# The Three Indistinguishable Dice Puzzle

This week I have a puzzle for you that I haven’t solved myself. But, I still know there is definitely a solution. I just don’t know if it’s a nice solution or not. When my maths buddy Lucas showed me this dice… Basically it’s one clear cube which has three tiny cubes Inside it and the idea is instead of rolling three dice which could scatter in all different directions, you just roll one cube and there you are, three dice, eleven. When Lucas showed me this three dice in one, he said he was wondering “let’s say you were at home and you want to play a game that requires a single dice” [ha] well actually he would have said “a single die”, because he’s more correct than I am whereas personally I never say die, and so he said “how could you use the three-in-one dice to simulate a single six-sided dice”. So that’s the first half of the puzzle How could you use this three dice in one to simulate a single D6 dice and I don’t want any cheating by which I mean thinking outside of this tiny box, right, this is not a lateral thinking problem It’s a maths problem. I want a mathematical solution of course in the real world if you had this and you wanted to have a normal six sided dice you just get a marker and write the numbers one to six on the outside, and then roll it normally. No, I want a maths solution. To state it a bit more precisely, you have three indistinguishable dice. You roll them simultaneously, can you have an unambiguous way that everyone playing the game can agree on to look at the three results and immediately know, “oh!, that’s equivalent to a” you know four or six or something, right? How can you do that? I do have a solution to this first half of the puzzle and I’m going to give it to you in just a moment. If you want to have a go yourself pause the video now and, uh, have a go because Lucas and I stood out there for a while rolling the dice and he also had a solution to this simple case where you’ve got to take three down to one and what you do is after you’ve rolled it you simply add the values together. Now, of course that might give you a number bigger than 6 and if you ever get a number bigger than 6 You just take whatever the remainder is once you’ve divided 6 into the number. We say it’s the answer Mod 6 or Modulus 6. Ok, so you roll the dice, and you have a look at the result and you go look at that I’ve just rolled a 13 which mod 6 is one piece of cake. After a while you can get pretty quick because used to ignore the sixes. I’ve just rolled a three, ignore the fact there’s a dot missing on the two. That’s the problem with a sealed container You can’t get in there to fix any of the dice. On to the slightly harder second half of the puzzle that I don’t have a solution to and It’s exactly what you’d expect. How would you use this to completely fairly simulate… two dice being rolled. So you roll three indistinguishable dice how can you collapse that down, fairly, to two dice the same that you would get if you [just] rolled two originally and again, no physical solutions. I mean, I guess in reality if you had to roll three you know to you could just say oh I’ll take whichever two are closest together or you could mark a dot or a line in whichever ones are closest to that. There are ways around the problem. I want to know mathematically, how do you do it in theory? How do I know there is definitely a solution even though I haven’t found it. Well, I know there’s definitely an ugly, crude, solution because when you roll three dice you’re going to get one of 216 possible results, and you could in theory make a giant list of one of those 216 results, and then you can use that as a kind of lookup table. So you roll the dice, you go “oh!, I just got a 4 1 1”, you look up 4 1 1 in the massive list and go “Oh, that’s a 3” or something. Now, there are some slight problems here. You need to put 216 possible results into 36 different categories and some of them are indistinguishable, so this one 1 1 4, that I just rolled, well I mean, I don’t know which dice is which and so I need to put all the results I can’t distinguish between in the same category. So if I put them all in buckets where all the indistinguishable ones are in the same bucket and then I assign each bucket to one of the possible 36 normal results from two dice. And then you end up with this massive table that you and your friends can just look up while you’re playing the game. But you’ve got to admit, that’s a bit disappointing. Just listing all the possible results of matching them up. We need a better way, and that’s what we did for the one dice solution. When you add them together and mod six there is a shortcut to effectively memorize the way you’ve put all the values in buckets. You could then use that to write out the complete list of how every possible result matches to a new result but because you can add and divide by six fairly quickly you don’t need to. And so the true puzzle here is, is there a way to take the three results from the indistinguishable dice and map them quickly and easily onto the possible results from just rolling two dice, so every result has the same probability of coming up. So there you are have a go I have shown this puzzle to a few other people I did a talk at the university of the west of England in Bristol last week and a few of the guys there got very excited about this they’ve been emailing me their findings so far, And it looks like there are better ways than just memorizing a lookup table So have a go. If you think you’ve got a solution just put it in the comments underneath don’t worry about spoiling if everyone else we can all work together I wanna find the best way to solve a problem. Which isn’t really a problem. And by the way, I haven’t forgotten and to announce who won the, guess how many times I had to flip a coin to get ten in a row, From this video here when I passed ten to the five subscribers. I will get that done soon. I guarantee it it’s just I’ve been ridiculously busy recently partly, very excitingly, because I’m writing a, uh, Radio program for BBC radio four, vestal the Spoken Nerd which I’m a part of. We’ve got a four part series On Radio 4 which will be coming out in july. Which is very exciting but it means I’m now brain deep in writing math jokes for the radio so, but that’ll be out soon. It’ll be great I’ll keep making videos just I’m sorry everyone who sent me an email or a message And I haven’t responded in a timely fashion. I will get right onto it, hopefully soon I’ll have more time to do more videos, but I will announce the winners to the competition soon

