Can you solve the cursed dice riddle? - Dan Finkel

[@ConorShea-m9p]

Can we all just appreciate how Demeter is just playing Settlers of Catan with mortals?

[@mellow3995]

Haha that's what I thought too

[@gkjaerby2441]

Nice.

[@lilyandryankoolkid1994]

my exact thoughts!

[@pabloricco8093]

Pog catan

[@mimumi3723]

Can't wait for mortals to find the Fire progression card and Prometeus to fall

[@patrickmitter1874]

Imagine you're a scientist and while others like Einstein come up with relativity, you're inventing a special dice that does the exact same thing as a normal dice. George Sicherman is a legend.

[@just-apt]

Here before this blows up

[@circle4602]

As long as you find something new...

[@saininsa98]

Sound fun

[@Alepfi5599]

What a lad

[@cat_sandwich]

lol

[@rohitsaralaya1449]

Oh my god this is the prequel to the Ragnarok riddle. Everything really is connected.

[@lilyandryankoolkid1994]

omg your right

[@Infernape1998]

In other words, damn it Loki you jump started Ragnarok.

[@kingwolf3044]

Riddleverse confirmed!

[@zesnowpea6347]

Can’t wait for the tcu Ted Cinematic Universe

[@RelatedSnow8420]

So there IS a TED-Ed Riddles Cinematic Universe!

[@zakwinduss]

I’m so proud of myself, I almost didn’t attempt this riddle and was just going to watch the solution but decided to try. I didn’t know my strategy but I “followed by nose” and drew the table . I then copied the table and removed the sums. I realised I was only allowed one “1” and one “12” so I filled them in exactly how the solution did and then worked my way back, ticking off each sun from the original table. In the end my work sheet looked exactly like the solution tables and I’m kind of stoked bc I’m usually not smart enough for TedEd riddles. Sorry I must sound a little boastful but I’m pleased as punch.

[@ofureotegaaigbomian1677]

Yessss🎉🎉🎉🎉

[@chiragchandan9135]

I wish i had ur experience.

[@KumaKaori]

f'real congrats! I mentally got as far as "we need the results map to match" but 1st guessed 1,2,3,3,4,4 as our low dice, had I bothered to work through the guess I'd see my mistake soon enough.

[@Szokynyovics]

Look, if you do something smart/great like this, feel free to boast!! Good people will be happy to hear your success! ;) glad you were able to solve it! :)

[@mikalayklimovich7878]

Nothing satisfies like a solved riddle). I solved it too, wasting some time on 0-hypothesis (I hadn't known you couldn't a side without dots) and then accidentally stumbled upon the solution. Btw, I found that you can actually create another pair of dice, if you include 0 and forget about 4-on-one-side limitation. First dice is 0,1,2,3,4,5, second one 2,3,4,5,6,7. I'm curious if those are called somehow too

[@SonicLoverDS]

I read about Sichermann dice by chance in a book of math puzzles. Fun fact: one critical difference between normal and Sichermann dice is that it's harder to get doubles on the latter (1/9 chance instead of 1/6). I wonder what it'd be like to play backgammon with Sichermann dice.

[@GoPieman]

A good reason for the animation to choose catan, given doubles don't do anything.

[@heatshield]

Okay that’s a great question. It’s been awhile since I played, so without going to refresh my memory of the rules, I’m assuming it would be much more likely for someone who gains a small early advantage to clean up, possibly resulting in it becoming acceptable and common to forfeit matches early. Might have to make some new dice and play test it.

[@KitagumaIgen]

@@heatshield Maybe it wouldn't make small advantages last, but it surely would make big and rapid swings more rare, so at least we'd get a calmer more sedate game?

[@trapjohnson]

Means you go to jail less often in Monopoly, but you also travel less far.

[@michaelmoody8793]

A lowered chances of doubles - and not being able to get double 2s, 5s, or 6s at all - would be a huge impact. It would also wildly change things to have a 9 available on one die, changing the strategy around priming.

[@HayTatsuko]

I enjoyed the animation in this a bunch! I'm no mathematician, but even I can understand the reasoning behind this solution, because it was so well explained... and the denouement at the end was entirely delicious!

