@@BolasMinion same here, I also have 3 distinguishable dice in a cube (as opposed to 3 indistinguishable dice in a cube) as well as a couple sets of the "full" dice array (by ttrpg standards at least)
Cyanoacrylate adhesives are really hard to use on acrylic without clouding generally, and this was the perfect storm of conditions for that to happen because one of the main ways they cloud (in addition to just incautious application beyond the joint) is blooming, where the C-A monomers evaporate and then deposit on surfaces - since the inside of the cube was completely unventilated the monomers had nowhere else to disperse and so anything that evaporated inside the joint deposited back on the nice clear acrylic...
After you roll, flat spin the cube so that the three smaller dice move to the corners. Add the two that end up in opposite corners for a two dice solution. Ignore the two in opposing corners for a one die solution.
+standupmaths I wish UA-cam made frame scrubbing easier, so I could see the frame where you cut. As far as I can tell without that, you just had your hand perfect.
I'm so sad I can't ask a friend to watch twenty minutes of maths videos just to share that one moment of laughter with me. But I'm glad some folks on the internet enjoyed it with four years of time gap !
I can't even imagine the amount of time editing this took. It was all spot on. This channel deserves all the views and then some. I'm a fairly new sub, but I already love this community, and the engaging conversation that is somehow able to take place in the comments. I'm already looking forward to the next puzzle.
+Michael Schmitz it may be "13xx", or he wanted to say "3xy" times, so the resoult can be from 3xy*1 to 3xy*9. So, don't rely on that beginning "th" (or at least, I hope it is different from that because I would have got the wrong answer ;_;)
and figure out how to take the three indistinguishable pairs of simulated distinguishable dice we can get from that to then simulate three distinguishable dice
Matt Parker! In addition to your obvious numeracy and charm, you're a KILLER video editor / graphics creator. The time and attention it must take to make the extremely helpful animations that pepper your videos is not lost on me - I'm impressed. Thank you!
"I've arranged them in a 10x10 grid... Of course missing the 3 in the corner..." Looks like you arranged them into a bit of a Parker Square then ;) Jokes aside, loving the videos, and while I couldn't really take part in this puzzle thanks to exams, I really enjoyed hearing the solutions people came up with and I'm looking forward to wracking my brain over the next one!
You could have just taken those 97 dice, turned 16 to each face (1-6), and then used those 16 to form a surface to create a large signing of your name. You would only have to sign your name six times, each person would know which dice was which based off the portion of your signature, and the side the portion was on, and the last die could have been tacked on at any point to extend a rather dapper tail to the end of one of the signatures.
me "Teacher, when am i ever going to use this in real life?" teacher "if you ever have too much time on your hands and want to make a popular youtube channel"
Those two videos of yours and the responses they caused are the reason I subscribed. A very worthy puzzle, and it produced a lot of interesting stuff. Congratulations.
I know I'm a bit late and someone might've had the exact same solution (891 comments is a tad too much to read), but after hours of twiddling around I came up with (and if I may say so myself) a rather beautiful solution to the problem. It has simple rules and only four specific exceptions and, most importantly, the result is ordered i.e. (1,2) is as probable as (2,1).First die:Basic one die result from three dice (i.e. sum of values mod 6 with 0=6).Second die:Case 1) All three dice have different values- Select the ONLY even (2,4,6) or odd (1,3,5) value which is not present (e.g. 125=>3, 146=>2).- Exceptions: If all are odd (135), select 1; If all are even (246), select 4Case 2) Two dice have same value, third has a different value- Select the single different value (e.g. 455=>4, 113=>3)- Exceptions: 144=>5, 114=>2Case 3) All dice have the same value- All odd => 3; All even => 6The exceptions may occur only when first die is 3 or 6, and actually they concern only results (3, 1), (3, 5), (6, 2) and (6, 4). All other results can be derived directly by the rules.The case 1 forms a nice diagonal arithmetic pattern in a 6x6 result table. Also the exceptions may be changed to following without breaking anything: 135=>5 with 225=>1 (144=>1 normally) and/or 246=>2 with 255=>4 (114=>4 normally).
Imo indistinguishable dice only exist in theory. You can use dice order (the order you see them in) or position to randomly choose a dice to exclude or include
Well most of this kind of mathematical question exists in the abstract. Of course you could just do something like the two dice that are closest to a certain point or something like that, it's more about using the constraints to derive an answer to that specific question, which may shine a light on other parts of the ever-connected map that is mathematics or just act as a way to practice.
The isometric axometry (forgive me for misspelling) of a cube is an exagon, so there's that. Instead of printing the immage you could look at the transparent dice from an angle and visualizing it that way. Nice video!
POV: you watch the original question the moment it releases, search for solution and find none. Then forget about it for a long time. *6 Years later, youtube recommends you the solution*
Mathematician: Here's my algorithm for mapping 3 indistinguishable dice to 2 distinguishable dice. Engineer: Roll the cube then jostle them onto the edge, read the first two dice left to right.
Neat little side thing I discovered while playing around with this stuff: if you take two dice and say 1, 2, and 3 are 0, and 4, 5, and 6 are 6 on one die, and then keep the other die as is, you get an even distribution for the numbers 1-12 (like dolling a D12). I haven't figured out how to generalize it for other multiples of 6 yet.
