Can you steal the most powerful wand in the wizarding world? - Dan Finkel
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- Опубликовано: 25 янв 2023
- Practice more problem-solving at brilliant.org/TedEd
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The fabled Mirzakhani wand is the most powerful magical item ever created. And that’s why the evil wizard Moldevort is planning to use it to conquer the world. You and Drumbledrore have finally discovered its hiding place in a cave, but the wand is hidden by a system of 100 magical stones. Can you figure out how to get to the wand before Moldevort? Dan Finkel shows how.
Lesson by Dan Finkel, directed by Igor Coric, Artrake Studio.
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怎么只有14个赞？！
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?
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.
Friggin genius
That's a proper answer to a proper riddle
Yessss!
Even better, since the Henchman basically just ruined it for Moldevort 😂
Congrats on having 1999999999999999999 iq
If only the Elder Wand was hidden as securely as this...
Lol
Ikr but weird this is about the horcrux, people. not the elder wand…
Ya
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
It is weird how Dumbledore never trained harry...
Like he was a sacrifice. All planned out...
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!
He'll soon be benchman
I assume he knows the only needed pedestal.
@@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
If the henchman accidentally sealed away the wand, then they ironically saved the day
Before apparating was invented
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?
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
@@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).
@@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.
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.
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.
Now I noticed that. cool.
I didnt even know about a Mirzakhani but now I'm interested
An Iranian mathematician
@@Pickled_Poet She is a fields medal winner , once in a lifetime gem that humanity lost, Maryam Mirzakhani.
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.
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.
LOL
You forgot to cut the werewolf antidote into 5 squares!
@@tezsaw486 I didn't forg- *turns into werewolf*, oh.
We also can’t forget to make the scientist and the janitor cross the bridge together.
Let's do this for every time Ted-Ed makes a new riddle video.
I love how Ted ED shows us that Mathematics are so important it can literally destroy the world.
E=mc²
🔝🌟
Eh, typical Ted Ed.
V+F = E+2
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.
FINALLY A RIDDLE
tom riddle
YESSS the moment we been waiting
@@powerllesss2672 LOL
@@powerllesss2672 tom riddle 2
Finally
i would totally watch a movie with Moldevort and Drumbledrore in it. Oh wait a minute....
Yeah... there's a "wait a minute" when you realize that it's drumbledrore and moldevort but not dumbledore and voldemort.
Harry Potter and the Half Blood Prince
@@aarushiyadav7101 Fortunately drumbledrore didn't die in this
@@aarushiyadav7101 no no no its parry potter
No look at the video there’s an M on the blue girls robe- I MEAN TUNIC. It’s clearly Mary potter
_“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_*
214 likes and no comment. seriously?
The wizards at Hogwarts only get normal education till 11 so that makes sense
lol
Finally understood the plot of “Mary Plotter and the Deadly Platforms”
Lol
"Harry Potter and the Deathly Hallows " and "Mary Plotter and the Deadly Platforms" hmm sounds familiar
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🤣
As he always does😂
And to leave the problem to ungraduated teenaged students
@@iaditiagrawal Lol
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!
Step 5 is wrong, the correct step is to say at least one of you have green eyes
The amount of references in this one comment is more than i have money
@@ZFroZenHail Done 👍
@@k3ose455but somehow we know where all the references came from😂
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?
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.
It doesn't change the end result, but placing them makes it easier to keep track of which pedestals have been already identified.
@@Stratelier i mean if u actually write down some schemes on paper just watching which pedestals lights on u can do it easier
@@Stratelier since you sign all the combinations and watch which one of them overlap and similiar
@@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.
@@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.
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.
You can't
@@bananaforscale1283 Why not?
@@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.
@@0mathgaming rule 5
@@bananaforscale1283 Rule 5 only says that you can't cast a spell on a given stone more than once.
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 😭
Child: I want to watch Harry Potter!
Parent: We have Harry Potter at home, sweetie.
Harry Potter at Home:
Child: "Wow! This is better!"
@@caelincoolz5814
The “child” would say the opposite tbh
@@youyaku-music Harry Potter is pretty awesome.
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%.
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.
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.)
@@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.