Can you please prove thet with the "reminder" solution, all the six numbers have the same probability?

Problem 1 (turn 3 into 1)

Find the mean

Problem 2 (turn 3 into 2)

Find the mean and times by two,

If it isn’t whole then round it (aplys to both)

Problem 1 eg

2+3+3 = 8

8/3 = 2.66

2.66 is roughly 3

Problem 2 eg

6+3+5 = 14

14/3 = 4.66

4.66×2 = 9.33

9.33 is roughly 9

The only problem is that in 1

1 and 6 is less likely

3 and 4 is most common

For problem 2

1 and 12 is least likely

6 and 7 is most likely

Divide each by three

I would have the biggest number dice subtract the 2nd highest and add the lowest for the first problem and that can result in any number between 1-6 like 6-5=1+3is 4 ect i think that works maybe idk 😐

Maybe I'm being stupid here, but couldn't you just remove the with the middle value? If you roll A<=B<=C, remove B, and A and D are your dice? Or wouldn't this work?

And similarly, for the 1 die problem, keep only the middle-most value; B from the last paragraph?

You could have the 216 results then just put 1-12 like 1 1 1 = 1. 1 1 2=2 1 2 2=7 1 2 6=11 1 3 5=2 etc

Add the lowest and highest and mod-6 the sum. Count that as one die along with the other die that was neither the highest or lowest.

Hash the values so that the probabilities are the same …there done…give me my cookie now

what if you roll an 18? 18 mod 6 is 0 right?

…wait, what was the problem again? Lol

could you not just do mod 12..?

Heck yeah modular arithmetic I called it

I came back to this video to comment on something I thought about a while after watching it, but this has nothing to do with the question that was asked. I was thinking about the mention of writing a number on each side of the outer cude, and it occurred to me that a much more useful device could be made as a slight modification of the one in this video. Imagine a hollow transparent dice with markings for 1 through 6 on it, and with three dice inside of it each in a different color, such as a red one, a green one, and a blue one. Rolling that would be rolling 4 dice at once. Pick one to ignore and you've got 3 at once. Pick 2 to ignore and you have 2 at once. Pick three to ignore and it's like rolling just one. You could also agree to apply a bit of specific math to each roll such as subtracting one from each dice and multiplying the result of that subtraction by a unique power of six. So this got me wondering… how hard would it be to make a mechanical device that could be rolled like a dice to internally produce a long binary number, perhaps strung together from a set of 8 sided dice, or even to produce a random base ten number of a given length from which a given number of digits could be agreed upon in advance to be ignored? Something to think about. 🙂