I investigated why the triangle numbers stop in the first puzzle, and it really is pretty interesting. Rolling two dice can be represented by a 2-dimensional table, with the first die's result on one axis, the second die on the other, and the grid being the sum. You'll notice if you do this that it creates diagonal lines of the same number. Similarly, you can step it up to three dimensions for three dice. This gives a cube of results. Also similarly, it creates diagonal planes of the same result. With a little imagination you can see that the first few planes are going to have triangle numbered size. However, the diagonal cross section of a cube does not stay triangular! Past 8 the cross sections become irregular hexagons. The tips of the triangles are being chopped off, so to speak. That's why the pattern doesn't hold all the way through.
I like how you point out that some of the solutions only give you the sum which is great for games like "Monopoly". Except that, for "Monopoly" it is important whether you rolled doubles or not. Parker squared the video there, Matt.
Even though this video is now over 4 years old, I decided to give it a go and try finding my own solution anyway. I spent a good couple hours working on it and I thought I found a novel solution, only to realize that my solution required distinguishing between the *indistinguishable* dice :'(
Here I am 7 months after you stumbling on this video. I learnt from the "simulate 1 die" warm up that trying to distinguish it is definitely not the correct path, but I am wondering, how many simply suggested to just "roll and mod 6, twice!" as a solution.
First off, thanks for taking such a great puzzle and bringing it to this community where we could all share our findings, intuitions, etc.. Also, I think you did an excellent analysis of a variety of solutions to this problem! I will however assert that we have missed out that you have not done an analysis of the 3 to 2 dice solution I posted, as I it contains what appears to be a very powerful mathematical generalization! I stress this, not because "its mine" (I do not even attempt to take ownership! It is for any and all.), but because its really cool, and very few have approached it so far, presumably because it looks intimidating due to a perhaps a lack of emphasis of elementary algebra in modern mathematics. Thank you for bringing this problem to such a large forum! I am very happy that I got the opportunity to participate socially about what would typically be a personal musing the results of which would be buried in a pile of notebooks. All the best! :-)
+myName Thanks! I'm very proud of the community here. Sorry I didn't get to include your solution. There were so many good ones, I had to cut it down a bit arbitrarily.
It's not really mathy solution, and I am 5 years late, but my very quick solution would be to discard the "odd" dice. "Odd" means not the number, but position. 1. If two dice are stacked, discard the one on the bottom (you could not read it anyway). If tree dice are stacked, discard the middle one and read the bottom dice from the bottom (through the floor). 2. Else if only one die touches walls (not counting floor) or if only one die do not touch walls, discard that die. 3. Else if only one die is in the corner (touches two walls), or is the only one that is not in the corner, discard that die. 4. Else if three dice are touching corners, discard the middle one (the one that does not have a partner in the opposite corner). 5. Else if all die touch walls, discard the middle one. I they are about equal distance apart, look at walls they are touching, one should be middle. 6. Else if you it cannot be decided what die to discard (for example no die touches walls- I think this is the only case, but I have not proven that :-) ), roll again. The only drawback for this is that from looking, we cannot prove in all cases that die is touching the walls or not. So there may be some pushing around if players see the result they don't like.
Best video I've seen. Completely absorbing! So cool how so many people get involved! Love your videos. Gutted the signed dice ran out already. (Thanks for the signed card though 😊). Trying to pluck up the courage to go to a maths club night - not quite up to speed yet though. Keep the videos coming - they're great. Thank you 😄
The way I thought about the warm up puzzle was just to note that it doesn't matter what the "first two" dice are - the third dice will always move you uniformly through the six possible remainders. You don't have to actually be able to *identify* the dice for this - it's a thought process. No matter which two you choose to "call" the first ones, that third still is completely independent and it will have a uniform probability distribution, and it does move you through the mod 6 cycle. QED. For the two-dice problem, you did not say we had to accept every single roll. You didn't say we couldn't re-roll. So my solution there is to roll until at least two dice match. There's a 4/9 probability of this happening, so it's unlikely to take many rolls. Then count the doubled number ONCE and count the odd die as the second one. I haven't simulated this, but as far as I can see by just thinking about it this should yield a valid two-dice probability distribution. The fun thing about that two-dice solution is that it works for the single-die puzzle as well. It's less practical though - you'd reject all rolls for which all three dice didn't match, and that mean you were rejecting 35/36 of all rolls. Not quite practical. But, it does work. And I think it would actually work for getting any M-dice roll out of N indistinguishable dice; you'd just have to roll until N-M+1 dice matched. Practicality varies, but... it is "a solution."
Easy way to do it without the hexagon: 1. No matching dice: 1.1 All even/all odd; lowest number & lowest number + 3 (ascending order). Ex: 2, 4, 6-> 2, 5 1.2 Straight; middle number twice, a straight can wrap around. Ex: 5, 6, 1-> 6, 6 1.3 Everything else (with no matching dice) check the two dice that are opposite (1 is opp. 4, 2 is opp. 5 and 3 is opp. 6) the first die is the remaining one, the other one is the closest one from that, going up (it can wrap around) (ascending order). Ex: 1, 3, 6-> 1, 3 2. Pair; it's simply the two values shown (descending order) Ex: 3, 3, 6-> 6, 3 3. Three of a kind; it is always 3, 6 (ascending order)
I haven't watched the original video yet. But I did solve it the way I considered not neat, started like 11:12. But the bit where the triplets go into the grid I simply did it (in my head) as a map table: 123 -> 11 124 -> 12 125 -> 13 126 -> 14 134 -> 15 135 -> 16 136 -> 21 145 -> 22 ... 456 -> 42 112 & 122 -> 43 113 & 133 -> 44 114 & 144 -> 45 ... 556 & 566 -> 65 three equal numbers -> 66 The reason I thought my solution wasn't neat is because I just did individual mapping rather than some kind of a rule to calculate the two simulated dice. The programming one just blew me, simply a formula with tetrahedral and triangular numbers? The hexagon one is neat, too.