The placement spell can only be use once on a stone which was used by the henchmen
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
you forgot about asking tee whether the alien overlord on the right is arr
@@desihirohamada You also forgot to toss the gems out the window to see if the keystone would survive the same fall.
@@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?
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
This is truely a RUclips moment.
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
How did Moldevert escape the chessboard??
he didn't, that's his horcrux body
He confirmed he has green eyes and asked the guards to leave XD
magic
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.
You can use magic only once per stone.
@@bananaforscale1283 You would only need to use it once per stone. The only difference is that you wait before placing them.
@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.
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.
@@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
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!
For me the nuclear blast would have probably destroyed it
The keystone is immune to all spells, remember?
@@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...
@@The.Intersection they didn't say when this was set. we can just blow it to pieces with the death star
Thanks for finally posting another riddle! I'm subscribed just for these.
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!
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.
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.
I missed your guys' riddle videos! they're always so good!
I always patiently wait for these riddles and Ted-ed never disappoints ❤️❤️
When you get the right answer but only because you didn't understand the riddle correctly
please never change your intro sound. it just feels so familiar listening to it in the beginning.
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.
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
First Ted-ed riddle I've gotten right, ever
If you are with Dumbledore there is no problem.
FINALLY A NEW RIDDLE
you forgot to mention that after one placement is glowing you *must* place your stone in it, which completely changes the question
Love how the wand is named after Maryam Mirzakhani, an Iranian mathematician. ❤
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.
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%.
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
Why is it that the riddles are less and less about logic and more and more about math
idk man
That's the same thing.
@@tavern.keeper no, it really isn’t.
@@tavern.keeper how dare
Math is just logic, but more so.
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.
Watching other Ted-Ed riddles, and this riddle appeared. How convenient! 😅
What I want to know is how Moldevort escaped that darn 5 x 5 checkered board
Everyday my desire to be a Ted ed video concept Writer increases.
Finally I understand the solution!! I love all kinds of Ted ED's video.🥰 they are very interesting.
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.”
RIDDLE!!! Love when you guys post these!!!
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
Hands-off the best Ted-ed series is obviously the riddles
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.
Rule number 5
Nope, rule #5 each stone only reacts to a spell once.
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
The Moldevort Ted-Ed-ematic Universe is expanding. Can't wait for the live adaptation!
“Felush fe-lucious potion”
_BRILLIANT!_
I'd place each stone next to the platform that lights up, so i would know when 2 stones light up the same platform
This is a famous problem in AoPS, like the flight seat and think about whether the last person will get to the right seat or not.
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.
Someone had fun drawing moldevort!
"HAROLD DIDJA PUT YER NAME IN THE CUP OF FLAMES?!?!?!?!" drumbledrore questioned peacefully,
*"JUST DO IT!!! WE NEED TO FILL IN THE STONES NOT THE ODDS"*
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.
I like how Ted ed is making part 2 riddles
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).
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!)
Really supporting the “it either happens or it doesn’t therefore it’s 50/50” theory
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
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.
The series continues!
NEW TED ED RIDDLE!! this is a great day :D
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.
This was a cool lesson on statistics. Thanks.
This is one of the only TED-Ed riddles I've gotten right lol
seal the other 99 stones in random spots as well to deny the wand to moldevort, we dont need the wand, plus leaving the keystone without function there is a solid taunt in his face.
Just cast the spell on the placed stone. Or the same stone twice
Drumbledrore , Moldevort and Myself Potter Harry...🙂
*hotter parry
@@eshitasahu Better..😌
@@eshitasahu 😂😂😂
Bro the TED-Ed riddle verse is deepening
IVE MISSED THESE
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
Mentioning Maryam Mirzakhani‘s name was excellent! Well done
Won't moldevort also have the feleush fe-leush lucky potion spell? It would be a matter of who gets there first now won't it
But Moldevort has the advantage because the problem was only caused by his minion in the first place. If he already knew where the cave was located than he could have gone there first and solved it without the puzzle being an issue.
@@AZ-rl7pg since moldevort got there first, imagine when you solve the puzzle it's empty since moldevort probably solved it and just wanted to be a Menace lol
interesting how a normal probability problem with some storytelling can get me hooked to a video
Cast the spell twice, since the stone meant to go on the now occupied platform will light another platform at random then it will be the only one that doesn't light the same platform twice in a row.
