Dice are cubes with pips (small dots) on their sides, representing numbers 1 through 6. Two dice are considered the same if they can be rotated and placed in such a way that they present matching numbers on the top, bottom, left, right, front, and back sides.

Below is an example of two dice that can be rotated to show that they are the same if the 2-pip and 4-pip sides are opposite and the 3-pip and 5-pip sides are also opposite. https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0

How many different dice exist? That is, how many ways can you make distinct dice that cannot be rotated to show they are the same? Note: This problem does not involve rolling the dice or the probability of roll outcomes.

I'm having trouble understanding exactly what is being asked in this question. I understand that I have to find how many different ways the dice can be placed to show that they are the same, but saying they cannot be rotated confuses me.

Could somebody make an attempt at rewording this? Or walking me through how to solve this?

  • $\begingroup$ The question is to find out how many “different” ways are there to label dice with the numbers 1 through 6, where “different” means you can tell one from the other by the relative positions of the numbers. For example, if one die has the 3 opposite the 5, and another has the 3 opposite the 6, they are definitely different. Another kind of difference might be between two dice that both have the numbers 1, 2, and 3 on sides that share a single corner, but on one of the dice, the numbers increase clockwise when you look at that corner and on the other, they increase counter-clockwise. $\endgroup$ – Steve Kass Aug 18 '15 at 16:42
  • $\begingroup$ So the best way to find this is by finding the total number of assignments for a six sided die, which is 6 * 5 * 4 * 3 * 2 * 1 = 7,560 and subtract the number that are the same, leaving a number of different die. $\endgroup$ – Connor Aug 18 '15 at 16:44
  • $\begingroup$ @Connor 6! = 720, not 7560. :) $\endgroup$ – wltrup Aug 18 '15 at 17:08
  • $\begingroup$ Ok, so that says there is 126 different dice types. But you wouldn't be able to tell this is there was a missing side, you technically only have 4 sides to work with. So wouldn't the answer be 24? $\endgroup$ – Connor Aug 18 '15 at 17:11
  • $\begingroup$ Should we consider the d20? $\endgroup$ – PyRulez Aug 18 '15 at 17:17

I won't give you an actual answer, and this post is not a hint to solve the given question, but I will suggest a way that you can simplify the problem in order to understand better what's being asked.

(By the way, that's often a very useful approach, that is, simplifying the problem to get a better understanding of it)

Say you have a square and you assign numbers to the edges, in a clockwise manner, starting from the top edge. There are 4! = 24 different assignments you could make. Here are a few examples: 1234, 3241, 1324, 4123, 4321, and so on.

Now, note that the assignments 1234, 2341, 3412, and 4123 don't actually represent different squares because you could rotate them to match one another. You could put them into a grouping of their own. On the other hand, 1234 and 1324 are different assignments in the sense that no matter how you rotate either of the resulting squares, you won't be able to make them match. So, 1234 and 1324 must belong to different groupings.

What the question about the dice is asking is how many such groupings there are, but with a cube and its faces, instead of a square and its edges. Note that you can rotate a cube in more ways than you can rotate a square.

Edit: Based on the comment exchange below, here's the brute-force solution for the analogous problem with triangles. It's brute-force because I'm actually listing all the possible assignments and explicitly testing if they're pairwise equivalent or not. You will not want to do that in the dice problem. That's not how you want to solve that problem. I'm only brute-forcing the solution here to make it easier for you to understand what the problem is asking.

You can divide the 3! = 6 possible assignments into only 2 groups where every assignment in one group is equivalent to the other assignments in the same group but different from all assignments in the other group. The solution to "how many such groups are there?" is, then, 2. One is coloured blue, the other red in the figure below.

enter image description here

Hint #1 In the case of 6-sided dice, if all the dice have opposite sides adding up to 7 (1 opposite to 6, 2 opposite to 5, 3 opposite to 4) then there are only 2 groups. Try to prove that result, then look at the case when opposite faces don't have to add up to 7. How many of those cases are there in total?

Hint #2 So, if you fix the top and bottom numbers, you get 2 groups (that's the result of hint #1 above). In how many ways can you fix the top and bottom numbers?

  • $\begingroup$ So out of those 24 different assignments, there is a number of assignments that can't be rotated to match. So this is what I need to find with the dice? The number of assignments that no matter how I rotate I cannot make them match? $\endgroup$ – Connor Aug 18 '15 at 16:23
  • $\begingroup$ Essentially, yes. I'd suggest you play with the square example a bit to get a better feeling for what's involved. Better still, try a triangle instead of a square. It's easier, since there are only three edges and 3! = 6 possible assignments. Try and find the number of collections of squares (or triangles, if you're doing that) that are all equivalent to one another but different from everyone else. That is the number the question is asking for, in the case of the dice. $\endgroup$ – wltrup Aug 18 '15 at 16:27
  • $\begingroup$ Ok, so I did what you recommended. I literally drew out all six possibilities for a triangle, 3 of which CAN be rotated to be the same thing. I then drew out all 24 possibilities for a square, 4 out of the 24 can be rotated to be the same. $\endgroup$ – Connor Aug 18 '15 at 16:42
  • $\begingroup$ Nice, but I think you're still a little confused about what's being asked. See my solution for the triangles above. (I edited my answer to include that solution). $\endgroup$ – wltrup Aug 18 '15 at 16:48
  • $\begingroup$ Ok, so I also brute-forced for a square, and there are SIX different groups. So it's the total number of possibilities divided by the number of sides, which will equal your different types? $\endgroup$ – Connor Aug 18 '15 at 16:52

Let us sit down at a table. Whatever die we have, it can be put on the table with the $1$ face down.