[@felixlee9645]

ditto

[@SuperDestroyerFox]

Yeah. I was thinking something similar of what they were saying but I didn’t think to use a graph and never thought of it before I gave up.

[@gabrieljohnson6304]

the blacksmith having accurate mouth movement with the narration is both cool and oddly uncanny

[@JLvatron]

Looked pretty creepy to me, lol! So the Blacksmith is really Ted-Ed!

[@YoungGandalf2325]

People who own land that's a 2 or 12 can never catch a break.

[@KOSJ153]

They'll catch a break 2.8% of the time lol.

[@SwedeRacerDC]

​@@KOSJ153 Except for times where the favor of the dice seems to be going there way. But then they'll all of a sudden go on a deadly drought...

[@teamcyeborg]

I really appreciate that you can hear the cut after "Hephaestus'" at 1:01

[@xavierburval4128]

I tried solving this by starting with a regular dice, and trying to rearrange the numbers in a way that worked. I tried but couldn’t find anything that worked, until I realized that I’m not allowed to have zero dots, which I though I was. Once I had that, it was way easier to build the dice step by step. Very interesting riddle

[@featherflight3493]

I also immediately went for the zero dots face, and kept running into problems in the middle. Kind of funny what assumptions we make about a problem.

[@AllenKnutson]

Indeed, in the verbal description there is no positivity requirement. (I count this as a mistake in the writing.) It is clearly written at 1:26.

[@cjfeinberg7613]

I think that would permit another solution, (0,1,1,2,2,3) and (2,4,5,6,7,9). Start with their solution, and move one dot from each side of the small die to each side of the large one. Given that in the story we start with blank dice, I think this should have been a permitted answer.

[@xavierburval4128]

@@cjfeinberg7613 yeah that would work, but they probably wouldn’t want two solutions so it makes sense

[@_wewww]

@@cjfeinberg7613 how do you even get 11 with that?

[@michaelsparks1571]

It's only mentioned in passing and not on the Rules screen, but I think it's fun to note that the Sichermann dice also use the same number of dots as two standard dice. You don't end up with extra or remaining dots after they fall off and are put back.

[@fatih3806]

Well it has to use the same amount of dots, otherwise it can’t have the same frequency table.

[@PragmaticAntithesis]

@@fatih3806 Only if you're required to keep the same number of sides. For example, the 4-sided die 1,2,2,3 and the 9-sided die 1,3,3,5,5,7,7,9 have an odd number of dots in total, despite being two dice that have the same odds of rolling any given number as 2d6.

[@psy8428]

2:28 "Assuming we have a 4..." I felt like the video brushed over considering having numbers less than 4 or larger than 8 on the dice (respectively). But it can actually be walked through relatively easily. You must have exactly one 1 on each die. If you had an 11, you could have only 1s on the other (without getting sums above 12), but this would make at least 6 ways to roll 12. Same sort of pattern continues for looking at 11, 10, and 9, with each leading to too many rolls of the high numbers. In this way, you can prove that the highest number you can use is an 8, meaning you must have exactly one 4 on the small die.

[@logansuits2406]

Great explanation. Thank you!

[@akay7877]

Thank you! I'd been stuck on them saying that but was struggling to work out that logic myself.

[@_whaet]

Guess sicherman will never die get it cause dice is plural of die

[@Buphido]

You can also go about it from the small die's perspective. It's clear that the highest side of the small die must be the only side of its number on that die. Also, both dice must have exactly one 1. That means, if you had at maximum a side with 3 eyes on the small die, all others save for the 1 would have to be 2, which would lead to at least 4 ways to form a sum of 3, which is too many. And a lower number of eyes is impossible due to the aforementioned requirements (more than one 1 or more than one maximum side). So the small die must have exactly one 4, and the high die must thus have one 8.

