Can you steal the most powerful wand in the wizarding world? - Dan Finkel
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- Опубліковано 20 вер 2024
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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
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
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...
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.
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.
_“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
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
@@Arushi701 Fortunately drumbledrore didn't die in this
@@Arushi701 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
FINALLY A RIDDLE
tom riddle
YESSS the moment we been waiting
@@powerllesss2672 LOL
@@powerllesss2672 tom riddle 2
Finally
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?
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
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.
I changed the comment, so your will never know how I got this many likes...
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 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 😭
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.
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 UA-cam moment.
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
Well worth the wait, I love ted-ed riddles, I do wish they were more consistent though, they are always banger videos every time 😎👍🏻🔥
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.
Gotta love the continuity - the wizard representing us is wearing the M logo for the Magnificent Marigold’s Magical Macademy like in that previous sorting hat/house video.
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.
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
The question no one asked, but everyone wanted a question.
My father was trying to solve this and had a huge sheet of paper out in preparation to do the calculations, then I came in, took one look at the riddle and just went: "It's 50/50." The look on his face when I turned out to be right... Impeccable.
Nuke the puzzle to eliminate it to 0 choices and a big crater
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
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!
you forgot to mention that after one placement is glowing you *must* place your stone in it, which completely changes the question
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
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.
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.
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%.
If you are with Dumbledore there is no problem.
Who tf is Dumbledore? Do you mean Drumbledore?
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 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.”
"HAROLD DIDJA PUT YER NAME IN THE CUP OF FLAMES?!?!?!?!" drumbledrore questioned peacefully,
Thanks for finally posting another riddle! I'm subscribed just for these.
Really supporting the “it either happens or it doesn’t therefore it’s 50/50” theory
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.
“Felush fe-lucious potion”
_BRILLIANT!_
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 😂
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
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...
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.
Here's an alternative way of making sure it isn't the wrong slot: pick each stone up twice. If it lights up the same twice, it's the correct one. If it lights up different slots each time, it's the occupied slot, so dont fill that in yet. Finish the rest of the stones until 2 slots are left, 1 and 100, then drink the luck potion and try the last stone again. Only thing remaining after that is getting the wand.
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).
First Ted-ed riddle I've gotten right, ever
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 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
I missed your guys' riddle videos! they're always so good!
I gave up at 0:25....my little brain can't
Drumbledrore , Moldevort and Myself Potter Harry...🙂
*hotter parry
@@eshitasahu Better..😌
@@eshitasahu 😂😂😂
1:08 imagine he does this but he actually puts it on the right one
moldrvort:did u stick the stone on a random platform
Henchman:goes into cave and sees the wand taken
moldrvort:did u pit the right stone on the right platform
Henchman:y..yyes
moldrvort evada kedavra
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.
What I want to know is how Moldevort escaped that darn 5 x 5 checkered board
FINALLY A NEW RIDDLE
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.
Everyday my desire to be a Ted ed video concept Writer increases.
To succeed, one must always have a perfect mentor, or be the perfect mentor. The ultimate mentor is perfection.
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.
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
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.
Voldemort after seeing thumbnail : Avada Kedavra 🐍
The Moldevort Ted-Ed-ematic Universe is expanding. Can't wait for the live adaptation!
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
*Moldervolt casually taking out his UNO reverse card*
Moldivort is truly the worst, most despicable villain we as a society will ever know
... Read more
i fell for it
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.
Please make a video about Lyndon b Johnson vs history.
Every Harry Potter reference: Moldevort->Voldemort,Drumbledrore->Dumbledore,The cave-> the cave where Harry and Dumbledore go in Half-Blood Prince,and you have a lightning shaped scar on your cheek.
HOLD UP!
You never said we HAD to place the stones there. So don't place them. We have a 1/100 chance of winning anyway if the randomly placed stone is placed correctly, and a 1/100 chance of losing if the stone occupies the keystone slot.
Now let's say the stone which we will label as 1 is placed on 2 instead. This means that stones 3-99 will glow correctly. We can also use the placement spell on the already placed keystone which will glow up pedestal one. So let's say we scanned the actual #2 last. A random spot will glow. However 98 spots have already glown, 1 is taken and 1 is for the keystone. So this means that #2 will pick a random spot to shine (because as we have not actually placed any of the stones they are all available)
98/99 it will light up a platform that has already lit. Meaning that the platform that has not glown once is the keystone spot. However 1/99 times the keystone spot will glow. In which case we have no info and it is random.