Here's a simplified rendition of the compressed look-up table I posted some time ago for converting an un-ordered roll of three dice to a number representing a roll of two dice (assuming 6 sides each, numbered 1 through 6)…

2 iff {1,1,1}|{2,2,2}|{3,3,3}|{4,4,4}|{5,5,5}|{6,6,6}

3 iff {1,1,2}|{1,1,3}|{1,1,4}|{1,1,5}

4 iff {1,1,6}|{1,2,2}|{1,3,3}|{1,4,4}|{1,5,5}|{1,6,6}

5 iff {2,2,3}|{2,2,4}|{2,2,5}|{2,2,6}|{2,3,3}|{2,4,4}|{2,5,5}|{2,6,6}

6 iff {3,3,4}|{3,3,5}|{3,3,6}|{3,4,4}|{3,5,5}|{3,6,6}|{4,4,5}|{4,4,6}|{4,5,5}|{4,6,6}

7 iff {5,5,6}|{5,6,6}|{1,2,3}|{1,2,4}|{1,2,5}|{1,2,6}|{1,3,4}

8 iff {1,3,5}|{1,3,6}|{1,4,5}|{1,4,6}|{1,5,6}

9 iff {2,3,4}|{2,3,5}|{2,3,6}|{2,4,5}

10 iff {2,4,6}|{2,5,6}|{3,4,5}

11 iff {3,4,6}|{3,5,6}

12 iff {4,5,6}

In this table, the order of the rolled numbers is given as lowest to highest but represents all arrangements of those three numbers. I really don't think a simple mathematical formula solution is possible, because there is no reliable way to distinguish between the same numbers rolled in different orders.

Relaunch till you have all the dice on the same side? Quite simple no?

As all 3 dice have 5 sides which you aren't using, you could select a any second side of each die as your second number. Which you then calculate using your given solution. Technically speaking, you can have up to 6 dice solutions using this.

ignore the highest and lowest die. if there is a tie for highest or lowest select which die to ignore at random as they are functionally identical. look at the remaining die. would that not have the effect of randomly selecting a single die that has a random number on it?

use a coin as a die

I think i found a way by taking the first dice to be x + y + z mod 6 and the second dice to be p(x) + p(y) + p(z) mod 6 where p is carefully chosen permutation from [0;5] to [0;5].

Subtract the higest two

how about:

if (more odd rolls) ignore largest score

else if (more even rolls) ignore smallest score

should behave like 2 D6, with minimally shifted probabilities..

Applying the relevant line of the sexnomial triangle here, we have 1, 3, 6, 10, 15, 21, 25, 27, 25, 21, 15, 10, 6, 3, 1 respectively as the ways to roll sums of 3, 4, 5, and so on. Of course, these numbers add up to 6^3, or 216, which means if we can group the different events such that they all have 36 ways of occurring, we have our single die roll. I have therefore determined that rolling a sum of 3, 6 or 9 is a one, a 4, 5 or 10 is a two, a 7 or 8 is a three, a 13 or 14 is a four, an 11, 16 or 17 is a five, and a 12, 15, or 18 is a six.

Simple enough. It would have been a lot harder if you couldn't arrange the numbers to get probabilities of 1/6. In that case, you would have to do something else.

You goonie!

Roll the cube and always drop the highest die. you will always be left with a two die result.

(Value mod 6)+1

I paused at 2:00 to create a solution. I like puzzles and brain teasers.

the results of the three dice are viewed as low(1-2), mid(3-4) or high(5-6).

1) 3low, or 2low +1mid

2) 2low + 1high

3) 1low + 2mid

*) low + mid + high, or 3mid

4) 2mid + 1high

5) 1low + 2high

6) 3high, or 1mid + 2high

* If a low-mid-high or 3-mids are rolled, we now have to observe the values of each die, to determine if they are the higher or lower of their range. For example a 3 or 4 is a mid, but since 4 is higher it is the higher mid. In this way any neutral roll of the three dice can be given a high or low value, placing it as a 2, 3, 4, or 5 accordingly, with 2 being all low-end and 5 being all high-end.

add the results and divide by three, or use the die on the right most side of the container after it lands. of course the direction you chose (right, left, up, or down) would be based on the position of the individual rolling the container.

double then amount rolled, then mod 12

I am surely missing something here. Because the single die solution seems unsatisfactory. Surely the modular 6 solution gives unequal probabilities for rolling each of the numbers on the single die. 3 & 6 are more probable than 1,2,4 and 5.