[@quest_wizard]

You are the reason I learnt English, your videos are so good it made me learn the language in two months

[@matteocappugi4115]

Very cool video. The way it's done rigorously in mathematics (which can lead to many other interesting results, like having two dice with a different number of faces, and still getting the same probabilities) is with generating polynomials. A normal six-sided dice is represented by x+x²+x³+x⁴+x⁵+x⁶, where the exponent represents the number of dots on a face and the coefficient, which in this case is always one cause all numbers from 1 to 6 are each displayed exactly once on the die, represents the number of sides you can find that precise number of dots on. The product between two of these polynomials gives you the amount of different combinations of numbers that can give you a certain sum, namely the polynomial representing their combination. If P(x) represents a die, computing P(1) gives you the total number of faces of that die. Notice that the polynomial for a standard six-sided die can be broken down into x(x³+1)(x²+x+1) -> x(x²-x+1)(x+1)(x²+x+1). If you plug in one, the first two operands yields one, x+1 yields two and x²+x+1 yelds three. Therefore, the presence of the first two doesn't affect the number of sides of the die, while the other two operands are fixed in place, as the two dice must be six-sided (but you could do it with four-sided and nine-sided ones as well, etc). Moreover, each face has to have at least a dot on it (must have x as a factor). Therefore, we can only shuffle x²-x+1: one die will have no copies of that factor and the other will have it squared. After all calculations, you'll get x⁴+2x³+2x²+x and x⁸+x⁶+x⁵+x⁴+x³+x, which are exactly the two resulting dice from the video. Remember to eliminate, if there were any, all results which have one or more negative coefficients, as negative sides don't seem to exist in geometry (I think the only viable way is just to eliminate them afterwards, but if you have a better method in mind I'll be happy to hear it). All of this looks scary at first, but actually it is fairly straightforward when applied, especially for harder-to-find combinations (like the one with four and nine sides respectively, which was a question on an entrance examination test for the Normale di Pisa dating back a few years ago). Hope I explained it somewhat clearly 😂

[@johannesvanderhorst9778]

I solved it in a similar way (apart from that once you wrote x⁴+2x³+2x²+1 rather than x⁴+2x³+2x²+x.) I just removed the factor x, and I calculated (1+x+x²+x³+x⁴+x⁵)² = ((1+x+x²)(1+x³))((1+x)(1+x²+x⁴)) = ((1+x+x²)(1+x))((1+x³)(1+x²+x⁴)) = (1+2x+2x²+x³)(1+x²+x³+x⁴+x⁵+x⁷)

[@matteocappugi4115]

@@johannesvanderhorst9778 True, I mistyped, thank you!

[@voyant_3]

An equivalent method disguised as something easier is creating a system of equations from a discrete convolution of the pmfs. The advantage is it doesn’t require generating functions, which are quite a leap for the average person

[@theybuck_]

First ever puzzle I've actually decided to sit down and solve myself!! I even made a google sheet to calculate the different sums and their counts, very satisfying!!

[@duruffio]

I missed the instruction that required us to only select numbers >1, so I had a die with a blank face (0,1,1,2,2,3) and added 1 to all the values on the other dice. After seeing the correct result, it makes sense that if you subtract one from each face in Die A and add one to each face on Die B, the results don't change.

[@TsetBlue]

Yes. I was thinking the same, getting a die with 234567 and another 012345, but stucked.

[@BeaDSM]

Except you didn't miss that instruction as it wasn't there! 011223 and 245679 is a much more pleasing dice combo in my opinion.

[@Tumbolisu]

@@BeaDSM it says it right there at 1:22?

[@seuri678]

@@Tumbolisu I am not a deep mathematician, and I don't view 0 as positively as I do not write -0, and this does make sense in the story.

[@AllenKnutson]

The instruction isn't stated (a mistake IMHO), only written.

[@william155]

I have never paused a riddle video

[@amateur-madman3047]

I tried a couple times but eventually I decided I wouldn’t fight the urge to be taught😂

[@just-apt]

Here before this random comment blows up for some reason

[@jessicafuller6474]

Me neither brother

[@L17_8]

Jesus loves you ❤️ please turn to him and repent before it's too late. The end times described in the Bible are already happening in the world.

[@iwams1]

You want me to give you a cookie or what?

[@desihirohamada]

step 1: confirm that you have green eyes step 2: ask loki, "if i asked you whether i had to roll three 4s, would you say ozo?" step 3: walk anti-clockwise across one block of land to add one coin to your balance step 4: calculate which lockers have numbers with perfect squares step 5: work backwards from zahra's answer to get the hallway required step 6: choose the gaussian and miss on purpose step 7: lock loki in pythagoras' cursed chessboard

[@alanherlan3429]

was looking for the obligatory "confirm you have green eyes" plagiarism of the solution to a different puzzle Congrats though you plagiarized 7 different puzzles (very slow claps)

[@xavierburval4128]

Nah your forgot step 8: head into the simulation mainframe and communicate which server is infected via binary code

[@eddycabreja3332]

​@@xavierburval4128and also step 9: Activate thrusters D & E

[@kingwolf3044]

I always love seeing the answers from the other riddles pop up in the comments.