So the 1/100 chance to fail from the initial placement and the 1/99 chance that the keystone pedestal is lit up. In every other scenario we place the keystone on the pedestal that has not lit up and is not occupied.
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
This riddle seems wrong in my eyes.
It is never said that all stones must be placed, only the keystone.
The platforms glows when you cast the spell on a stone, but you are not required to move the stone (it glows before you move it and the riddle never specifies that you must place them)
Let's say you test every stone by casting the spell on it. 97 stones will tell the correct platform and one will show a random platform.
In this scenario, it is most likely that one single platform will glow twice (once for its correct stone, and once for the stone that belonged to the sealed platform).
Therefore, at the end, just pick up the platform that you NEVER SAW GLOW. The only way to loose is if the stone that belonged to the sealed platform randomly selects the keystone platform when it need to select a random platform to glow. In this unlikely case (1/99 chances), the platform that did not glow is not the keystone but the one that belonged to the sealed keystone.
The odds of loosing are therefore 1/100 (henchmen sealed the keystone platform) + 99/100*1/99 (random glow on the keystone platform) or a 2% rate of failing and a 98% rate of success with this method.
Tl;dr : this riddle is only correct if you need to place the stone once you cast a spell on it, which the riddle never specifies
A similar line of logic, there's no rule which says you have to place a stone once you know which platform it belongs to. Identify where each stone wants to go, but don't place them. 1/100 times, the random stone will be where it belongs, and you will win by default. 1/100 times, the random stone will be on the key stone platform, and you lose by default. 98/100 times there will be a stone whose platform is taken, and you will end up with either a duplicate glowing platform (and a platform that has not glowed where the keystone goes), or all platforms having glowed. In the former case you win. In the latter case, you still have a 1/99 chance of winning. It's a 50/50 which case happens.
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
I love how Ted Ed just causally rips off Harry Potter right down to the small details like Felix falecios or whatever the potion is called to felush felusious it’s a masterpiece 👌 4:34
*"JUST DO IT!!! WE NEED TO FILL IN THE STONES NOT THE ODDS"*
1. Number stones and platforms 1 to 100.
2. Cast magic on every stone, but don't place it and write a pair of stone-platform numbers.
3. Search for a platform that was highlighted twice (99% chance), if it was, you have 100% chance to win.
4. If it doesn't (1%) chance, you have 50% chance to win.
The total chance to win is 99.5%
Step one: Confirm Moldemort has green eyes and ask to leave.
Step two: Pick the weakest wand and miss on purpose.
Step three: Say ozo.
Step four: Get magic dice and win the game.
I found this to be an excellent explanation (thanks, TED-Ed!), but if anyone found this confusing, I want to share one of the simplest possible examples of this dilemma: Rock, Paper, Scissors.
When you play Rock, Paper, Scissors, there are nine possible match-ups that can happen between you (P1) and your opponent (P2):
(P1) Rock = Rock (P2)
(P1) Rock → Paper (P2)
(P1) Rock ← Scissors (P2)
(P1) Paper ← Rock (P2)
(P1) Paper = Paper (P2)
(P1) Paper → Scissors (P2)
(P1) Scissors → Rock (P2)
(P1) Scissors ← Paper (P2)
(P1) Scissors = Scissors (P2)
Regardless of which choice you make, there are only three possible outcomes - or rather, there are two final outcomes: a) you win, b) you lose, or c) you tie, and you repeat the challenge until you end up at outcome a) or b). In the end, you have a 50% chance of winning, and a 50% chance of losing.
What the video says about the riddle applies to this simple game as well: "You're playing a game where you have equal chances to win and lose, and some chance to delay the decisive moment. No matter how many times this process repeats, you'll inevitably [make the choice that causes you to lose] or [make the choice that causes you to win]. That's all that determines whether you succeed or fail, and critically, the chances of those events are equal... It might take a while, but the delays don't give an advantage [to winning or losing]." This riddle is basically a bigger version of Rock, Paper, Scissors.
...I was going to add something else here, but honestly, I think that covers it. I hope that helps anyone who was looking for some extra explanation.
P.S. Yes, I do have green eyes. No, none of my passwords are "Ozo" - no website or program will let you use a password that short. And in my real life, I would succeed at a lot more challenges and logic puzzles if I got to press a pause button to think for a little longer.
Correct solution: "Hey, Dolby! Put this stone over there. We'll be waiting outside."
Wow. You just made me realize something so simple… I really appreciate this.
I'd place each stone next to the platform that lights up, so i would know when 2 stones light up the same platform
What a nod and wordplay to The Boy Who Lived! (I mean, dead Sirius!)
