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@Chsieyfnd

Жыл бұрын

怎么只有14个赞？！

@carealoo744

Жыл бұрын

I never really got the concept of luck. You don't know if the keystones platform is already taken, so is the luck potion bounded by what you know?

@mysteryman7076

Жыл бұрын

What happens if i scan each stone more than once.

Forget the potion, having solved the logic behind the riddle, the correct choice of action is just to take the keystone. Our stated goal isn't to use the wand, just to keep it out of the villain's hands while not dying in a cave-in.

@sirnikkel6746

Жыл бұрын

Friggin genius

@theunclecappy4846

Жыл бұрын

That's a proper answer to a proper riddle

@miriamrosemary9110

Жыл бұрын

Yessss!

@destroyerofworlds4663

Жыл бұрын

Even better, since the Henchman basically just ruined it for Moldevort 😂

@Evyton

Жыл бұрын

Congrats on having 1999999999999999999 iq

Moldrvort: Did you stop them from getting the wand? Henchman: Yes. I stuck a random gem to a random pedestal. They now might accidentally cause a cave-in. Moldevort: Wait... how am I supposed to get the wand now? Henchman: ... Moldevort: ... Henchman: ... Moldevort: Kavada Edavra!

@Yildirim.Bayezid

11 ай бұрын

He'll soon be benchman

@ForeverEpsilon

10 ай бұрын

I assume he knows the only needed pedestal.

@mystrenula3911

8 ай бұрын

​@@ForeverEpsilonbut there's a 1/100 chance that the henchman put the stone on the keystone pedestal and in that case the wand is stuck forever

@TheVoicesOfTheBeyond

6 ай бұрын

If the henchman accidentally sealed away the wand, then they ironically saved the day

@user-fd9bx4ki9l

5 ай бұрын

Before apparating was invented

If only the Elder Wand was hidden as securely as this...

@malafanai1357

Жыл бұрын

Lol

@ankcx

Жыл бұрын

Ikr but weird this is about the horcrux, people. not the elder wand…

@rachnamookerjee1386

Жыл бұрын

Ya

@RawFishHeader

Жыл бұрын

Well, the whole point was for Voldemort to find it since he was being tricked. So it's good that he was able to find it easily otherwise Harry wouldn't have been able to defeat him

@koharumi1

Жыл бұрын

It is weird how Dumbledore never trained harry... Like he was a sacrifice. All planned out...

Hold it! Nowhere in the rules does it state that I have to place the stone down, only that a random pedestal will light up. If I keep a pen and paper on me, I can track exactly where the stones need to be placed, and which ones potentially overlap. Let’s say you have a ten by ten grid, and mark down each one with I for just one stone, and II for the overlap. The overlap tells you two things in particular: One, the previously placed stone didn’t take the winning spot, so you still have a chance; and Two, this means you have two pedestals left, effectively narrowing it down to one coin flip. However, we can rig this flip in our favor; recall how the henchman only sealed the gem to a pedestal rather than using the spell to find it’s true home. The placement spell only works once per stone, and since this stone hasn’t had the spell cast on it, you can check then and there where it was supposed to go, putting the superfluous stone in that spot and securing your victory. This turns a 50/50 coin flip into a 99% victory chance, with the only way you could fail being if the henchman had put the stone on the winning pedestal, netting a loss. A 1% chance on their end. That is how I would go about solving this puzzle. Thoughts?

@skandragon586

Жыл бұрын

my thoughts essentially... though i heard the spell put the stone on it's pedestal, so i was thinking just slide it onto the floor beside its pedestal, leaving it vacant (to find the overlapping stones) but yes, i agree the true probability was 99%. i only lose if the sealed pedestal is the keystone's

@majorjohnson8001

Жыл бұрын

@@skandragon586 Actually is a 50-50 shot if all the other stones match 1 to 1 with no duplicates. If that's the case, the first stone was glued to *either* its own pedestal *or* the keystone's, leaving the same 50-50 odds on a 1.9% chance outcome (so 2% of the time you drink a potion and the other 98% you keep it for later).