[@landongreen7510]

Always something like this in the comments. MAGNIFICENT

[@BananenBoerBob]

I like how the concept of a set of 2 dice that would have the same relative chance of rolling results as regular dice do, would be such an alien thing to come up with. Once you turn it into a problem like this video has (and add the constraint of max 4 eyes per side on one die, which was a massive hint) it suddenly becomes a trivial thing to solve

[@shr07i]

I love this riddle! It looks complex as my initial thoughts were getting a piece of paper and solve probability for each roll but in reality its very simple, straightforward and easy to understand.

[@fallouthirteen]

Yeah, like you can just do a lot of it in your head. You know for a fact each die must have one and only one 1. On that same track when solving the first die you know you can't have more than 2 of any number (since it'd probably result in too many 3s and 11s). And you can only have 1 of the largest number. So that's just 1,2,2,3,3,4 easily. Other die then already has the 1 and 8 solved (to match with 1 and 4). Leaves you with 4 faces between 2,3,4,5,6,7. Even without proving it to yourself by doing the math/chart you just go "if you have to leave out 2 of those numbers, probably the ones at the ends so you get a more middle distribution of results".

[@vito_far]

Solving these riddles makes me feel good. Thanks Ted

[@maxhatterschannel5140]

4:07 Jörmungandr be like: Sorry Dad, but I owe Dementer a favour

[@r.pizzamonkey7379]

Dice with sides {0, 1, 1, 2, 2, 3} and {2, 4, 5, 6, 7, 9} also work. I guess the stipulation said that "every side must be a positive whole number" rather than "every side must have a natural number", but there's no conceptual reason why a die couldn't have a blank side to indicate zero. Every other constraint was justified by the narrative, so I just assumed this was an oversight in the wording.

[@qqma4791]

I love riddles that challenge you but aren’t impossible to understand without background knowledge of the subject. One of the few riddles I was able to solve and I’m glad I was able to do so :D. Thanks Ted-ed

[@owenkariuki4438]

I love these riddles as they were the first kind of videos that got me interested in this channel. Please never stop uploading them ❤

[@shinyagumon7015]

Everybody gangster until you show up with Sickerman Dies to family game night.

[@Xenx2000]

Before the video starts, i just want to express this first: I've never tapped so fast on a video before. But when it comes to New Ted-ed riddles, I AM HERE!!! Keep them coming 😊

[@紀阿霸]

So basically, the goddess of harvest plays Catan.

[@CalliopePony]

I took some time to just sit down and think about the various possibilities, and I was able to figure it out. I used some trial-and-error, but I got figured it out mostly the way it was shown in the video.

[@sisiwen]

Finally, another riddle! Have been waiting for 2 months for one. Can we have one every week? I just love these riddles that you come up with. 🙏🙏🏼🙏🏻🙏🏾🙏🏽💖🥰💗💓😍💕

[@user-zj5ff3ms4d]

Last year, as a senior in high school, our class participated in a maths contest. We would get three riddles during the year. This is actually one of those riddles. fun to see the explenation of that riddle by some one else. sidenote: we didn't manage to win the contest but got to the finale.

[@Sunflowersarepretty]

This video explained it better then my teachers ever could.

[@user-hh2is9kg9j]

Your Teacher isn't a big multimillion-dollar corporation that has a big team of experts and researchers?

[@Sunflowersarepretty]

@@user-hh2is9kg9j well education in my country is ridiculously expensive however the quality is poor. So yeah they should do a better job of hiring good teachers when they charge a lot.

[@NadavBenNatan]

Wow that was a great video, one of my personal most beloved tricky questions!! I also want to share that after thinking of other ways to create such dices, I discovered that if you allow one dice to have a zero on it (I know it’s not in the original rules, but hear me out) - you can create dices that produce the same frequency, in this way: Dice 1: 0-1-1-2-2-3 Dice 2: 2-4-5-6-7-9 Now that I think about it I just added one to every number in the second dice and subtracted one from each in the first😅😅 * Continuing with that logic you can create the same affect if you allow a minus 1😮🙃 Any way that was a really nice riddle, thank you😁!!

