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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.
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!
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
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.
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.
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.
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.
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!
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🤣
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 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.
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 😭
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
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 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.
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
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
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!
@@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...
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.
@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
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...
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.
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
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.
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%.
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.
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.
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.
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
Okay, let’s assume that in order to make sure moldevort will at least have a chance to get it right, being a good minion, let’s say the minion stuck stone 1 on a pedestal that the keystone doesn’t belong to, with the pedestal for the keystone being immune to having a stone sealed to it, so it’s pedestal 1-99, this does give you some info, but otherwise, the rules are the same, all stones will go to their correct spot, once you use the spell, unless a stone occupies it, where it will be random
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 😂
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.
Wow. Never thought I'd see the day. A TED-Ed video using one of the most broken franchises in terms of internal logic, to use as an example for logic. Well done.
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 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
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).
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
Idk if this was a mistake in the wording of the 5 rules... On the "Pause here" bit, but I would just cast the spell which tells you where it goes and keep note of which platforms light up with which stone, and then if a spot lights up twice, one of the stones the revealed it must belong on the one that was placed by the henchman If none light up twice, then you can play it safe by repeating this until you find the "faulty" stone pair, and then you can confirm it by checking both of them until one gives a varied result... Effectively eliminating chance (though there's a 1/50 chance you won't find anything despite repeating things, if the keystone's spot was taken or if the henchman placed his stone correctly, and you can basically confirm this through enough repetition to make the odds impossibly in your favour
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.
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
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.
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
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.
Can I just memorize where stones 1-99 would be placed and place none of them until I find the overlapping stone before placing the Keystone last? Then place the remaining stones after that and place the keystone on what hasn't been highlighted/occupied as stone 100? Since the sabotaged stone only makes random empty platforms glow, the stone it was before (29 on 45 in the video for example) would "overlap" one other random stone. Once the spell makes a pedestal glow twice (Would be 29 in the video) it's safe to place all the stones after that by placing stone 45 at pedestal 29's spot. Doing this allows the rest of the stones after 45 to not require memory to know their places for when you place the Keystone. If you got a magical dude who can make numbers appear, then he can help with the memory thing too. That's my thought... Edit: I know they put a 1 on the first stone placed and numbered the rest, I'm just explaining this as if they hadn't done that yet.
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
At first I thought the answer was 1/2 because it’r very similar to a well-known problem usually phrased involving airplane seats. That’s how the video treats it, in fact, making me think there was a mistake in re-theming the question. However, the problem is actually different, because we can cast the placement spell on everything *before* making another platform occupied. Unfortunately, I can’t actually see a way to *use* that advantage; the answer is still 1/2 (without the potion). Even if the “random unoccupied” result is memoryless, in which case we can identify which stone’s platform was taken (unless we’re in one of the two special cases where it’s 1 or 100, in which case we know that we’re in one of those but not which), I don’t see a way to use that; there’s still no way to distinguish platforms 1 and 100 from each other even when we know which two they are.
Think of it this way, 98 stones will reveal their correct platform, and one stone will lie. If the first stone is not on the keystone's platform, then the only way to lose is if the lying stone points to the keystone platform. Then you have no information and can only guess at where the keystone goes.
@@tavern.keeper That isn’t the only way. The lying stone might instead point to the same platform as another stone, or tell inconsistent lies so you know it’s lying, and either way you still can’t tell.
@@danielrhouck If it points to the same platform as another stone, then there will be exactly one platform that was not pointed to by any stone. That is the keystone's platform, and you place the keystone there to win.
@@tavern.keeper If it points to the same platform as another stone, there will be *two* remaining platform. There are 99 non-key stones; 98 after one is glued down. If two point to the same platform, 97 platforms glowed. 100 platforms - 97 glowing - 1 glued = 2 remaining. It all comes down to the fact that, once stone 1 is glued down, there is nothing that will honestly point to platform 1. This leaves no way to distinguish platform 1 from platform 100.
