Can you solve the cursed dice riddle? - Dan Finkel

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

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.

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

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.

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.

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!

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

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

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

I have never paused a riddle video

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.

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

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.

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 😂

Imagine having a Norse and a Greek god at the same time😂💀☠️

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.

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!!

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

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

This video explained it better then my teachers ever could.