[@ArcySparky]

My solution doesn’t work under the rules, but I used three dots to create a binary notation, corner middle side representing 1, 2, 4 respectively. This allows you to make an ordinary six sided dice using a maximum of three dots. The rules state the sum of the dots, but if different dots have different values, as in the binary notation dice, then the sums can still match a normal dice.

[@KimiiiRaikkonen]

I did spend too much time on solving it, but I'm happy I did, when I got the logic and it worked. Great riddle, loved it. Thank you :)

[@trillionbones89]

i realised the 1 on each dice and the 4 - 8 relationship and decided to continue the video instead of filling up a table. but i hope i would have done similiarily as well as Demeter in a life/death situation

[@dobrinbashev250]

Another way to find the second die after the observations about the first are made is simply by division: 12,345,654,321÷1,221=10,111,101

[@0-Elias-0]

*Dice A* = [4,3,3,2,2,1] *Dice B* = [8,6,5,4,3,1] This is a nice riddle. One which just takes some knowledge of probability theory (specifically: *combinations* ). Plus, it helps to use a spreadsheet/chart. An algebraic way to solve this would be to set up 12 different equations (easier with *matrix multiplication* )

[@MatthewMorris6148]

I actually managed to do one of these for once, and I can actually explain every single step I took to reach it. As such, I’d like to explain all my reasoning. 1. Since there’s only a single two, (and every die face must have

[@Vetrical]

congrats on being a silksong waiter sane enough to solve this

[@CT-7731]

Me : _brings sicherman dice to my first DND campaign_ The table : what in the ungodly creation have you brought upon us?

[@foxbat1766]

Sicherman dice for: Monopoly? Yahtzee? Backgammon? Any other board games these dice could really mess with?

[@evanrosman9226]

It took me a lot of thinking but I was able to get it right.

[@TheMonkeyGrape]

Nice mix of specific calculation and logical thinking. Liked this.

[@Dude-xb3xh]

I’m now afraid of getting on Demeter’s bad side

[@shreyasshenoy7215]

that graph of probabilities helped me solve this fun riddle since there were only four values for one die i could solve with brute force but this tabulation is faster and convenient

[@threethousandbees7260]

When you are looking for a fun riddle and you end up having a math lesson explained at you.

[@CascadeHope]

This is the first riddle from this channel that I actually managed to solve!

[@sanjanapanchal8563]

0:27 Reminds me of the Catan Boardgame

[@wesleybantugan5604]

Loki getting absolutely demolished by a dragon is fair compensation I'd say

[@charlesmasonberg3552]

Now if only I can use those dice at the casino at the craps table.

[@daikiri3]

Wow, I have last seen these riddles about 5 years ago. I am so amazed that the narrator is still the same!!!

[@lifedeather]

Finally after months another riddle ❤

[@PorterB]

My answer was simply to symbolize the missing 5 and 6 with a different pattern of dots, then if that side rolls, even if it only has 4 dots, you treat it as a 5 or 6 just make sure they are in a different pattern to the regular 4th side.

[@JH-cp8wf]

Wow, Loki's pranks have got /really/ thinky since the Edda.

[@mindlessmeat4055]

I learned about this in abstract algebra. I even have a pair of d6's that answer this question. According to my dice, one is a 1, 2, 4, 3, 3, 2 and the other is 6, 5, 3, 4, 1, 8. That should yield the same results as a standard 2d6. I paused to before the answer to grab my dice. Didn't realize it was answered.

[@danielrhouck]

2:27 Why do we need positive numbers? I was also trying 0/2 for that spot.

[@abhijiths5237]

the rule is that every face needs to have a positive number

[@danielrhouck]

@@abhijiths5237 That rule isn’t part of the scenario as presented. I missed if it was on the rules screen but there’s no reason for it in the story.

[@lenoncerqueira8308]

​@@danielrhouck But have you found a valid solution which includes a 0? Something tells me no such solution exists (even though I don't feel like proving it rn). In this case, you don't actually "need" that rule, it would be more of a "hint" to get you on the right track faster.