@ittiponkhamrangsi3627

Жыл бұрын

@@majorjohnson8001 Another case is the gen belonging to the glued pedestal when you cast a spell to it and it randomly shows its place on a keystone pedestal. This produces a match 1 to 1 with no duplicates.

@elSethro

Жыл бұрын

Yes, this was exactly my thought. However, this will work 98% of the time. There is a 2% chance that you could have 99 pedestals lighting up once each (no duplicates): (1) the bound stone is on the keystone's platform already (guaranteed failure: 1% chance overall, 50% chance given 99 lights) (2) the bound stone is on its correct platform AND the keystone platform was randomly selected when you cast the spell on the bound stone (0.01% chance overall; 0.51% chance given 99 lights) (3) the bound stone is on a common platform, and the stone that is SUPPOSED to activate the bound platform randomly activated the keystone platform instead (0.099% chance overall; 49.5% chance given 99 lights) If you see 99 pedestals light up, then I don't think there is any strategy that does better than a 0.51% chance of success. There would be a 50% chance that the keystone pedestal is bound (no chance of success), a 0.5% chance that the keystone pedestal randomly lit up when you cast the spell on the bound stone, and a 49.5% chance that the keystone platform was randomly lit up by the normal stone that is supposed to go on the bound platform. I spent way too long on the maths for this one lol.

@beautifulnova6088

Жыл бұрын

Why not just cast the spell on each stone twice and see if it changes its pedestal.

I love the fact that you named the wand Mirzakhani Really cool, maybe name everything and everyone after underappreciated scientists and artists. It would another layer to your videos and it would be cool to Google some of them and find amazing stories that people just don't talk about.

@ganiti_314

Жыл бұрын

Now I noticed that. cool.

@Pickled_Poet

Жыл бұрын

I didnt even know about a Mirzakhani but now I'm interested

@claradashti860

Жыл бұрын

An Iranian mathematician

@Morphysince94

Жыл бұрын

@@Pickled_Poet She is a fields medal winner , once in a lifetime gem that humanity lost, Maryam Mirzakhani.

@farbodlv8294

Жыл бұрын

R.I.P Maryam Mirzakhani, she was a national treasure for iranians and a source of inspiration for a lot of women in and out of iran.

I love how Ted ED shows us that Mathematics are so important it can literally destroy the world.

@islandmaster5064

Жыл бұрын

E=mc²

@diminikolova

Жыл бұрын

🔝🌟

@leanna5733

Жыл бұрын

Eh, typical Ted Ed.

@ugendranr3011

Жыл бұрын

V+F = E+2

@nicholasvlamis5603

9 ай бұрын

But also that their math is always rigged in some way, like in this case there was an unmentioned luck potion that made the situation almost certainly play out in the favor of the unnamed wizard.

i would totally watch a movie with Moldevort and Drumbledrore in it. Oh wait a minute....

@varshinilolla3090

Жыл бұрын

Yeah... there's a "wait a minute" when you realize that it's drumbledrore and moldevort but not dumbledore and voldemort.

@aarushiyadav7101

Жыл бұрын

Harry Potter and the Half Blood Prince

@varshinilolla3090

Жыл бұрын

@@aarushiyadav7101 Fortunately drumbledrore didn't die in this

@justrandom8344

Жыл бұрын

@@aarushiyadav7101 no no no its parry potter

@foxbox.9277

Жыл бұрын

No look at the video there’s an M on the blue girls robe- I MEAN TUNIC. It’s clearly Mary potter

1: Take the keystone so moldevort can't get the wand. 2: Confirm you have green eyes 3: Ask the guard if you can leave 4: Steal the secret sauce recipe 5: Lick the male frog 6: Pick the Churrozard disk 7: Cheat death 8: Get your guitar from the drumset box 9: drop the worthless egg from story 34 10: Separate the Fire dragons from the Ice dragons 11: Write down the jousting tournament scores 12: Ask "If I had a burrito for lunch, would you say Ozo"? 13: Light all of the giant's birthday candles 14: Put in the charged batteries in the giant iron 15: Stop going on youtube because you watch WAY too much Ted-Ed riddles. 16: Have a nice day! Edit: 17: MOM COME OVER HERE I'M FAMOUS.