Just watched this 4 year old video and haven't seen the simple (geometric) solution to eliminate one of the dice independent of the results by just discarding the one closest (or furthest) from the center (when you look from the top). In the rare cases where you can't decide just reroll.
So much crazy math and convoluted solutions, heh. Even the simplest one is pretty crazy. You've got a way to map them to a single dice, right? Roll it twice.
+logicalfundy That wasn't the puzzle. If you can change a puzzle it most likely becomes way more easy to solve. Also those crazy math and convoluted solutions, as you call them, are (at least for me) fun and way more interesting if you are in a group (of math-interest people).
+logicalfundy Roll it twice?! If you're going to cheat, at least cheat efficiently! Label each side of the outer cube 1 to 6 to get a second die in one roll.
In the original video, he said a way to count 3 dice as 1, and that is to take the sum%6 (remainder when sum is divided by 6). In the case when you get 0, treat it as 6.
Instead of super glue, you should use acetone for acrylic. It dissolves the acrylic then evaporates and welds the two pieces together. Leaves an optically clear finish, won't stick to your fingers, and is REALLY strong after letting it sit for a day or so.
the solution with the hexagon is really good. It is the only one I have seen so far, that is as fast as a lookup table and can be used while playing with children. (Or, if you do NOT have children (like me): while being a little drunk.)
Wow! I was wrong... Sometimes we learn more from being wrong than right. A good reminder for us to question everything we THINK we know. Cheers to Matt and all his subscribers
An updated version of my solution can be found at github.com/vlent/dice The code is simplified slightly and there are versions in several programming languages, not just Python. There is also a suggestion on how to use the method to do the calculations by hand (or in your head). Matt, thanks again for posing this interesting problem. jan
Simply ignore the outsider. Chances are that two dice are obviously closer to each other than any other pairing. Use those 2 dice and ignore the one that's further out. Reroll if it's not obvious. You can use the left to right distance as a fall back before re-rolling, but the 2 dimensional distances are more table friendly. The dice faces are also fair, but I calculated a few hundred thousand results to prove that it works for the sum. Here are the counts of each result starting with rerolls, then 2-12 in order: 8896 14632 29043 43548 57806 72846 86871 72759 58336 43332 29118 14575 Here that is for 2-12 in 36ths of the total, next to the theoretical ideal: 1.007432 1 1.999648 2 2.998336 3 3.980018 4 5.015541 5 5.981181 6 5.009551 5 4.016509 4 2.983464 3 2.004812 2 1.003508 1 That's a total error of 2.22x10^ -16 Method: for each random die value I also generated center weighted X and Y values from 0-100. I divided and rounded each result by 5 in a pointless simulation of things being too close to call. Then I compared distances between dice in pairs and checked for a unique lowest distance. If that failed, then I check just checked X axis differences for a unique lowest. The first number in the first list is the number of times both tests were too close to call, forcing a re-roll.
Is there a way to describe the amount of entropy (randomness?) in a system? Obviously three indistinguishable dice have more of this property than two distinguishable dice, but how is that quantifiable?
New solution based on this video: glue a die into the corner. If the acrylic box is sufficiently heavier, you can convince yourself that the off-centered weight doesn't matter. Need to focus on two: use the non-glued ones :)
I have what might be a solution, not sure quite yet, if you want to check it, be my guest, but this is what i came up with: take the 3 dice, take the sum, and mod 6. Then you get the fair result of a single dice. next, take the highest and lowest die values and sum them and mod 6, then take the remaining die and the "virtual die" (value obtained above) and sum and mod 6. This method, if fair, was my technique of coming up with a method that doesn't involve a computer, table, complex math, or cardboard cutout. It requires you to do: 3 sums. 3 modulus. remember 1 value of 1-6. and finally, distinguish the lowest and highest values of 3 values. This is interesting, but it does not work, i just tested it with some code.
Dice A, B, and C, in ascending order: ( (A + B) mod 6 ) + 1 = Die 1 of 2 ( (B + C) mod 6 ) + 1 = Die 2 of 2 The solution to the one die problem from the first video, just done twice. Quick simulation in R seems to show it's good. Anyone disagree?
Matt accidentally made a paradox while saying : "... for going to far which is exactly the right distance to go ..." If to far is right then it is not to far so it is not right
A contradiction isn't a paradox. Besides, statements in this format generally mean that what an individual might think is going too far is, in reality, going just far enough.