As there are 56 different ways in which three indistingushable dice can fall isn't it impossible to replicate the throw of a single die because 56 divided by 6 doesn't give an integer number – therefore they can be no way of allocating equal probability to rolling each of the six numbers on a single die? (without assigning particular throws as a null result and rethrow option, so as to achieve a multiple of 6 – for example ignore 111 and 666 and now there are 9×6=54 possible results)

My solution is use the clear cube and a sharpie to make 1 die.

Sum of 3 dice / 3 then round the number ez solution

The first solution is correct. He hasn't overlooked anything.

divide the total by 3 and round it

My solution (equiprobable, "proved" by python code over 1,000,000 rolls):

.) there are 56 combinations, so if dice ares (1,2,3) or (4,5,6), roll again

.) if there are two or more identical dice, take their number

.) else it's (sum – 7) assuming 0 is 1 and 7 is 6.

Ex: 344 => 4, 325 => 3, 511 => 1, 635 => 6 (7)

At 3:45, if one can't differentiate the dice, it's 56 combinations, not 216 ( (3+6-1)!/(3!*(6-1)!) ). So the intuitive lookup table isn't that big. But either way, because 56 isn't dividable by 6, the table wouldn't be balanced, so I believe it impossible to mimic a single die in a single roll (it needs 3, then maths) without excluding two combinations from the table.

DIVIDE THE TOTAL NUMBER BY 3 AND ROUND TO THE NEAREST WHOLE NUMBER

Consider whichever value is the largest!

Enjoy your life

I have a solution to your second problem (as well as a slightly different solution to the first). In short if we can determine the outcome of three dice to produce the effect of a single die then all we need to do is do this twice and we have the effect of two dice. The probabilities are precisely correct and you have a solution (although perhaps not the one you were looking for). For a somewhat more in-depth look at my process look below.

For the first problem It's just a matter of combining the necessary outcomes together in such a way as to produce 6 separate equally likely cases.

for three dice

3 and 18 have 1/216 chance of being rolled each

4 and 17 have 3/216

5 and 16 have 6/216

6 and 15 have 10/216

7 and 14 have 15/216

8 and 13 have 21/216

9 and 12 have 25/216

10 and 11 have 27/216 chance

This process is fairly straightforward actually. 10 and 11 each require an additional 9/216 chance in order to produce the necessary 36/216. The only way to create this is by combining the outcomes (5 or 4) and (16 or 17). Do similar strategic combinations for the other numbers and you get the sum combinations to be.

[10 or 4 or 5] (1/6 chance) = 1 (it doesn't have to be a one I just chose it arbitrarily)

[11 or 16 or 17] (1/6 chance) = 2

[9 or 6 or 3] (1/6 chance) = 3

[12 or 15 or 18] (1/6 chance) = 4

[7 or 8] (1/6 chance) = 5

[13 or 14] (1/6 chance) = 6

tada problem 1 solved although admittedly sum mod six is easier to calculate without needing to remember which sums correspond to which outcomes. However I believe this approach to the problem can be used to solve the other more challenging problem you posed. let's give it a go.

for two dice

2 and 12 have a 6/216 chance of being rolled each

3 and 11 have a 12/216

4 and 10 have a 18/216

5 and 9 have a 24/216

6 and 8 have a 30/216

7 has a 36/216 chance

therefore we need to combine our 16 outcomes from the three dice case in such a way as to create the 11 outcomes for a two dice case (preserving respective probabilities).