@catoctober8005

Жыл бұрын

LOL

@tezsaw486

Жыл бұрын

You forgot to cut the werewolf antidote into 5 squares!

@ehtacoguy4079

Жыл бұрын

@@tezsaw486 I didn't forg- *turns into werewolf*, oh.

@lacyomsberg1235

Жыл бұрын

We also can’t forget to make the scientist and the janitor cross the bridge together.

@ehtacoguy4079

Жыл бұрын

Let's do this for every time Ted-Ed makes a new riddle video.

FINALLY A RIDDLE

@powerllesss2672

Жыл бұрын

tom riddle

@nandluv

Жыл бұрын

YESSS the moment we been waiting

@nandluv

Жыл бұрын

@@powerllesss2672 LOL

@YoureRatharStewpidMate

Жыл бұрын

@@powerllesss2672 tom riddle 2

@sir_albaxious1909

Жыл бұрын

Finally

_“This isn't magic-it's logic-a puzzle. A lot of the greatest wizards haven't got an ounce of logic, they'd be stuck in here forever.”_ *_Hermione_*

@gomshom67iscool23

10 ай бұрын

214 likes and no comment. seriously?

@jiyoonjun5783

6 ай бұрын

The wizards at Hogwarts only get normal education till 11 so that makes sense

Finally understood the plot of “Mary Plotter and the Deadly Platforms”

@bigkristian1891

11 ай бұрын

Lol

Finally a riddle I understand! Of course you can only pick the switched pedestal or pick the one the Keystone should be one, every other number just delays the inevitable. Thankfully Drumbledraw knows the secret to win against any odds: Cheat🤣

@iaditiagrawal

Жыл бұрын

As he always does😂

@satwiksahu486

Жыл бұрын

And to leave the problem to ungraduated teenaged students

@catoctober8005

Жыл бұрын

@@iaditiagrawal Lol

@metal_pipe9764

Жыл бұрын

I'd just nuke it

Step 1: Give one coin to Charlotte and Eliza to secure their vote! Step 2: Make the Janitor and the Scientist cross the bridge together Step 3: Lock Moldevort in the Magical chess board Step 4 : Choose the Bannekar and skip the first turn Step 5: Say at least one of you have green eyes Success!

@ZFroZenHail

Жыл бұрын

Step 5 is wrong, the correct step is to say at least one of you have green eyes

@k3ose455

Жыл бұрын

The amount of references in this one comment is more than i have money

@Anish_Deshmukh

Жыл бұрын

@@ZFroZenHail Done 👍

@saininsa98

Жыл бұрын

@@k3ose455but somehow we know where all the references came from😂

@HorseEater

Жыл бұрын

You forgot to flip 20 random coins

If Moldevort wants the super-wand so badly, doesn't it go against his plans to sabotage the cave so it can potentially never be found?

@gmanplaysgames256

8 ай бұрын

maybe he didn't know where it was, and sent the henchman to follow you to it?

What if you casted the spell on all 99 stones, without placing any of them? That would more intuitively show where the possible keystone spots are, and you don't have to worry at every step wether or not the stone you just placed is correct or will mess up the whole process.

@Stratelier

Жыл бұрын

It doesn't change the end result, but placing them makes it easier to keep track of which pedestals have been already identified.

@LL-di6yy

Жыл бұрын

@@Stratelier i mean if u actually write down some schemes on paper just watching which pedestals lights on u can do it easier

@LL-di6yy

Жыл бұрын

@@Stratelier since you sign all the combinations and watch which one of them overlap and similiar

@jacky7204

Жыл бұрын

@@Stratelier It does change the end result, because when the spell fails, it specifically illuminates an unoccupied pedestal. If you leave pedestals unoccupied, you can highlight a pedestal twice (much more than 50% likely) leaving the keystone pedestal unlit.

@TheFinalChapters

Жыл бұрын

@@jacky7204 Except two such pedestals would (most likely) remain unlit: the keystone pedestal and the one for the original "sealed" stone. Unless you are allowed to cast the placement spell on the sealed stone, your odds will still be 50-50 in the end.