+ST-Leo-TS GodGoneRogue It was awesome and unexpected! For anyone who doesn't know what it is about, chekc Numberphile's channel, video: "The Parker Square"
I have a solution: All numbers are the same has priority over their actual sum. ie rolling 3,3,3 gives 7 not 5. 2 dice;3 dice 2; 5 3; 4 or 6 4; (8 or 10) where two dice are the same number 5; 9 6; (8 or 10) where all the dice are different numbers 7; 7 or 14 or all dice are the same number 8; (11 or 13) where all the dice are different numbers 9; 12 10; (11 or 13) where two dice are the same number 11; 15 or 17 12; 16 Its not a too complicated exchange of values. Found by mapping types of combinations with (whether dice values are repeated) in 'buckets' with small extra conditions to fit for probability.
“Hey, does anyone have a couple 6 sided dice handy?”
“No, but I do have three dice sealed in a box and a printed out hexagon if that will work?”
I used to have large, hollow, transparent d6's with a miniature d6 within each. Don't know what happened to them.
@@BolasMinion same here, I also have 3 distinguishable dice in a cube (as opposed to 3 indistinguishable dice in a cube) as well as a couple sets of the "full" dice array (by ttrpg standards at least)
10:35
Gluing one of the dice to the corner;
most elegant solution
You inadvertently solved it!!! simply glue a dice in the corner, and then ignore it! problem solved!
lol
As was said earlier, "no physical solutions allowed."
A dice?
you mean a die?
A parker square of a solution!
+Broken Wave no. Singular dice is also correct and to many people sounds better.
Cyanoacrylate adhesives are really hard to use on acrylic without clouding generally, and this was the perfect storm of conditions for that to happen because one of the main ways they cloud (in addition to just incautious application beyond the joint) is blooming, where the C-A monomers evaporate and then deposit on surfaces - since the inside of the cube was completely unventilated the monomers had nowhere else to disperse and so anything that evaporated inside the joint deposited back on the nice clear acrylic...
After you roll, flat spin the cube so that the three smaller dice move to the corners. Add the two that end up in opposite corners for a two dice solution. Ignore the two in opposing corners for a one die solution.
Okay. I admit it. When you grabbed the graphic, that was awesome.
+Teddy Boragina I'm disproportionately proud of that.
You should be
+standupmaths I wish UA-cam made frame scrubbing easier, so I could see the frame where you cut. As far as I can tell without that, you just had your hand perfect.
I think it was hanging behind the graphic and there was no cut.
satisfiction What was it hanging on?
"Turns out, simulating 1 from 27 is trivial." Brilliant :)
+Penny Lane Indeed. That's the best Parker's Square I've ever seen.
*rolls* ...five
*rolls again*
*the dice fall everywhere*
...three
For a few brief seconds he became an engineer.
@@markenangel1813 I died.
I'm so sad I can't ask a friend to watch twenty minutes of maths videos just to share that one moment of laughter with me. But I'm glad some folks on the internet enjoyed it with four years of time gap !
My brain exploded when you plucked the hexagon off of the screen.
+LittleMikey _head asplode_
+LittleMikey Your head exploded‽ Imagine how I felt!
standupmaths Nah just my brain. Thankfully my skull contained the mess.
+Popcornio Strong Bad!
+LittleMikey Are you still alive?
That Parker cube quote was priceless
He has submitted to his fate. Loving it. Parker cube ftw!
He said he didn’t want to name the Parker Square because he knew this would happen.
I can't even imagine the amount of time editing this took. It was all spot on. This channel deserves all the views and then some. I'm a fairly new sub, but I already love this community, and the engaging conversation that is somehow able to take place in the comments. I'm already looking forward to the next puzzle.
"Well I can confirm it actually took me..."
*Video ends*
GODDAMMIT MATT!
>Rolls big die
>Explodes on table
"..........Three!!"
Didn't he ask for a math solution?
You're providing a physics solution.
That's not what Shadowrun meant by exploding dice.
This kind of dedicated community involvement is something we need more of these days. Thank you! Keep up the good work!
You stopped fighting.
You accepted your legacy.
Your resistance was a classic #ParkerSquare, and it was brilliant.
Thanks for the Parker Cube quote.
+Melody Williams "Parker, the (almost) Great Mathemagician!"
loved the last seconds.
Seems like the answer begins with a three.
+Michael Schmitz it may be "13xx", or he wanted to say "3xy" times, so the resoult can be from 3xy*1 to 3xy*9. So, don't rely on that beginning "th" (or at least, I hope it is different from that because I would have got the wrong answer ;_;)
The rules to the hexagon method should be on the backside.
Now we need a bigger dice that contains 3 dies which have 3 dies inside.
+StereoBucket soo meta
and figure out how to take the three indistinguishable pairs of simulated distinguishable dice we can get from that to then simulate three distinguishable dice
you don't seem to understand how the words die and dice work...
Die is singular.
Dice is plural.
Dies isn't an applicable word.
so there you go.
+Broken Wave Ikr
Yo dawg, I heard you like dice...
Matt Parker! In addition to your obvious numeracy and charm, you're a KILLER video editor / graphics creator. The time and attention it must take to make the extremely helpful animations that pepper your videos is not lost on me - I'm impressed. Thank you!
BTW, I loved the 10x10 Parker Grid :)
+Antonio Barba
good one xD
+Antonio Barba cute!
"I've arranged them in a 10x10 grid... Of course missing the 3 in the corner..."
Looks like you arranged them into a bit of a Parker Square then ;)
Jokes aside, loving the videos, and while I couldn't really take part in this puzzle thanks to exams, I really enjoyed hearing the solutions people came up with and I'm looking forward to wracking my brain over the next one!