Unfortunately when we try to combine the outcomes there is no way to create the precise distribution. It occurred to me that this was in part due to the fact that the distribution of two dice has an odd number of element whereas the distribution of 3 dice has even elements. I've thought about it some and I'm convinced that this is part of the problem because an odd distribution can never map to an even one. To make matters more difficult even distributions also do not necessarily map to other even distributions. I made a valiant effort to do so with six dice believing that the solution might be to instead roll the set of three dice twice and sum up all of their outcomes (effectively giving you six dice). This distribution however is impossible to map to the case of two dice (believe me I tried). I believe it may still be possible with some other unknown number of dice however I have no inclination to try because the calculations get very ugly very quickly.

Finally I had a bit of an epiphany. We can easily map three dice to one dice (as we did with the first problem) why not just do that twice. Essentially just roll your group of three dice once, determine the value of that roll then do it again and sum up the results together. Three dice to one dice to two dice. Perhaps not the simplest solution but it does work.

Can't this be solved by adding up the values of all the dice and doing a modulo by 6? I.E. dividing by 6 and taking the remainder?

Of course 0 in the solution would mean 6.

Youst dived by 3 or the higest number

Add the dice, and mod 3. Use the result to determine which die to remove from the set. If you get a 0, remove the lowest dice. If you get a 1, remove the middle dice. If you get a 2, remove the highest dice. This works because the result of the mod is proportionately random, and randomly removing one die yields the same results.

Update:

I tested this with 100,000 iterations and it … doesn't work. 🙁 … Removing a random die is flawed. The randomness has to be biased so it gives the middle die an advantage … but I'm on to something here. 🙂

So I'm not sure if I'm cheating or not, but I think I have a solution so you will be only looking at one dice. This is the method, reletive to the person who rolled the dice (in my example, I'll pretend they are facing North), whichever one in the box is furthest in the North direction is the one you look at, if there is a tie, whichever one is furthest (of the dice that are tied) in the East direction, if their is a tie for that, whichever one is higher up is the one you look at.

Note: You don't have to use the directions I did, just have it agreed upon before you start playing with them

Just roll it twice and add up the results mod 6

3 dice = max of 18

18/6=3

Therefore

1-3 is 1

4-6 is 2

7-9 is 3

10-12 is 4

13-15 is 5

16-18 is 6

For 2 dice… well, this is difficult since you can’t divide 18 by 2 (9) and then by 6 and get a “pretty” number. Therefore, I suggest just rolling twice thus meaning the highest amount you could get is 36. This means your new number ranges will be:

1-6 is 1

7-12 is 2

13-18 is 3

19-24 is 4

25-30 is 5

31-36 is 6

Idk this just seemed the easiest for me and WAY easier than that number chart or whatever the heck it was.

Sum then mod 3

Ya know what's funny? His first solution for one dice is wrong.

If you take the mod 6 of the sum for three dice, end up with the following probabilities of getting each number:

1: 9/56

2: 9/56

3: 10/56

4: 9/56

5: 9/56

6: 10/56

As you can see, you have a slightly higher chance of rolling a 3 or 6, so this is not a fair representation, and thus is not a valid solution to the problem.

Surely you could just use the median

your solution for 1 dice isn't correct as it doesn't maintain the natural distribution of 1 dice.

i.e. there are less outcomes of 1 and 2, as you can't make 1 and 2 with three dice. So you can only rely on 6,7 12,13

Pick the 2 dice that are furthest apart in value. if you roll 1,3,6 you have 1+6=7 if you roll 4,5,3 then 5+3=8

for doubles and triples: 1,1,6 = 1+6 = 7 4,4,4 = 4+4 = 8

Would the harmonic mean, rounded, be decent?

A four is missing a white dot

The easy answer is to divide the result by 3, however some results are more probable than others.

If you did make a table, list all possibilities in order, so 4,1,1 and 1,4,1 would just be listed as 1,1,4.