Child: I want to watch Harry Potter! Parent: We have Harry Potter at home, sweetie. Harry Potter at Home:

@caelincoolz5814

Жыл бұрын

Child: "Wow! This is better!"

@youyaku-music

Жыл бұрын

@@caelincoolz5814 The “child” would say the opposite tbh

@caelincoolz5814

Жыл бұрын

@@youyaku-music Harry Potter is pretty awesome.

Well worth the wait, I love ted-ed riddles, I do wish they were more consistent though, they are always banger videos every time 😎👍🏻🔥

Or you could cast a spell on the stone that was already placed and see where it's supposed to go. Then you cast spells on all of the other stones and see which pedestal lights up twice. One of those two belongs where the first one was placed, so it goes where the first one was supposed to go, leaving you with the one you need the keystone to be placed on.

@bananaforscale1283

Жыл бұрын

You can't

@0mathgaming

Жыл бұрын

@@bananaforscale1283 Why not?

@tavern.keeper

Жыл бұрын

@@bananaforscale1283 That is allowed within the stated rules, but I'm not sure it actually helps your chances. If the first stone was placed correctly, then it's platform would be occupied and per the rules a random one revealed. You have no way of knowing which case is true. After thinking about this more, I think this is the correct strategy. 1/100 chance that the keystone platform is already occupied, and you've already lost. Otherwise, 98 stones will reveal their correct platform and one will lie. If the lying stone points to the keystone platform (1/99 chance), then each platform glows once, and you just have to guess (1/99 chance to win). If the lying stone points to some other platform (98/99 chance), then one platform will remain unluminated, and that will be the keystone platform (you win). So overall chance to win is 99/100 * 98/99 = 98/100.

@bananaforscale1283

Жыл бұрын

@@0mathgaming rule 5

@0mathgaming

Жыл бұрын

@@bananaforscale1283 Rule 5 only says that you can't cast a spell on a given stone more than once.

What about this though: Cast the placement spell on the stone already placed. Another platform will light up. Take any random stone and place it there, then cast the placement spell on the stone you just placed. Take another stone and put it on that platform, then repeat. By the end, you should eventually have every stone placed except the keystone, since each stone you place is telling you where to place the next one. It doesn't matter if all the stones are in their allocated places, just that the keystone eventually finds its home. The only way this method fails is either if the stone that already got placed is sitting on the keystone spot, or if it's already on its own spot. If its already on its own spot, there's still a chance of success anyway since it follows the same 50/50 chance as the rest of the video. Overall, that puts the odds of success at > 98%.

@tavern.keeper

Жыл бұрын

It could still fail if the first stone was placed on another stone's platform. Because then when you eventually place that stone and cast the placement spell, it will reveal a random platform which might be the keystone platform.

@Stratelier

Жыл бұрын

My analysis of this method does not match yours ... who went wrong where? - By definition, cast a spell on a stone to highlight an EMPTY pedestal, preferably the one the stone should go on. - The first stone was placed either on its (a) correct pedestal or (b) an incorrect pedestal, but either way let's refer to this pedestal as "X". By definition, casting a spell on this stone will identify either (a) a random empty pedestal or (b) its correct location, with (b) being far more likely (99/100). - Assume for now that (b) was the case. Picking any stone and placing it on the highlighted pedestal will guarantee it is placed _incorrectly_ and thus highlight its correct pedestal -- *EXCEPT IF* it belonged on Pedestal X, forcing the spell to highlight an empty pedestal at random (with some probability of being the Keystone Pedestal). But for now it is still more likely (98/99) that the home pedestal will be empty, and get correctly highlighted. - Assuming the more likely outcome was also the case, the above step repeats, now with a (97/98) probability of highlighting a correct pedestal, and a (1/98) probability of highlighting a random pedestal. - Iterating again, we have a (96/97) probability of highlighting a correct pedestal and a (1/97) probability of highlighting a random pedestal. Yet another iteration yields a (95/96) probability of a correct pedestal and a (1/96) probability of a random pedestal. - If we ultimately never find the stone that correctly belongs on pedestal X, this is a (n-1)/(n) probability multiplied over a sequence of n=[1 - 100] which conveniently simplifies to a (1/100) overall probability. And if it does succeed then _by definition_ the initial stone must have been placed correctly all along, which itself was a (1/100) probability to begin with. These numbers sync up! - Thus, we know there's an overall 99% chance that at _some_ point your process will highlight _at least one_ pedestal at random, every random pedestal having a probability of being the Keystone Pedestal and failing the puzzle as a whole. (I do not know how to compile the probabilities of randomly picking Pedestal X over the entire sequence.)