That hexagon solution was brilliant.
Straight to the point
Did it take you 2,835 tries? (Video is 28:35)
or maybe an average of 28.36 times LOL
+DiCasaFilm That would be cool!
+DiCasaFilm That would be very clever of him. I have to admit I literally yelled "Matt no!!!" when he didn't give the answer at the end though....
Imagined if he never showed his face again. The mystery of the coin flips, forever shrouded in... more... mystery, yeah.
For the 2 dice simulation, couldn't you just role the 3 dice cube 2 times and use the 1 dice simulation? That seems like the simple answer to me.
I love the transitions, I love the physical hexagon, and I mostly love the ending
„simulating one from 27 is trivial“ - rolls mother of all dice hahahaha
+Cr42yguy Aaaaaaaa... Three. xD
The shear amount of gusto with which he says, "5!" Is just fantastic.
You could have just taken those 97 dice, turned 16 to each face (1-6), and then used those 16 to form a surface to create a large signing of your name. You would only have to sign your name six times, each person would know which dice was which based off the portion of your signature, and the side the portion was on, and the last die could have been tacked on at any point to extend a rather dapper tail to the end of one of the signatures.
me "Teacher, when am i ever going to use this in real life?"
teacher "if you ever have too much time on your hands and want to make a popular youtube channel"
Those two videos of yours and the responses they caused are the reason I subscribed.
A very worthy puzzle, and it produced a lot of interesting stuff. Congratulations.
Daaahhhhhhhh! You scoundrel, leaving us on a cliffhanger!
+bdot02 Now ask yourself why he specifically showed the dice labeled "9, 7" and "6, 2".
+Cristi Neagu I am too stupid to understand ;-;
+bdot02 Video length is 28:35 for a reason.
I know I'm a bit late and someone might've had the exact same solution (891 comments is a tad too much to read), but after hours of twiddling around I came up with (and if I may say so myself) a rather beautiful solution to the problem. It has simple rules and only four specific exceptions and, most importantly, the result is ordered i.e. (1,2) is as probable as (2,1).First die:Basic one die result from three dice (i.e. sum of values mod 6 with 0=6).Second die:Case 1) All three dice have different values- Select the ONLY even (2,4,6) or odd (1,3,5) value which is not present (e.g. 125=>3, 146=>2).- Exceptions: If all are odd (135), select 1; If all are even (246), select 4Case 2) Two dice have same value, third has a different value- Select the single different value (e.g. 455=>4, 113=>3)- Exceptions: 144=>5, 114=>2Case 3) All dice have the same value- All odd => 3; All even => 6The exceptions may occur only when first die is 3 or 6, and actually they concern only results (3, 1), (3, 5), (6, 2) and (6, 4). All other results can be derived directly by the rules.The case 1 forms a nice diagonal arithmetic pattern in a 6x6 result table. Also the exceptions may be changed to following without breaking anything: 135=>5 with 225=>1 (144=>1 normally) and/or 246=>2 with 255=>4 (114=>4 normally).
Another Monopoly error: You do need to know the individual dice because otherwise, you wouldn't know when you've rolled a double.
But you don't need two distinguishable dice.
Imo indistinguishable dice only exist in theory. You can use dice order (the order you see them in) or position to randomly choose a dice to exclude or include
Well most of this kind of mathematical question exists in the abstract. Of course you could just do something like the two dice that are closest to a certain point or something like that, it's more about using the constraints to derive an answer to that specific question, which may shine a light on other parts of the ever-connected map that is mathematics or just act as a way to practice.
@@fisheatsyourhead hmm true
I hoped when signing them you used a permanent marker.
Matt; "Yes: it'll always be a marker."
that has got to be the.best ending to a UA-cam video I have ever seen. seriously. many props!
God, the end was awesome!
I didn't know, that you really flipped that coin so often!
The isometric axometry (forgive me for misspelling) of a cube is an exagon, so there's that. Instead of printing the immage you could look at the transparent dice from an angle and visualizing it that way. Nice video!
THAT ENDING?! DON'T DO THIS TO ME, I'VE WAITED SO LONG! :'(
edit: Such a Parker Ending
+Fettklomp A Parker Square of an ending*
+888SpinR Well done, good sir!
+Fettklomp Parker square of a comment I might add ;P
Very parker square
POV: you watch the original question the moment it releases, search for solution and find none.
Then forget about it for a long time.
*6 Years later, youtube recommends you the solution*
Mathematician: Here's my algorithm for mapping 3 indistinguishable dice to 2 distinguishable dice.
Engineer: Roll the cube then jostle them onto the edge, read the first two dice left to right.
That was my thought too
or just roll it twice
That cheeky solution of gluing one of the dice in place was really the most elegant one. Never saw it coming!
since when did parker squares turn into parker cubes?
+Agent M The glue made it a parker square cube ;)
+Agent M 10/10 bought the shirt
Amr Idrees it's the parker square cube law
parker hypercubes, coming soon to a dimension near you
Misterlegoboy but I can't see 4D
Neat little side thing I discovered while playing around with this stuff: if you take two dice and say 1, 2, and 3 are 0, and 4, 5, and 6 are 6 on one die, and then keep the other die as is, you get an even distribution for the numbers 1-12 (like dolling a D12). I haven't figured out how to generalize it for other multiples of 6 yet.