[@danielrhouck]

@@lenoncerqueira8308 i have neither come up with one nor shown it’s impossible; if I remember I’ll try tomorrow or over the weekend.

[@kohwenxu]

⁠@@danielrhouckYou can just subtract one from each face of the 1,2,2,3,3,4 die in the Sichermann dice and add it to the other die, giving 0,1,1,2,2,3 and 2,4,5,6,7,9. I.e. the two dice will have faces of 0,1,1,2,2,3 and 2,4,5,6,7,9 respectively, if 0 is allowed, still keeping to the less than 4 dots rule on one die.

[@bluehoodie0205]

New solution, use 12 pieces of paper, two have 1, two have 2, and so on*

[@loszhor]

Making Catan more difficult is all part of the fun!

[@Vega_Voice]

1:23 A hint for those who may be struggling: The rules don’t state this, but you would have a limited number of dots to work with, this number is (dots on dice)*2, which is 21*2=42. This would be a “catch” of the puzzle.

[@TaranAlvein]

Obviously. You're reattaching the dots that fell off of the two dice, so the number would have to add up to the same amount as a normal pair of six-sided dice. It's probably also the only way your two new arrangements can still come out with the same distributions.

[@leotimotian8990]

I never thought i without pay so much attention more to a riddle video

[@Apersonl0l]

the Goddess really be playing Catan on the poor people 💀

[@lilyandryankoolkid1994]

so close to my thoughts!

[@kittyaya3425]

Ah yes, Loki being his original self. Good for him for cursing that dice🙃

[@Skarpo89]

It would be exactly the same mechanically, but for some reason I'd find playing Catan with this set of dice to be more exciting

[@signsoflife]

On normal dice 1 is always opposite 6, 2 opposite 5, 3 opposite 4. You could have totally blank faces for 4, 5 and 6 and still end up with perfect dice simply by checking the opposite face.

[@k3ose45]

Pips. The dots on the dice are called pips

[@TheVASgamer]

Really? Like the lil dots from Wizard 101 aren't some made up term but a term used for dice dot!?!??!

[@davecorry7723]

That was such a complicated question that I didn't care what the answer would be.

[@slorgdulschmodus]

agreed - I wouldnt call this a riddle at all. Its like a math olympiad style question, and only the most enthusiastic of number crunchers could find it even remotely interesting

[@ThAlEdison]

Classic, my favorite solution method: f(x)=(x+x^2+x^3+x^4+x^5+x^6)^2 the coefficient represents the frequency of the outcome, and the exponent represents the number on the face. So 1d6 is the number in parenthesis and squaring gives 2d6 So now we factor f(x)=x^2(x^5+x^4+x^3+x^2+x+1)^2 f(x)=x^2(x^6-1)^2/(x-1)^2 f(x)=x^2(x^2-1)^2(x^4+x^2+1)^2/(x-1)^2 f(x)=x^2(x+1)^2((x^2+1)^2-x^2)^2 f(x)=x^2(x+1)^2(x^2+x+1)^2(x^2-x+1)^2 So the sum of the coefficients of each factor tell us what we can combine. We want 6 faces on each die, and the sum of coefficients works out to (1)^2(2)^2(3)^2(1)^2, so the factors that have coefficients adding to one can go on either die. They don't have to be spread evenly. x(x+1)(x^2+x+1)=x^4+2x^3+2x^2+x x(x+1)(x^2+x+1)(x^2-x+1)^2=(x^4+2x^3+2x^2+x)(x^4-2x^3+3x^2-2x+1)=x^8+x^6+x^5+x^4+x^3+x f(x)=(x^4+x^3+x^3+x^2+x^2+x)(x^8+x^6+x^5+x^4+x^3+x) So one has sides 4,3,3,2,2,1. The other has sides 8,6,5,4,3,1. {{3,2,2,1,1,0},{9,7,6,5,4,2}} also works.

[@ThAlEdison]

Physical dice with no way to show negative pips, can still show 0 by having a blank face, so the constraint should be non-negative whole numbers, not positive whole numbers.

[@msolec2000]

I'll leave you with this follow-up riddle. If you shuffle a deck of nine cards and then draw two of them, you have 36 possible hands (since drawing card A and card B is the same thing as drawing card B and card A). Can you put a positive number on each of the nine cards so that when you draw two their sum has the same distribution as the pair of dice?