@tavern.keeper

Жыл бұрын

@@Stratelier The issue is that stone x might be the last one before the keystone. In that case, there is a 50:50 chance of it indicating the keystone platform. The better approach is to not place any stones. Just take note of which platforms are lit up. That way, the one random indicator has the least chance of targeting the keystone platform.

@dhruvaggarwal9755

Жыл бұрын

The placement spell can only be use once on a stone which was used by the henchmen

@UltraLuigi2401

Жыл бұрын

Alternatively, cast the placement spell on every stone (including the one that was already placed), keeping track of what pedastal lights up for each but not placing any. If a pedastal lights up twice, then the pedastal that didn't light up must be the keystone pedastal. If every pedastal lights up once, then either the already placed stone is on the keystone pedastal, which is a failure no matter what, or the random stone hit the keystone pedastal, in which case you can follow your strategy, except that the placement spell already has been cast on every stone. I'm not sure the exact probability of success with this method, but I think it's greater than yours.

Step 1: say at least one of you has green eyes Step 2: wait 100 days for the gems to confirm they all have green eyes Step 3: all the gems leave the island all having asked to the night before Step 4: miss your shot on purpose Step 5: wait for either of the wizards to be turned into either fish or stone Step 6: coat the outer layer red Step 7: profit

@desihirohamada

Жыл бұрын

you forgot about asking tee whether the alien overlord on the right is arr

@GTron13

Жыл бұрын

@@desihirohamada You also forgot to toss the gems out the window to see if the keystone would survive the same fall.

@desihirohamada

Жыл бұрын

@@GTron13 good point, but did anyone remember turning on the first unlit platform we see and looping back to see if we've done the loop correctly?

@notinsideyourwalls

Жыл бұрын

you forgot about splitting the team up to find the exit before the temple collapses and two random team members are free of the curse

@gilmulth

Жыл бұрын

This is truely a KZread moment.

I was so madly writing a comment on how wrong the calculation is at 4:00 because it seemed like you're forgeting the elimination of any number being picked between 1-100 but then I realised it doesn't matter. Probability and possibility is always so hard man 😭

How did Moldevert escape the chessboard??

@potatoheadpokemario1931

Жыл бұрын

he didn't, that's his horcrux body

@gdfreezerburn9250

Жыл бұрын

He confirmed he has green eyes and asked the guards to leave XD

@arcengal

Жыл бұрын

magic

The question no one asked, but everyone wanted a question.

You don’t need to place all the extra stones after you drink the luck potion, just place the keystone on the platform you vibe the best with. The incredible luck the potion brings will either guide you to the right pedestal, or you were already screwed

If you are with Dumbledore there is no problem.

There is another way. You can cast the spell which highlights the platform for each stone before placing them. There is a high chance that there will be a time when one platform will light up twice. That's when you know which stone was randomly glued to which platform and the problem is solved. It has a more favorable chances of winning than 50/50.

@bananaforscale1283

Жыл бұрын

You can use magic only once per stone.

@kingdelune

Жыл бұрын

@@bananaforscale1283 You would only need to use it once per stone. The only difference is that you wait before placing them.

@pinesmoke618

Жыл бұрын

@UCJAb17yEai2_WzyVuVF_-KQ If a platform glows twice, there will be 2 left unglown: one for the keystone and one for the stone that was glued down. You’ll have to guess, making it a 50/50.

@zwergesel

Жыл бұрын

I thought the same thing. The rules say that the correct platform glows when casting the spell. They don't say that we have to actually place the stone there. But actually it doesn't really help because you'll be left with two platforms that never glow: the one where the glued stone belongs and the correct one. So you're left with a 50/50 again.