Oh my god the trickery with the hexagon my brain hurts
I investigated why the triangle numbers stop in the first puzzle, and it really is pretty interesting.
Rolling two dice can be represented by a 2-dimensional table, with the first die's result on one axis, the second die on the other, and the grid being the sum. You'll notice if you do this that it creates diagonal lines of the same number.
Similarly, you can step it up to three dimensions for three dice. This gives a cube of results. Also similarly, it creates diagonal planes of the same result. With a little imagination you can see that the first few planes are going to have triangle numbered size. However, the diagonal cross section of a cube does not stay triangular! Past 8 the cross sections become irregular hexagons. The tips of the triangles are being chopped off, so to speak. That's why the pattern doesn't hold all the way through.
The simplest solution: use the one dice solution, twice.
Winner well awarded, the solution was very true to the spirit of the challenge.
This is so simple. Just exclude the one you glued to the corner.
I like how you point out that some of the solutions only give you the sum which is great for games like "Monopoly". Except that, for "Monopoly" it is important whether you rolled doubles or not.
Parker squared the video there, Matt.
Even though this video is now over 4 years old, I decided to give it a go and try finding my own solution anyway. I spent a good couple hours working on it and I thought I found a novel solution, only to realize that my solution required distinguishing between the *indistinguishable* dice :'(
Here I am 7 months after you stumbling on this video. I learnt from the "simulate 1 die" warm up that trying to distinguish it is definitely not the correct path, but I am wondering, how many simply suggested to just "roll and mod 6, twice!" as a solution.
I mean, if they were distinguishable you could just ignore the same one every time lol.
Thank you for taking the time and effort to making these videos. It's a joy watching them. Thank you.
21:38 wow
+mark van dijken Thank you. Very proud.
+standupmaths that cut at the end
+standupmaths first the magic lemons, now this Magic Parkergon?
Even at 0.25 speed it's impressive how you made that work!
Hey Matt!
Just wanted to say that I really enjoy and appreciate your videos!
Greetings from Finland. Thanks!
If this was to be made as a marketable dice, you could have a semi transparent hexagon on all of the faces
First off, thanks for taking such a great puzzle and bringing it to this community where we could all share our findings, intuitions, etc.. Also, I think you did an excellent analysis of a variety of solutions to this problem!
I will however assert that we have missed out that you have not done an analysis of the 3 to 2 dice solution I posted, as I it contains what appears to be a very powerful mathematical generalization! I stress this, not because "its mine" (I do not even attempt to take ownership! It is for any and all.), but because its really cool, and very few have approached it so far, presumably because it looks intimidating due to a perhaps a lack of emphasis of elementary algebra in modern mathematics.
Thank you for bringing this problem to such a large forum! I am very happy that I got the opportunity to participate socially about what would typically be a personal musing the results of which would be buried in a pile of notebooks.
All the best! :-)
+myName Thanks! I'm very proud of the community here. Sorry I didn't get to include your solution. There were so many good ones, I had to cut it down a bit arbitrarily.
I cant imagine trying to wade through all this! This has been awsome though! Thank you! :-D
well I would've just thrown the dice twice
It's not really mathy solution, and I am 5 years late, but my very quick solution would be to discard the "odd" dice. "Odd" means not the number, but position.
1. If two dice are stacked, discard the one on the bottom (you could not read it anyway). If tree dice are stacked, discard the middle one and read the bottom dice from the bottom (through the floor).
2. Else if only one die touches walls (not counting floor) or if only one die do not touch walls, discard that die.
3. Else if only one die is in the corner (touches two walls), or is the only one that is not in the corner, discard that die.
4. Else if three dice are touching corners, discard the middle one (the one that does not have a partner in the opposite corner).
5. Else if all die touch walls, discard the middle one. I they are about equal distance apart, look at walls they are touching, one should be middle.
6. Else if you it cannot be decided what die to discard (for example no die touches walls- I think this is the only case, but I have not proven that :-) ), roll again.
The only drawback for this is that from looking, we cannot prove in all cases that die is touching the walls or not. So there may be some pushing around if players see the result they don't like.
My favorite thing about these solutions is that it's way easier to just roll two dice.
The easiest solution is to just pick the number 1 twice. (Sure, it's the least random, but it's definitely easiest.)
@@johnnye87 Meh, you should choose 4 and 2. :p
Best video I've seen. Completely absorbing! So cool how so many people get involved! Love your videos. Gutted the signed dice ran out already. (Thanks for the signed card though 😊). Trying to pluck up the courage to go to a maths club night - not quite up to speed yet though.
Keep the videos coming - they're great. Thank you 😄
Loved the Parker cube joke :D
The way I thought about the warm up puzzle was just to note that it doesn't matter what the "first two" dice are - the third dice will always move you uniformly through the six possible remainders. You don't have to actually be able to *identify* the dice for this - it's a thought process. No matter which two you choose to "call" the first ones, that third still is completely independent and it will have a uniform probability distribution, and it does move you through the mod 6 cycle. QED.
For the two-dice problem, you did not say we had to accept every single roll. You didn't say we couldn't re-roll. So my solution there is to roll until at least two dice match. There's a 4/9 probability of this happening, so it's unlikely to take many rolls. Then count the doubled number ONCE and count the odd die as the second one. I haven't simulated this, but as far as I can see by just thinking about it this should yield a valid two-dice probability distribution.