@@danielrhouck Cast the locator spell on the first stone as well. This is allowed within the stated rules. There is a restriction that the spell can only be cast once for each stone, but it does not say that the spell cannot be cast on a stone that has already been placed.
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).
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.
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.
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.
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.
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.
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
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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.
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
@@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
_“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
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.
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 👍
@@k3ose45but somehow we know where all the references came from😂
You forgot to flip 20 random coins
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
Well worth the wait, I love ted-ed riddles, I do wish they were more consistent though, they are always banger videos every time 😎👍🏻🔥
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
@@tamsteev4391Never heard of that Harry Potter whatever
I meant that they sound likeeach other @@josedaniel4810
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.
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 😭
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
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.
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
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.
The question no one asked, but everyone wanted a question.
Thanks for finally posting another riddle! I'm subscribed just for these.
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
please never change your intro sound. it just feels so familiar listening to it in the beginning.
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 keystone may be immune to magic, but are the platforms? Blow 'em all up!
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
I missed your guys' riddle videos! they're always so good!
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
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...
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!
0:14 bud has an interesting hairstyle, where have I seen that before 💀
Yeah😂
Nah 💀💀💀💀💀
you forgot to mention that after one placement is glowing you *must* place your stone in it, which completely changes the question
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.
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 always patiently wait for these riddles and Ted-ed never disappoints ❤️❤️
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.
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%.
Someone had fun drawing moldevort!
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.
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.
Wow. You just made me realize something so simple… I really appreciate this.
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.
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
FINALLY A NEW RIDDLE
Okay, let’s assume that in order to make sure moldevort will at least have a chance to get it right, being a good minion, let’s say the minion stuck stone 1 on a pedestal that the keystone doesn’t belong to, with the pedestal for the keystone being immune to having a stone sealed to it, so it’s pedestal 1-99, this does give you some info, but otherwise, the rules are the same, all stones will go to their correct spot, once you use the spell, unless a stone occupies it, where it will be random
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!
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.
And you forgot to program the multiverse teleportation robot.
First Ted-ed riddle I've gotten right, ever
If you are with Dumbledore there is no problem.
Who tf is Dumbledore? Do you mean Drumbledore?
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 😂
Love how the wand is named after Maryam Mirzakhani, an Iranian mathematician. ❤
The series continues!
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
RIDDLE!!! Love when you guys post these!!!
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.
The Moldevort Ted-Ed-ematic Universe is expanding. Can't wait for the live adaptation!
Everyday my desire to be a Ted ed video concept Writer increases.
Wow. Never thought I'd see the day. A TED-Ed video using one of the most broken franchises in terms of internal logic, to use as an example for logic. Well done.
Looking for mistakes in HP is a meme. It's as consistent as any other book series.
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.”
Really supporting the “it either happens or it doesn’t therefore it’s 50/50” theory
When you get the right answer but only because you didn't understand the riddle correctly
Watching other Ted-Ed riddles, and this riddle appeared. How convenient! 😅
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.
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
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).
Hands-off the best Ted-ed series is obviously the riddles
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
Idk if this was a mistake in the wording of the 5 rules... On the "Pause here" bit, but I would just cast the spell which tells you where it goes and keep note of which platforms light up with which stone, and then if a spot lights up twice, one of the stones the revealed it must belong on the one that was placed by the henchman
If none light up twice, then you can play it safe by repeating this until you find the "faulty" stone pair, and then you can confirm it by checking both of them until one gives a varied result... Effectively eliminating chance (though there's a 1/50 chance you won't find anything despite repeating things, if the keystone's spot was taken or if the henchman placed his stone correctly, and you can basically confirm this through enough repetition to make the odds impossibly in your favour
Rule 5, you can only cast a spell once per stone
@@mambodog5322 that refers to the placement spell, if I don't move anything I should be able to cast as much as I want
@@VeetiM The placement spell is what causes the platforms to glow.
@@JoeThomas-lu6fy yea probably. But that's why I said it might be a mistake in the wording, because the rules on the pause screen don't say that
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
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
You're a mathematician, Harry
Moldevort is back!