[@johannesvanderhorst9778]

I can't. Because if a solution exists, two of the cards chould have a 1 because otherwise we can't draw "2." To be able to draw two possible "3"s, we need the third card to be a 3, and no other card. To achieve some "4"s, we need a number of cards with a 3. For each such card, we get two possible "4"s as the 3 can be paired with both 1s. It is impossible to achieve precisely three possibilities to draw "4", because that number must be even. So my conclusion is no solution exists.

[@msolec2000]

@@johannesvanderhorst9778 Good try. But I can assure you there is a solution.

[@johannesvanderhorst9778]

My bad. I only proved there is no solution involving positive integers rather than positive numbers. A solution involving fractions is 1/2, 3/2, 5/2, 5/2, 7/2, 9/2, 9/2, 11/2, 13/2 I also believe this solution is unique.

[@msolec2000]

@@johannesvanderhorst9778 Indeed! That is the solution. :)

[@nayutaito9421]

I was amazed how you did not mention polynomials and get the perfect solution

[@icarusty6481]

Demeter casually starting Ragnarok because of something Loki did is very on character for both of them.

[@ThewOrldIssqUare]

Normal brain: answer from video Big brain: recreate the normal, non-cursed die and roll it twice Planetary brain: just make/buy some new dice lmao Galaxy brain: make a 36-sided die with those values from 2 to 12

[@Anonymoususer44569]

Everyone: “Another fun riddle! Thanks Ted-Ed!” Me: *struggling to even understand the rules*

[@kingwolf3044]

First you have to ask the demon of logic if at least one of the sides has green eyes. If the answer is ozo you count the number of trees that the founders of the houses can see to find where the baniker is buried. If ulu then you trap nym in the magic checkerboard to prevent the ai from releasing the robot ants. Afterward you must give the dino nuggets the magic tarot cards you stole from fate so they can lite the correct number of candles on the giants cake. The giant will then agree to use the tri source to make you a new set of dice.

[@LPW4BC]

I tried this and it worked

[@sherlock1854]

One of the best animations by ted-ed ❤

[@darkner2390]

Real answer: Get a marker or some charcoal and just draw the numbers on all sides. Or just get a new set of dice.

[@thatguystar817]

Demeter being the cause of ragnarok is really jarring yet funny

[@DrRank]

What makes the Sicherman Dice even cooler is that they not only have the same sum distribution as standard 2D6, but also the exact same number of dots (42)

[@JasonAFlintham]

Fun Fact I noticed when crafting my solution and mapping out the potential combinations (cos apparently I have absolutely nothing better to do on a Saturday afternoon): You know how numbers on opposite sides of standard dice always add up to 7? Well Sicherman dice can be mapped out in a similar way. To use this riddle as an example, the cursed die can be mapped so that opposite sides always add up to 5 (4+1, 3+2, 3+2), and opposite numbers on the non-cursed die always add up to 9 (8+1, 5+4, 6+3)

[@dayfordays]

Am I the only one who falls asleep to these videos? His voice is just so calming.

[@corvididaecorax2991]

Technically the scenario as presented has a second answer, which is only ruled out by the addition of rule 3 in the summary page. But the described scenario would allow a face with zero dots.

[@davidx9426]

Me: just buys new dice

[@Basil-the-Frog]

This is a great demonstration of how dynamic programming (DP) works. Filling out the die roll table is the algorithm for dynamic programming. And, the dual to that is the method used by George Sicherman to solve the missing dot problem. I believe studying this idea in terms of machine learning, AI, and neural networks will result in at least one computer science PhD. If you are looking for an idea, you are welcome.

[@pianistluna]

my first ted-ed riddle, proud of myself that i solved this in 15 mins with no help!

[@Hellion_Dragons]

4:07 SNAKE ATTAAAAACK 🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍🐍

[@kainevins9927]

A solid puzzle; nice callback to the ragnarok one, too.

[@TerenceTeri-xu3jc]

What a solution! There is also a non obvious solution that involves factoring the polynomial (x+x^2+x^3+x^4+x^5+x^6)^2 in two different factors. This solution also shows that Sicherman is the only other option.