The fun thing about that two-dice solution is that it works for the single-die puzzle as well. It's less practical though - you'd reject all rolls for which all three dice didn't match, and that mean you were rejecting 35/36 of all rolls. Not quite practical. But, it does work. And I think it would actually work for getting any M-dice roll out of N indistinguishable dice; you'd just have to roll until N-M+1 dice matched. Practicality varies, but... it is "a solution."
I see that Matt Parker has come to terms with the Parker Square ( 22:22)
Easy way to do it without the hexagon:
1. No matching dice:
1.1 All even/all odd; lowest number & lowest number + 3 (ascending order). Ex: 2, 4, 6-> 2, 5
1.2 Straight; middle number twice, a straight can wrap around. Ex: 5, 6, 1-> 6, 6
1.3 Everything else (with no matching dice) check the two dice that are opposite (1 is opp. 4, 2 is opp. 5 and 3 is opp. 6) the first die is the remaining one, the other one is the closest one from that, going up (it can wrap around) (ascending order). Ex: 1, 3, 6-> 1, 3
2. Pair; it's simply the two values shown (descending order) Ex: 3, 3, 6-> 6, 3
3. Three of a kind; it is always 3, 6 (ascending order)
You need to know the individual values for Monopoly. If they're both the same you get another go, get out of jail free, etc.
Two indistinguishable dice will do, you don't need to know which one is first and which one second.
I haven't watched the original video yet. But I did solve it the way I considered not neat, started like 11:12. But the bit where the triplets go into the grid I simply did it (in my head) as a map table:
123 -> 11
124 -> 12
125 -> 13
126 -> 14
134 -> 15
135 -> 16
136 -> 21
145 -> 22
...
456 -> 42
112 & 122 -> 43
113 & 133 -> 44
114 & 144 -> 45
...
556 & 566 -> 65
three equal numbers -> 66
The reason I thought my solution wasn't neat is because I just did individual mapping rather than some kind of a rule to calculate the two simulated dice. The programming one just blew me, simply a formula with tetrahedral and triangular numbers? The hexagon one is neat, too.
D'oh! Just watched the original video, and it specifically does not want a lookup table, haha.
Lost it when he said parker cube.
+Marcel Robitaille
Parker cube confirmed.
Parabolati Confirmed.
Parker Cube and Parabolati start with the same letter.
Coincidence?
I think not!
The parker cube is infinitely many parker squares stacked, so it must be an even bigger fail than the parker square
vg fabi An infinitely bigger one. There's probably a definition for how much more of a fail it is.
Just watched this 4 year old video and haven't seen the simple (geometric) solution to eliminate one of the dice independent of the results by just discarding the one closest (or furthest) from the center (when you look from the top). In the rare cases where you can't decide just reroll.
Hows this for a solution:
Write the numbers 1 through 6 on the outside (clear) die, then take the mod(6) of the sum of inner dice
+elvishfiend Genius!
+elvishfiend I guess this would be an outside the box (like writing the numbers on the cube and rolling twice) solution which was excluded
Elchi King Literally! (As the ink from the marker you used would be on the outside of the box.)
+elvishfiend This method was mentioned in the orignal video and is not a valid solution. This is a maths problem, not a party trick.
+Tumbolisu Maths is srs bzns.
rly srs bzns.
As soon as you mentioned it was acrylic and you talked about gluing your fingers together (implying a CA glue), I wept for the clarity your megadie.
So much crazy math and convoluted solutions, heh. Even the simplest one is pretty crazy. You've got a way to map them to a single dice, right?
Roll it twice.
We choose to -go to the Moon- figure out these puzzles in this decade and do the other things, not because they are easy, but because they are hard.
+logicalfundy That wasn't the puzzle.
If you can change a puzzle it most likely becomes way more easy to solve.
Also those crazy math and convoluted solutions, as you call them, are (at least for me) fun and way more interesting if you are in a group (of math-interest people).
+logicalfundy Roll it twice?! If you're going to cheat, at least cheat efficiently! Label each side of the outer cube 1 to 6 to get a second die in one roll.
In the original video, he said a way to count 3 dice as 1, and that is to take the sum%6 (remainder when sum is divided by 6). In the case when you get 0, treat it as 6.
logicalfundy lol, as if he didn't make it very clear that wasn't allowed xD
I would just ignore the 1st die and keep the second two.
That acrylic cube assembly job really turned out to be a bit of a Parker square.
True
He made that joke in the video.
*cube
jgallantyt uh oh. Somebody didn't watch all of the video before commenting
22:19 Nice reference. I would've totally commented calling it a parker square of a trophy but this is much better.
Honestly, Matt, you may well be the modern-day Martin Gardner, posing these puzzles to us like this.
+doctorwhofan883 More like ferma XD
Instead of super glue, you should use acetone for acrylic. It dissolves the acrylic then evaporates and welds the two pieces together. Leaves an optically clear finish, won't stick to your fingers, and is REALLY strong after letting it sit for a day or so.
the solution with the hexagon is really good. It is the only one I have seen so far, that is as fast as a lookup table and can be used while playing with children. (Or, if you do NOT have children (like me): while being a little drunk.)
+Ceragon little children == little drunk: TRUE
+standupmaths Lol.