There are three possible interpretations of 'simulating rolling two dice':

i) You need to know only the total of the hypothetical 2 dice.

ii) You need to know the numbers on the dice (for example 1-6 is distinct from 2-5)

iii) You need to distinguish the two hypothetical dice (for example 1-6 is distinct from 6-1).

So there are probably methods which are solutions for i but not ii or iii, and/or for i and ii but not iii.

Of course, there's the trivial solution… roll a die mod6, twice, to simulate two dice. 😛

The mod 6 solution to simulating a single die roll isn't valid. In the case where you get something like 1,2,3 on the three dice, 6mod(1+2+3)=0. Rolling a die never gives 0 as a result. 0 is outside the sample space of the experiment.

simple roll the die twice and get the one die results then add them together

Take the sum the numbers rolled on the three dice and calculate the average (divide by 3), if the answer is a whole number, take that as your answer.

If not do one of two things:

If the average is lower than two of the dice value, round up

If the average is greater than two of the dice values, round down

mod 6 and div 3

JUST HAD A THOUGHT. Use the same numbers you had for the first one. 1-6 remainder based on the grand total of the dice. Now if more dice are odd, leave it. If more dice are even, add 6. This creates all numbers 1-12 in a fair and equal way. As there are three dice, there will always be either more even, or odd dice, and at random. As you can have EEE, OOO, EOO, or OEE

#3diceto2dice #3diceto2diceconvertion

For 1 die rolling, you could take the average of the 3, divide by 3, and round to the nearest integer. For example, 2 + 5 + 3 = 10, 10/3 = 3 1/3, rounds to 3.

1. Throw the cube.

2. The leftest dice is the one you ignore because it is the least right.

3. Unless you have 2 dice of the similar leftness, in which case you ignore the other dice.

Couldn't you just use the t-SNE algorithm?

Easy. Use the three dice two simulate one method once, then another time.

Where can I get one of those? They're neat but the MathsGear website says they're out of stock 😐

What yo do you do what for one die then do it again so then add

Take the ceiling of the number that comes out divided by 3

take the mod6 result and subtract it from the total sum

(Sum/1.5) then round

Just divide the result by 3 when you want to use 1/3 dice. Divide by 1.5 when you want the result from 2/3 dice. shouldnt be harder, right? 😛

Easy, take result mod 6, which is what you suggested as the first answer and subtract it from the total.

Does 3d6 mod6 give the same distribution as 1d6? I don't expect it would & think it's important for this puzzle

Median of the three

Before seening the whole video i would tell:

Solution for getting a 1 to 6 value: Round(Sum(A,B,C)/3).

But that is not well balanced, so a better solution is:

1+(Sum(A,B,C) MOD 6)

Yes, do not forget to add one, else the solution is 0 to 5, not 1 to 5; Y MOD X never gives X.

Make the sum between and then divide by three that's the way I'll do it

Roll the dice. Sum up two wichever you like that are not greater than six.

There you go.

Just divide by 3

just take 2/3rds of the total of all three dice?

I was sceptical. so sceptical. blown away.

1 has 15+21 = 36 ways

2 has 21+15 = 36 ways

3 has 1+25+10=36 ways

4 has 3+27+6=36 ways

5 has 6+27+3=36 ways

6 has 10+25+1=36 ways

I twitch every time he uses the word "dice" for a single die.

What about the sum of the dice less the sum mod6? I admit, I'm too lazy to check if this is correct.

Can't you just mod 12

2 out of three dice?

Sum the three dice, then mod 12 (the highest possible value for two dice) to the result.

There's a different probability of getting some of the numbers, such as it's (it think) twice as likely to result in 1-6 than 7-12 because 1/3 of the time is over 12.

It might be more complicated than that, but it's quick and somewhat easy.

Never say die

What about zero dice

The marker trick would potentially give you 4 dice in one.

First I thought of approximating the sum of the outcomes of the 3 dice to the nearest larger multiple of 6 from 6-36 and take the multiplier as the equivalent outcome from 1-6 , wrote a little program to run some number on it and it works

But it feels very dissatisfactory to use approximations.