@mambodog5322

Жыл бұрын

@@zwergesel At that point, you could cast the spell on the glued stone (they never said you couldn't do that, just that it can't be moved), and light up the pedestal it belongs to, leaving the only unlit pedestal to be the keystone's

Thanks for finally posting another riddle! I'm subscribed just for these.

I missed your guys' riddle videos! they're always so good!

the dumbest thing in the entire HarryPotter universe is the lack of "wristbands"(like the ones on the Wiimote) keeping the wand close & impossible to loose!

I always patiently wait for these riddles and Ted-ed never disappoints ❤️❤️

Honestly speaking, I simply would have picked the keystone and destroyed it. My job is not letting Moldevort get the all powerful wand. Destroying the keystone itself is the most sensible and easier option according to me as it would not only keep Moldyvort from the wand but also the upstart aspiring future Dark (read: Dork) Lords from the wand!

@metal_pipe9764

Жыл бұрын

For me the nuclear blast would have probably destroyed it

@pillypuppy666

Жыл бұрын

The keystone is immune to all spells, remember?

@The.Intersection

Жыл бұрын

​@@pillypuppy666 In the video it says it is immune to any form of magic, but no where does it say it is immune to any other forms of destruction...one can simply blast it into pieces using dynamite or rdx...non magical means can also be used to destroy it...

@gomshom67iscool23

10 ай бұрын

@@The.Intersection they didn't say when this was set. we can just blow it to pieces with the death star

When you get the right answer but only because you didn't understand the riddle correctly

You have no idea how much it means when you post a riddle omg like I spend an amz9ng time w my dad and I'm so grateful tysm

Wouldn't drinking a potion of luck /after/ doing the math be pointless? You've already eliminated the uncertain nature of the puzzle. You'd be much better off drinking it first, then picking up the keystone and randomly selecting the location. At that point you're working on nothing but luck.

@zmaj12321

Жыл бұрын

This doesn't make sense. If I drink a luck potion, then flip a coin, I should be more likely to get the result I want even though I already knew it was 50/50.

This is the first ever TED-ed riddle I've figured out in the time they give you to pause and do it. I was absolutely baffled when I pressed play again and they started saying what I had thought 😂

Wow. You just made me realize something so simple… I really appreciate this.

RIDDLE!!! Love when you guys post these!!!

Watching other Ted-Ed riddles, and this riddle appeared. How convenient! 😅

Everyday my desire to be a Ted ed video concept Writer increases.

Whenever there is a stone that is going to be randomly placed, it can do one of three things: Moved to the keystone's placement, moved to the placement of the first stone, moved to any other remaining placements. The first two scenario's have always an equal chance of occurring. When the first scenario happens, you lose because the keystone can no longer go the correct placement. However when the second scenario happens, then all the stones that follow will go to the correct placement, which includes the keystone. The third scenario which is usually the most likely one will just lead to a repetition of this whole setup except now another stone will be randomly placed. So to simplify there are three states: Win, Lose, Repeat. You start in the state Repeat and either go to this state again with some probably (can be 0% to 99%, but what it is doesn't matter) or you go from that state to the Win or Lose state with equal probability. Since it will always lead to Win or Lose and both are equally likely to happen at any time, we can conclude that you succeed with 50%.

First Ted-ed riddle I've gotten right, ever

Why is it that the riddles are less and less about logic and more and more about math

@SpacyPVP

Жыл бұрын

idk man

@tavern.keeper

Жыл бұрын

That's the same thing.

@crem-crem4070

Жыл бұрын

@@tavern.keeper no, it really isn’t.

@AlgerianRatt

Жыл бұрын

@@tavern.keeper how dare

@vylbird8014

Жыл бұрын

Math is just logic, but more so.

please never change your intro sound. it just feels so familiar listening to it in the beginning.

thank you god for giving me a blessing (a ted ed riddle that i canr solve that i watch for the plot) when i need it

Finally I understand the solution!! I love all kinds of Ted ED's video.🥰 they are very interesting.

