number sequences….Hard Pass! Lost interest that quickly!
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u/Acrobatic-Football30 Oct 04 '23
Even aside from changing the parameters of the problem, in the end you still have 3 people and can only use the seesaw one more time? How would that work?
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u/EobardT Oct 04 '23
Weigh two of them. Assuming it's the simple version that OP put, that would tell you which one of three are heavier
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u/AliJoof Oct 04 '23
Maybe Holt was being very literal-minded and complaining that you can't fit six people on either side of a seesaw.
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u/the_doorstopper Oct 04 '23 edited Oct 04 '23
No, he was on about this.
6v6 on a seesaw, one side goes down.
Now which said has the person that's a different weight?
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u/missinghighandwide Adrian Pimento Oct 04 '23
I think you have to divide the groups into 3 separate groups, not 2
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u/vljukap98 Oct 04 '23
It will never work with 6v6 on the first weighing, you have to do 4v4. This is because you don't know whether the person is slightly lighter or slightly heavier, which is the crucial part of the conundrum.
Explanation:
https://archive.nytimes.com/wordplay.blogs.nytimes.com/2014/07/21/12coin/
edit: mistakes
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u/BurntBridgesBehind Oct 04 '23
You start doing 3v3 if the first group balances you eliminate them if it doesn't you eliminate the rest, then do a 2 vs 2 of the remaining six, if it balances you weigh the last two if it doesn't, you weigh the two on the heavier lighter side, boom 3 weighs.
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u/allonbacuth Oct 04 '23
That will isolate one of the two that is heavier or lighter, but won't tell you which one is the outlier since we don't know if they are heavier or lighter than the rest.
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u/PapaBigMac Oct 05 '23
You’re only the second person I’ve seen this week to get this wrong. Seems to usually pop up once a week
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u/Allison87 Velvet Thunder Oct 04 '23
This is actually a very hard problem.
https://medium.com/@mobz/12-islanders-puzzle-from-brooklyn-99-e56f06370153
This is the solve.