Drumbledrore , Moldevort and Myself Potter Harry...🙂
*hotter parry
@@eshitasahu Better..😌
@@eshitasahu 😂😂😂
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.
Voldemort after seeing thumbnail : Avada Kedavra 🐍
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
Please make a video about Lyndon b Johnson vs history.
Correct solution: "Hey, Dolby! Put this stone over there. We'll be waiting outside."
Moldivort is truly the worst, most despicable villain we as a society will ever know
... Read more
i fell for it
"HAROLD DIDJA PUT YER NAME IN THE CUP OF FLAMES?!?!?!?!" drumbledrore questioned peacefully,
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.
*Moldervolt casually taking out his UNO reverse card*
Can I just memorize where stones 1-99 would be placed and place none of them until I find the overlapping stone before placing the Keystone last? Then place the remaining stones after that and place the keystone on what hasn't been highlighted/occupied as stone 100? Since the sabotaged stone only makes random empty platforms glow, the stone it was before (29 on 45 in the video for example) would "overlap" one other random stone. Once the spell makes a pedestal glow twice (Would be 29 in the video) it's safe to place all the stones after that by placing stone 45 at pedestal 29's spot. Doing this allows the rest of the stones after 45 to not require memory to know their places for when you place the Keystone. If you got a magical dude who can make numbers appear, then he can help with the memory thing too. That's my thought...
Edit: I know they put a 1 on the first stone placed and numbered the rest, I'm just explaining this as if they hadn't done that yet.
Mentioning Maryam Mirzakhani‘s name was excellent! Well done
1:27 You'd think best move would be purposely screw it up and teleport away so Mort is locked off from the wand.
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
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
At first I thought the answer was 1/2 because it’r very similar to a well-known problem usually phrased involving airplane seats. That’s how the video treats it, in fact, making me think there was a mistake in re-theming the question. However, the problem is actually different, because we can cast the placement spell on everything *before* making another platform occupied.
Unfortunately, I can’t actually see a way to *use* that advantage; the answer is still 1/2 (without the potion). Even if the “random unoccupied” result is memoryless, in which case we can identify which stone’s platform was taken (unless we’re in one of the two special cases where it’s 1 or 100, in which case we know that we’re in one of those but not which), I don’t see a way to use that; there’s still no way to distinguish platforms 1 and 100 from each other even when we know which two they are.
Think of it this way, 98 stones will reveal their correct platform, and one stone will lie. If the first stone is not on the keystone's platform, then the only way to lose is if the lying stone points to the keystone platform. Then you have no information and can only guess at where the keystone goes.
@@tavern.keeper That isn’t the only way. The lying stone might instead point to the same platform as another stone, or tell inconsistent lies so you know it’s lying, and either way you still can’t tell.
@@danielrhouck If it points to the same platform as another stone, then there will be exactly one platform that was not pointed to by any stone. That is the keystone's platform, and you place the keystone there to win.
@@tavern.keeper If it points to the same platform as another stone, there will be *two* remaining platform. There are 99 non-key stones; 98 after one is glued down. If two point to the same platform, 97 platforms glowed. 100 platforms - 97 glowing - 1 glued = 2 remaining.
It all comes down to the fact that, once stone 1 is glued down, there is nothing that will honestly point to platform 1. This leaves no way to distinguish platform 1 from platform 100.
@@danielrhouck Cast the locator spell on the first stone as well. This is allowed within the stated rules. There is a restriction that the spell can only be cast once for each stone, but it does not say that the spell cannot be cast on a stone that has already been placed.
Finally I understand the solution!! I love all kinds of Ted ED's video.🥰 they are very interesting.
NEW TED ED RIDDLE!! this is a great day :D
I'd place each stone next to the platform that lights up, so i would know when 2 stones light up the same platform
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).
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.
I like how Ted ed is making part 2 riddles
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.
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.
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.
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.
Look at Rule 5.
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