Who said that we have to place the stone on the pedestal? All the puzzle said the spell did was make its correct platform glow, no one said anything about HAVING to place the stone there immediately.
So instead of placing the stones, pull out some chalk, number the stones from 2 to 100, then cast the spell on each in turn and write the number on the pedestal. No chalk? Place the stone on the floor next to the pedestal.
Somewhere along the way either (a) we'll have the same platform glow twice (once for a correct placement, once for the random) and one doesn't glow at all (the keystone's correct placement, (you win!)) or (b) every pedestal glows exactly once except one (because the random stone was either placed in its correct position or the keystone's position, regardless of which the other pedestal won't glow).
Scenario (a) is *far* more likely meaning we can use logic to guarantee a win and (b) gives us the same 50-50 (if this happens, drink the potion).
this riddle is actually pretty funny. this logic proves that the frog riddle is actually correct.
Fun fact : Voldemort real name is Tom RIDDLE
well played Ted-Ed, well played 👏
Someone had fun drawing moldevort!
Bro the TED-Ed riddle verse is deepening
Just ask Drumbledrore to unseal the stone off the pedestal. What is the probability "a henchman" of Moldevort knows magic Drumbledrore doesn't?
In general, probability of stone n's spot being occupied = probability of n-1 occupying that spot + probability of n-2 occupying that spot...+ probability of n-(n-1) occupying that spot, with the base case that probability of space 1 being occupied is 0.
I love how they’re saying their names wrong to avoid copyright
"WE'RE SCREWED MAN!!!"
drumbledrore said calmly
Essentially it's just the same as if the henchman stole the first stone. After placing the other 98 stones correctly you'd have two pedestals left - the one that the first stone should've gone on and the one that the keystone should be on, and there would be an equal chance of either being the right one
I thought of exploiting a potential loophole: stone 1 is the only stone that cannot be removed from the pedestal it was placed on. All other stones except the keystone can have multiple placement spells cast on them. In the 98% chance that stone 1 was placed on pedestal 2-99, whichever stone was cockblocked by stone 1 will light up a random pedestal, and can therefore be detected by multiple placement spells that will most likely light up a different platform each time. For each non-keystone that was not cockblocked, casting a placement spell on the same stone will light up the same platform every time.
Unfortunately, this does not actually improve your odds. Even if you've detected the cockblocked stone, you can place all other non-keystones on their proper pedestals, after which there will be two empty pedestals (1 and 100), and two unplaced stones (the cockblocked stone and the keystone). Placement spells have no effect on the keystone, and casting it on the cockblocked stone will light up either platform 1 or 100 at random, but no amount of repetitions will tell you which is which. So your odds are still 50/50. 😥
Hmm the way I interpret the problem, you can get the right answer 97% of the time without guessing.
There is nothing in the rules that says you have to place a stone when you cast the spell or that you can't cast the spell on the stone that was placed randomly. So what you do is cast the spell on every stone, including the stone that was placed randomly, and record the results.
Let's say the randomly placed stone is on platform X but should be placed on platform Y. Our goal is to figure out which stone belongs on platform X, this is the stone that will give a random answer when we cast the spell. If we know that, then we know that the spell is giving the right answer for every other stone. So how do we know which stone belongs on platform X? If two stones cause the same platform to glow twice, then one stone belongs on that platform and the other belongs on platform X. There is a 98/99 chance of this happening. In this scenario, we know that the keystone belongs on the only platform that did not light up and the puzzle is solved. If we are unlucky, then every platform will glow once. The stone that belongs on platform X randomly chose the keystone's platform. There is nothing left to do but guess.
Using this strategy, there's a 98/100 × 98/99 ~ 97% chance to solve the puzzle without guessing. Not as good as using a luck potion, but much better than a 50% chance.
How did I do? Is my logic sound?
Whether or not you place each stone after casting the spell doesn’t change the result, it’s still 50/50. You are correct that casting the spell on the glued stone makes the probability higher, though. Based on the narration, this isn’t supposed to be a possibility, so Ted-Ed messed up by not including a rule preventing it.
@@PinesmokeArt You don't understand. The problem is entirely if you cast the spell without placing the stones because you can have two stones light up the same platform. That let's you test for the misplaced stone.
@@rose52152 No, you don't understand.
If you can't cast the placement spell on the glued stone, then after placing/removing the other stones you are still left with two remaining pedestals. One of these will be the keystone pedestal and one will be the pedestal for the glued stone.
50-50 odds.