Love how the wand is named after Maryam Mirzakhani, an Iranian mathematician. ❤

The spell tells what platform to place the stone on, but you don't actually have to place it. So you record for each stone without placing any of them and find the duplicate spot which tells you one of the two stones that showed that spot originally went on the locked platform. It still ends up at a 50/50 since you end up with two open spots to place the key, but somehow feels smarter.

Hands-off the best Ted-ed series is obviously the riddles

Y'know, you don't need to place the other stones. Just keep track of which platforms glow. Use the spell on all the stones, and either one will glow twice and one not at all, or each will glow once. In the latter case, you just need to use the spell on all of them one more time. Do this until you have one platform that didn't light up. That platform belongs to the keystone.

FINALLY A NEW RIDDLE

“Felush fe-lucious potion” _BRILLIANT!_

You could repeatedly remove and replace the stones in a random order leaving out the 100 that can have magic used on it doing this you would know the two pedestals in question because if you track where the stones normally light up vs the change where they light up you can narrow down the stone that was tampered with

What I want to know is how Moldevort escaped that darn 5 x 5 checkered board

Does the glued down gem become immune to magic? If not, then just cast the location spell on stone 1, and place the stone with the place value onto place 1. That way you are guaranteed a win

you forgot to mention that after one placement is glowing you *must* place your stone in it, which completely changes the question

The Moldevort Ted-Ed-ematic Universe is expanding. Can't wait for the live adaptation!

My dad taught me a solitaire game we called "4 kings" wich uses this same logic, just changing the pedestals and gems for face-down cards. It was fun but it doesnt really involve much ability, its mostly pure luck.

Drumbledrore , Moldevort and Myself Potter Harry...🙂

@eshitasahu

Жыл бұрын

*hotter parry

@hardiksnair

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@@eshitasahu Better..😌

@rafsanjanijarif369

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@@eshitasahu 😂😂😂

How does casting the placement spell on the glued stone change the answer? Obviously you can't move it but it would tell you where it properly belongs.

NEW TED ED RIDDLE!! this is a great day :D

I like how Ted ed is making part 2 riddles

Good video.

The series continues!

Cast the placement spell on each stone but remove it from the platform before the next one. That way you can find the one other stone that's randomly placed and have a 50/50 chance of finding the keystone platform

We can check every single stone and write it's number on the glowing pedestal. If there is any overlap, we will know which pedestals to avoid

I'd place each stone next to the platform that lights up, so i would know when 2 stones light up the same platform

is it exactly 50-50? The very first time you cast the spell and the gem is supposed to go on the pedestal that is taken, there is a small chance of going on the keystone spot but it can't go on it's correct spot so there is one play that has no winning result. I very well may be confused.

This is one of the only TED-Ed riddles I've gotten right lol

*"JUST DO IT!!! WE NEED TO FILL IN THE STONES NOT THE ODDS"*

Someone had fun drawing moldevort!

Wouldn't you have a 99% chance to win? The stipulation does not say you cannot cast the spell multiple times on the same stone. So if I were to cast it on stone 45 and it should light up pedastal 45. I the place stone 3 on pedastal 45 and cast the same spell on stone 45. It should light up a random pedastal and then take stone 3 off the pedastal and then cast the same spell on stone 45 and it should light up the pedastal 45 again confirming it was the right spot.

@bananaforscale1283

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Rule number 5

@Stratelier

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Nope, rule #5 each stone only reacts to a spell once.

Use the spell on the glued gem. And see where it glows. The others, use the spell and write down where they go. There will be exactly one random gem. When done, there will be one spot with two gems and one free spot, place the keystone there. Or No free spot, and you have to spell them all again, don't forget the glued one!. If they all go to the exact same place, your random stone went to the same spot, rinse and repeat.

Love these

Can't you cast the spell on the stone stuck on the pedestal? It hasn't had the placement spell cast on it yet.

I made a paper about this a couple years ago in high school, everyone, even the teacher told me I was wrong. I’ll never get over it

What a nod and wordplay to The Boy Who Lived! (I mean, dead Sirius!)