Wow! I was wrong... Sometimes we learn more from being wrong than right. A good reminder for us to question everything we THINK we know.
Cheers to Matt and all his subscribers
You really Parker Squared that Parker Cube...
+vipero07 savage
An updated version of my solution can be found at
github.com/vlent/dice
The code is simplified slightly and there are versions in several programming languages, not just Python. There is also a suggestion on how to use the method to do the calculations by hand (or in your head).
Matt, thanks again for posing this interesting problem.
jan
I almost flipped my desk at that ending.
+viralinfecticide How many times did you flip it? That's what we want to know!
+EHaraka 0.5 times, since it was "almost" flipped.
Simply ignore the outsider.
Chances are that two dice are obviously closer to each other than any other pairing. Use those 2 dice and ignore the one that's further out. Reroll if it's not obvious. You can use the left to right distance as a fall back before re-rolling, but the 2 dimensional distances are more table friendly.
The dice faces are also fair, but I calculated a few hundred thousand results to prove that it works for the sum. Here are the counts of each result starting with rerolls, then 2-12 in order:
8896
14632
29043
43548
57806
72846
86871
72759
58336
43332
29118
14575
Here that is for 2-12 in 36ths of the total, next to the theoretical ideal:
1.007432 1
1.999648 2
2.998336 3
3.980018 4
5.015541 5
5.981181 6
5.009551 5
4.016509 4
2.983464 3
2.004812 2
1.003508 1
That's a total error of 2.22x10^ -16
Method: for each random die value I also generated center weighted X and Y values from 0-100. I divided and rounded each result by 5 in a pointless simulation of things being too close to call. Then I compared distances between dice in pairs and checked for a unique lowest distance. If that failed, then I check just checked X axis differences for a unique lowest. The first number in the first list is the number of times both tests were too close to call, forcing a re-roll.
Cliff hangers... gotta hate them...
Is there a way to describe the amount of entropy (randomness?) in a system? Obviously three indistinguishable dice have more of this property than two distinguishable dice, but how is that quantifiable?
That's one Parker Square of a first place trophy.
Dang, I was looking forward to hearing your favourite solution.
Parker cube.... I love it! Haha
New solution based on this video: glue a die into the corner. If the acrylic box is sufficiently heavier, you can convince yourself that the off-centered weight doesn't matter. Need to focus on two: use the non-glued ones :)
lol!
first a square that does not work and now a damaged cube.
i am waiting for you to name a spoiled hyper cube.
Don't forget the Parker line and Parker dot for 1 and 0 dimensional objects!
Almost missed that one!
I have what might be a solution, not sure quite yet, if you want to check it, be my guest, but this is what i came up with:
take the 3 dice, take the sum, and mod 6. Then you get the fair result of a single dice.
next, take the highest and lowest die values and sum them and mod 6, then take the remaining die and the "virtual die" (value obtained above) and sum and mod 6.
This method, if fair, was my technique of coming up with a method that doesn't involve a computer, table, complex math, or cardboard cutout. It requires you to do: 3 sums. 3 modulus. remember 1 value of 1-6. and finally, distinguish the lowest and highest values of 3 values.
This is interesting, but it does not work, i just tested it with some code.
Matt, I fear that you did not actually count the number of flips it took. Please give me closure.
Dice A, B, and C, in ascending order:
( (A + B) mod 6 ) + 1 = Die 1 of 2
( (B + C) mod 6 ) + 1 = Die 2 of 2
The solution to the one die problem from the first video, just done twice. Quick simulation in R seems to show it's good. Anyone disagree?
Came here to say the same thing - beaten by a day! doh!
I liked the gluing solution the best. ;)
No idea how I got to this video, but I really enjoyed it.
Matt accidentally made a paradox while saying : "... for going to far which is exactly the right distance to go ..." If to far is right then it is not to far so it is not right
***** I just really like paradoxes
yay! Paradox
It is not a paradox, it is simply a quantum superposition of states
A contradiction isn't a paradox. Besides, statements in this format generally mean that what an individual might think is going too far is, in reality, going just far enough.
Beat Button I know what he meant I just like finding paradoxes and what is a paradox if it ins't a self contradiction ?
that was a kewl video. man the graphics in this show constantly surprise me.
Actually you almost get the easiest solution to the 2 dice from 3 question.... Just glue one into a corner xD xD
lolllll
I'm quite late to the party; just watched the video. Choose integers X, Y, and Z. Let Z, Y, and Z be coprime where 1
Parker cube!
10:26 That "quick calculation" is probably my favorite part of the video
i love the Parker cube reference XD
+ST-Leo-TS GodGoneRogue It was awesome and unexpected!
For anyone who doesn't know what it is about, chekc Numberphile's channel, video: "The Parker Square"
I have a solution:
All numbers are the same has priority over their actual sum. ie rolling 3,3,3 gives 7 not 5.
2 dice;3 dice
2; 5
3; 4 or 6
4; (8 or 10) where two dice are the same number
5; 9
6; (8 or 10) where all the dice are different numbers
7; 7 or 14 or all dice are the same number
8; (11 or 13) where all the dice are different numbers
9; 12
10; (11 or 13) where two dice are the same number
11; 15 or 17
12; 16
Its not a too complicated exchange of values. Found by mapping types of combinations with (whether dice values are repeated) in 'buckets' with small extra conditions to fit for probability.