Assuming the keystone platform is not occupied:
1) Pick all the remaining stones up, but don't place them.
2) Don't touch the platform that lights up twice.
3) Proceed to place all the stones until you have just 3: the keystone and two of the stones that lit the same platform.
4) So now we have just a 50/50 chance to place the keystone correctly, since there are 2 platforms that don't light up.
5) We still have a 50/50 chance, but we feel better, because we used logic. Time for a luck potion, because there is no solution. Even if the stone whose platform is taken changes the platform each time it's picked up, in that case we can place the stone that doesn't behave this way and still achieve nothing.
The rules say that a placement spell can only be used once on each stone, but it doesn't say that the spell has to be used before it's placed. It also says that the henchman sealed the stone without using a placement spell on it. There's no reason not to use the placement spell on the sealed stone and mark its pedestal to know which one it is.
Tfw I actually explained this problem to my fellow classmate the last year XD. But my variation is a crazy grandma on a plane, who sat on a random place, and other passengers choose their place if it is not occupied, or random is it is
So you cast the placing spell on all gems without actually placing them, the one platform that dont lit is the one for the key stone, if they all lit that mean that either the key sone's platform is taken or you are in the 1/99 chance that the randomness lost you which is a better odd than the 1/2 the video shows
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
Ted ed is my favorite way to study for my classes 😭
Oh. This video went VERY differently than I thought it was going to. Here is my strategy:
The question only said that the appropriate platform would glow; it never once said that you are committed to placing it there. You can greatly increase your chances by casting the placement spell on each of the 99 gems to see which platforms they belong on, WITHOUT actually placing them there.
There is also no reason that you cannot cast the spell on the one that is bound (it specifically said that you can cast the spell on ANY stone other than the keystone), so that the platform that the bound stone is SUPPOSED to be on will glow (unless it happens to be on the correct platform already, in which case a random one will glow). Either 98 platforms will glow (with one glowing twice), or 99 will glow. The platform the bound stone is on will never glow, as it is always occupied.
There is a 98% chance that one of the platforms glows twice (and 97 glow once). In this case, simply place the keystone on the one that doesn't glow and you are guaranteed success.
There is a 2% chance that 99 platforms each glow once - this means that either:
(1) the bound stone is on the keystone's platform already (50% chance; guaranteed failure)
(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 (0.51% chance)
(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 (49.49% chance)
If 99 platforms each glow once, then best you can do is a 0.51% chance of success if you EITHER select the platform that activated when you cast the spell on the bound stone (i.e., guaranteed to work ONLY in scenario 2), OR randomly guess among the 99 available platforms. Either strategy is slightly better than avoiding the one that activated when you cast the spell on the bound stone (i.e., counting on scenario 3).
If 99 platforms lit up, I'd probably just leave without placing the keystone. Too risky. But there is only a 2% chance of this outcome, so my strategy is much better than the 50/50 odds you get with the strategy in the video.
you can cast spell twice per stone too
problem tho isn't in finding which stone has its pillar occupied, but which pillar has its stone taken
and after determining 97 pillars with correct stones, you're still left with 50/50 where to place orphaned stone and where the magic gem
Good video.
I suppose I misunderstood the game. By using simplified games consisting of 3 and 4 stones, I figured everything up to 2:18. Then, for the third scenario (discussed onwards), I thought the every stone would find a random place because in this scenario one stone is taking another's place, forcing the one who no longer has a numerically logical place to assume a random position, taking the place of another stone and so on. If this was the case, the probability of escaping with the keystone could be expressed as 1/100+(1/99*98/100), or about 0.198989 to my calculations. Furthermore, we can create a formula to figure out the probability of finding the keystone for any natural number of stones (as long as the number is equal to or greater than 2, logistically speaking) as far as I can tell, the probability of being able to find the keystone is expressed as 1/x+(1/[1-2*1/x])*(1/[n-1). But alas, I was apparently solving the wrong riddle...
Love these
These are prime picks for Zelda-esc video game puzzles!