"HAROLD DIDJA PUT YER NAME IN THE CUP OF FLAMES?!?!?!?!" drumbledrore questioned peacefully,

If you pick up a key stone and the place lights up, you don't have to put it down right? You can just put it back and keep picking them up one by one until two of them are the same spot. After all, a time limit wasn't given

IVE MISSED THESE

Why can't you just use the placement spell and just not place the stones? Just keep a note of which platform lights up when a spell is casted. So basically in the start, the invisible numbering you said would now be visible. So I don't get why it would be a risk ?

What is it ends in a loop, thats when you get a higher chance in placing the keystone correctly?

Oh I had a clever 50/50 strategie. If we aren't forced to palace a stone once we see a pillar light up we can just look at all positions for the stones without placing them now only one pillar should have lighted up twice and we can safely place the rest now its only down to choosing the correct of the two which depending on how the randomization work can either be trivial or down to 50/50 chance

There is a bonus trick you can use: by casting the placement spell on the randomly placed stone, you can know which platform the stone was supposed to be on. Then, if another stone shows for that platform, you know you have succeeded, and if not, you know you have failed.

Alright, I solved it. Step one: get someone you don't like and who doesn't seem too bright (Beville) Step 2: tell him everything about the cave. Step 3: wait outside. That one is the tricky bit because Moldevort may have the same plan. You may need to bring along that chess board. Step 4: if there is a (very likely) cave in, good job protecting the staff. If there isn't, refer to step 5 Step 5: take the staff from Beville as he walks outside. As stated, Moldevort will have the same plan, so have him dealt with.

I got 50/50 a different way. If you put each stone in the FRONT of the pedestal that it glows for, you will arrive at two possibilities. One (more likely), two rocks are in front of one pedestal, which means that the keystone could be placed on the glued stone's original spot or the keystone's correct spot. Two (less likely) one stone is in front or each pedestal, meaning that the keystone is occupying the one remaining spot, either the correct one or the glued stone's.

@metal_pipe9764

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I got a 100/100 via a nuclear explosion

Really supporting the “it either happens or it doesn’t therefore it’s 50/50” theory

or you enchant each stone twice. The first time just to see the position and the second time to check if the position has changed. Because if the actual place of the stone would already be taken, it shows a random one. Which means that every time the occupied stone is enchanted it shows a random place.

@JoeThomas-lu6fy

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Look at Rule 5.

Just use magic to protect yourself from the cave-in and place the keystone on every pedestal, then use that magic wand to get out of the cave. Or, put every stone except the keystone on a random pedestal and leave, so Moldevort either gives up on finding the correct place for the keystone, or he causes the cave-in himself. Or, find someone willing to sacrifice themselves to do the riddle, either you get the wand, or nobody gets it. Or, join Moldevort's side (and maybe potentially assassinate him).

interesting how a normal probability problem with some storytelling can get me hooked to a video

i think there was another way , would take a few seconds extra , lift each stone twice before placing . you first lift it see where it glows and then again , if a stone glows in a different position that's the whose place was taken

Alternatively, take notes on which platforms light up without actually moving the stones. Inevitably you will reach a situation where one platform lights up twice and you can know for a fact that one of two stones has had it's spot taken. place one of those 2 one said platform and then place the rest. There are only 2 spots left, so you've got a 99 in 100 chance of having a 50/50 shot at winning.

I always misunderstood the rules. I’m over here like “just spell each stone twice, only one will change stands both spells because only one will be random.”

This was a cool lesson on statistics. Thanks.

You dont need to place the stones, you can just light it up and track it. In the end there will only be either 2 which are not lit or 1 which is not lit. The probability would be slightly more than 0.5 I’d say. Correct me if im wrong

I gave up at 0:25....my little brain can't

Mentioning Maryam Mirzakhani‘s name was excellent! Well done

Bro the TED-Ed riddle verse is deepening

Would using the identify spell on a stone that’s spot has been occupied give different locations each time? Can you use the identify spell on the first stone that was already fused? Why not just take the keystone and put it somewhere else? There are so many ways to circumvent the problem since the goal isn’t to actually get the wand.

Serious question. What if there are two three or n number of misplaced stones.