[@cheese_productions]

I wonder if theres a dice that lets every number have the same chance to roll

[@naonv]

I didn't spend much time on working the table although mine is completely different because I was thinking further out of the box I got 1,4,2,2,3,3 and 1,8,2,6,4,5 I got 1 mistake and I didn't have a chance to check it because I had enough of trial and error. I love these types of brain activities, this will serve as a lesson.❤

[@PaddyRoon7]

I can never be bothered to get a pen and paper to try these myself, and every time I see the solution I think "Damn, that would have been fun to figure out myself"

[@JordanBeagle]

You know, Loki was just getting Demeter to think outside the box

[@ameenulinsan]

Best educational channel ever

[@FF-tb2dt]

Finally another riddle video 🫶🏻

[@L17_8]

Jesus loves you ❤️ please turn to him and repent before it's too late. The end times described in the Bible are already happening in the world.

[@__lifeline__]

I thought that god doesn’t play dice

[@trilloff]

4:04 "Hey Loki, guess what" "Snake eyes!"

[@clementdato6328]

A general method is to see the original distribution as a result of (x6+x5+x4…+x1)^2 and factorize it into product of two polynomials one of which have degree not greater than 4 and both of which have sum of coefficients as 6.

[@GeFeS]

I know the rule 3 says very clear that each face must have a positive, whole number of dots. But if you allow a face to have no dots, which in practice would also work for Demeter, I found another solution: the cursed die with the numbers [0, 1, 1, 2, 2, 3] and the other die with the numbers [2, 4, 5, 6, 7, 9]. Furthermore, the number of total dots is still 42, so Hephaestus could work that out. (It is because this approach is the result of subtracting one dot per face in the cursed die and adding one dot per face in the other die.) Also fun fact: as @SonicLoverDS pointed out, the probability of getting doubles with the Sicherman dice is 1/9 instead of 1/6 with the normal dice. With this approach the chance is 1/18.

[@RohitKulan]

I want these dice to be used in an actual game. Imagine the happiness when you roll the first die and get an 8

[@the_blue_mark_detective]

your videos are incredible

[@silviujaglau]

You could also just create a new shape for 5 and 6 on the cursed dice, which seems much more obvious.

[@gabrielbuonomano]

I'm sure you could do some neat analysis with a 6x6 square representing the sums of the dice, but you can also figure this out in an ad-hoc manner just because of the amount of constraints you can place on your die. For example, each die must have exactly one face with 1 pip each, because there must be only one way to roll 2. You can make the same assertion about the highest face of the die that's limited to 4 pips; only rolling that highest face and the highest face of the other die (which must also be unique among its die's faces) should yield 12. At this point I made some educated guesses; you could continue to build up and say each die also needs a 2, so there are exactly two ways to roll a 3, but that's not sustainable. At some point the 4-pip die needs to double up (at least) on its numbers. I just assumed it was symmetrical and went with that, although you can prove that having more than three 2s wouldn't work (eg. with 3 or more 2s, if the main die rolls a 1, you'll get more than 2 ways to roll a total of 3). Once the smaller die is proved or assumed to be 1 2 2 3 3 4, you can build up the probability distribution. The way I modeled it is that selecting a face for the larger die adds to the distribution of possible totals, starting with all totals at 0 ways to make them. We now select the six faces of the larger die to build the distribution shown onscreen for a standard 2d6. We know the first face is 1, so the totals raise by the following amounts: 2: +1 3: +2 4: +2 5: +1 That 1-2-2-1 shape is added every time you select a face for the larger die, with the leftmost 1 starting in the column that's one higher than the number of pips on the face. The next necessary placement is to add 1 possibility of getting a total of 4, so the next pattern must be placed with the leftmost 1 in the 4 column, meaning the second face for the large die is a 3. The same logic can be used to place the third pattern with its leftmost 1 in the 5 column, so the third face for the large die is 4. The overall pattern is mirrored, so we can just mirror our placements for the last 3 patterns, resulting in an answer of: Small die faces: 1 2 2 3 3 4 Large die faces: 1 3 4 5 6 8 Edit: Just watched the official explanation, that's remarkably similar but as I suspected there was an easier method using the 6x6 table. It's interesting how two different methods ended up taking nearly the same path to get to the answer!