There are then $5$ possibilities for the up face (only one is legal, the face opposite $1$ is always $6$, but we will not worry about that). We make a choice for the up face, like $3$, count the dice that have up face $3$ and then multiply the result by $5$.

So now we concentrate on counting the dice with down face $1$ and up face $3$. Take the smallest face that has not been mentioned yet, in this case $2$. Rotate the die, keeping it with the $1$ face on the table, until you are looking at the $2$ face. Now there remain $3$ faces. Any two orderings of the numbers on these $3$ faces gives different dice, giving a total of $3!$ dice with down face $1$ and up face $3$.

Finally, multiply by $5$.


Put your dice on a table such that the $1$ is on the bottom for all of them. You still can rotate them freely around the vertical axis passing through the single dot on the bottom. Now begins the counting: You have five choices for the top face. The remaining four numbers can be paired in three ways to form pairs of opposite numbers on the vertical faces. For each chosen pairing you can realize two different orientations leading to two cubes which are mirror copies of each other. It follows that the total number of cubes is $5\cdot 3\cdot 2=30$.

By the way: There is a famous puzzle (Macmahon's colored cubes) using $30$ colored cubes instead of numbered ones. Take any of the $30$ cubes as model and put together a $2\times2\times2$ cube looking like the model, with the extra condition that in the interior only faces with equal colors may touch.


In the case there are no limitation on the way we sort the different values (i.e. it's not a normal dice which opposite sides sum up to 7), and there's also no rotation allowed (the cube is fixed) - then the answer would be 6! = 720 options.

(n-Factorial is the number of different unique combinations you can arrange all the numbers 1,...,n in a series)

Now let's consider 1-axis rotation is allowed. You can choose 6 values for the upside face, then you would be left with 5 values for the downside face. Sorting out the 4 faces of the sides would normally be 4! but since you're allowing rotation, this needs to be divided by 4, since each no-rotation combination is equivalent to 4 with-rotation combinations (1234 = 2341 = 3412 = 4123). So the answer would be 6!/4 = 6*5*4!/4 = 180.

(Note that it does matter here which value you choose for the upper and for the lower, since if you turn them around, the dices will not be the same - but will switch between clockwise, or counter-clockwise, is. a dice with 6 on the top, 5 on the bottom, and 1234 on the sides, is different than a dice with 5 on the top, 6 on the bottom and 1234 on the sides)

Now let's consider 2-axis rotation is allowed. We would need to take our last answer, and then divide it to take into account that now we're allowed to "play" even more with the dice. We now find out that each "unique" combination in the 1-axis version is equivalent to 6 different combinations in the 2-axis version. 6,1234,5 = 5,4321,6 = 1,5264,3 = 3,4625,1 = 2,5361,4 = 4,1635,2 (the notation I'm using is upper,4-sides,lower).

So our answer will be 6!/4/6 = 6*5*(4!/4)/6 = 30.

Now what would be the answer in the case we limit ourselves to normal dices, i.e. dices that follow the pattern that each opposite sides sum must be equal to 7?

This means that we can now only freely choose 3 faces of the dice, corresponding to its 3 dimensions space, since choosing one face number to be equal to a value, immediately sets the value of the opposite side.

So for no rotation - we could choose 6 values for the upper side, but then immediately the lower side is set, so you can choose 4 value for a another face, but then the opposite face is again set, and finally you can choose 2 values for the next face, which will immediately set the final face. I.e. the answer is 6*4*2 = 48.

For 1-axis rotation - we saw that we need to divide by 4: 48/4 = 12.

For 2-axis rotation - we need to further divide by 6: 12/6 = 2.

And that actually corresponds with the two normal 6-sided dices that exist in the world, called: clockwise (Also - left handed), and counter-clockwise (Also - right handed).


Assuming that the question refers to "normal" die, where the opposite sides must total seven, (why would you make any other kind?) I believe this is a spatial problem rather than a mathematical one.

Lets begin with the 4 side facing up, it is symmetrical around both the horizontal and vertical axis so it doesn't matter how you set it down. The addition rule means that the 3 side is on the bottom but we can't see that so we ignore it for now.

Next we arbitrarily choose to place the 1 side on the vertical face to the right of the 4. It is also symmetrical so this is trivial. The addition rule dictates that the 6 side must then be placed on the vertical face to the left of the 4. At this point there are two options, the dots on the 6 form two parallel tracks of three which can either be turned as if they were a continuation of the dots on the 4 face, or as if they were 'perpendicular' to the 4 face.

We then place the 5 on the front face of the die as we have set it down. It is symmetrical so it doesn't matter how it goes on. This forces the 2 onto the back face of the die in one of only two possible configurations with the dots placed on the face top-right bottom-left or, top-left bottom-right.

If we now flip the die over and place the 3 on top, we see we have the same options as the 2, only with a dot in the center.

So each of the choices for the 6, 3 and 2 faces gives us eight ways (2x2x2)to build our die. The arbitrary choice to place the 1 on the right and the 5 on the front is akin to chirality so that there are mirror images of the eight choices given above. Thus there are a total of 16 ways you can build a 'normal' die.


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