Can you steal the most powerful wand in the wizarding world? - Dan Finkel
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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.
Date: 2023-01-26
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Comments and reviews: 14
TheRealQwade
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%.
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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%.
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Street_BB
Isn't there a better solution, cast your spell on each stone 1 at a time, and mark which platform lights up each time but don't place the stone you cast on the platform, just near it to mark it.
Then because you are not using up locations, at least 98 different pedestals will glow, one glowing twice assuming the keystone platform wasn't taken. In that case the only platform that didn't glow is the keystone one. Only failure chances are the keystone one is already taken, or the 1/99 odds the keystone one randomly lights up when the correct stone for the randomly placed one is checked.
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Isn't there a better solution, cast your spell on each stone 1 at a time, and mark which platform lights up each time but don't place the stone you cast on the platform, just near it to mark it.
Then because you are not using up locations, at least 98 different pedestals will glow, one glowing twice assuming the keystone platform wasn't taken. In that case the only platform that didn't glow is the keystone one. Only failure chances are the keystone one is already taken, or the 1/99 odds the keystone one randomly lights up when the correct stone for the randomly placed one is checked.
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Van
Fun fact:
. .. .
If multiple stones get misplaced, the odds are still 50-50. The reason for this is that for every misplaced stone, there is a spot where it's good to put a stone. If 2 stones are misplaced, the next stone has a 2/100 chance to be a stone that can't land on the right spot. but it also has a 2/100 chance to land on either of those misplaced stones' correct places.
Notably, if you started from 2 stones being misplaced instead of one, that'd hurt your odds, since you need to get 2 stones correct instead of just 1, in order to guarantee victory.
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Fun fact:
. .. .
If multiple stones get misplaced, the odds are still 50-50. The reason for this is that for every misplaced stone, there is a spot where it's good to put a stone. If 2 stones are misplaced, the next stone has a 2/100 chance to be a stone that can't land on the right spot. but it also has a 2/100 chance to land on either of those misplaced stones' correct places.
Notably, if you started from 2 stones being misplaced instead of one, that'd hurt your odds, since you need to get 2 stones correct instead of just 1, in order to guarantee victory.
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Cadis
2: 29 But. let's say we are following scenario three. Isn't it possible that one number you check for from 1-100 is shown a random place? How can you just place every stone at their own place when it's basically 50/50 possibility for 100s of stones?
In simpler terms how would you know where the number 45 is marked on the place where you're placing stones?
Oh. i didn't realise that you can just check location of each stone twice to confirm it's not the special one.
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2: 29 But. let's say we are following scenario three. Isn't it possible that one number you check for from 1-100 is shown a random place? How can you just place every stone at their own place when it's basically 50/50 possibility for 100s of stones?
In simpler terms how would you know where the number 45 is marked on the place where you're placing stones?
Oh. i didn't realise that you can just check location of each stone twice to confirm it's not the special one.
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1mysterioues1
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.
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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.
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GD
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
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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
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0math12345
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.
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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.
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Field
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.
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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.
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The
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!
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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!
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Artiamus
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.
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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.
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samarendra
Why can't you just use the placement spell and just not place the stones?
Just keep a note of which platform lights up when a spell is casted.
So basically in the start, the invisible numbering you said would now be visible.
So I don't get why it would be a risk?
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Why can't you just use the placement spell and just not place the stones?
Just keep a note of which platform lights up when a spell is casted.
So basically in the start, the invisible numbering you said would now be visible.
So I don't get why it would be a risk?
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brahim
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
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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
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Saanvi
Felix Felicis to Felush Felucious lol. also if he had it the whole time why go through all that trouble doing the math when you can simply have the potion of luck? Of course, it's Drumbledore. the easiest answer's right there, yet you wanna go the other way
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Felix Felicis to Felush Felucious lol. also if he had it the whole time why go through all that trouble doing the math when you can simply have the potion of luck? Of course, it's Drumbledore. the easiest answer's right there, yet you wanna go the other way
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Shiny
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
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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
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