I could write “5” on the base, but if it lands on the base you will not see the figure…
That brings us back to the question raised in the OP. It’s not a regular or uniform polyhedron, to be sure. I am not claiming, by the way, that it is impossible to manipulate the roll of such a die, at least as far as the base vs one of the other 4 faces, but I tried to cast it spinning randomly.
Allthe tetrahedral dice I’ve seen have a number written along each side so that you can tell which side it’s resting on. No reason you can’t do the same.
I’ve also seen d4s where the numbers are arranged oppositely from that, with the number that counts being the uppermost instead of the lowermost, as well as d4s that have rounded corners with digits directly on the corners.
The commonality being “which digits are oriented ‘correctly’ for upright reading?”.
OK, that’s just weird.
ETA: I did find a vendor with these: “Truncated 4-sided dice”.
https://www.tarquingroup.com/media/catalog/product/cache/1/image/9df78eab33525d08d6e5fb8d27136e95/K/H/KH-S5Q7-SW1G_l_2.jpg
Anyway, that makes it technically an octahedron where you hope it doesn’t land on one of the smaller faces.
You can also make the truncations rounded enough that it can’t land on them, at least not stably. This is easier if the numbers aren’t indented, but drawn on (e.g., with a marker).
I saw this question many years ago in the British magazine Games & Puzzles.
The answer I came up with was a height of 1, the same as the length of the base.
I made one out of cardboard to test it out. Rather than rolling it I tossed it up, allowing it to fall and bounce on the carpet. It landed on the base 19% of the time, which I thought was a pretty good result.
My die was solid. I assume in your case it is a light shell? How did you derive H/B=1 in that case?
Yes, my model was a light shell rather than a solid.
I don’t remember how I did my calculation that led to H/B=1. It was something like
This was over forty years ago, so my memory is a bit rusty.
OK, but according to the theory in one of the above links, the sphere should be centered on the center of mass of the object (your thin shell), which will not coincide with the circumscribed sphere. Or am I missing something? The idea/approximation is that the model will be randomly spinning around the centroid.
Good point about the circumscribed sphere needing to be centred on the centre of mass of the pyramid.
But according to my quick calculation, they do coincide: the centre of mass of a square pyramid is one quarter of the distance from the base to the apex. For a pyramid with a height of 1, the centre of mass is a distance of 3/4 from each vertex, so it is also the centre of the circumscribed sphere. For any other height, they do not coincide. This makes me feel a lot more confident in my answer of H/B=1.
Make the die be a regular icosahedron (which is one with 20 even sides). Put the number 1 on four sides, the number 2 on four sides, the number 3 on four sides, the number 4 on four sides, and the number 5 on four sides. The five numbers will then come up the same proportion of the time when you throw that die.
This has already been suggested - twice - see posts #13 and #28. It’s also been suggested a couple of times to use a 10 sided die numbered 1-5 1-5.
Both of which are perfectly good solutions if you want a common die shape with equal probabilities of getting a result from 1-5, but that’s not what the OP was asking about. The interesting bit is making a true 5-sided die with equal probabilities, which requires some unusual shapes and experiments, some of them by our very own @DPRK (see post #35 and succeeding).
My apologies for not read the thread more carefully.
As an aside, it’s been mentioned that for a truly fair die, it’s not enough that the sides all be congruent; it also has to be isohedral, meaning that the shape can be rotated or reflected to match up any side with any other. But it turns out that it’s very difficult to come up with a convex shape that has all faces congruent but which isn’t isohedral. I’ve only managed to find one (well, one family; there are a couple of free parameters to it)-- Can anyone think of any others?
But, let’s say your centre of mass is at height H', and let x = H'/B. If you centre the sphere at height H', the angle subtended by the base is 4\sin^{-1}\left(\frac{1}{1+4x^2}\right). One-fifth of a sphere is 4\pi/5, so we can solve for x = 0.41872. So it seems that your H/B=1 pyramid is too low?
So you want convex dice that have all the faces congruent but are not symmetrical? There are some encylopedia of polyhedra type book where you can look them up; e.g
…and many more…
Hm, yes, those’ll do (though I think that last one is isohedral-- It’s the deltoidal icositetrahedron, a Catalan solid). The one I was thinking of was to construct a skewed tetrahedron from four congruent acute triangles, and then to glue it face-to-face with its mirror image.
The images are all linked to Wikipedia; I did not double-check the rendering, but some bits should basically be rotated with respect to the other one:
deltoidal icositetrahedron (octahedral symmetry)
pseudo-deltoidal icositetrahedron (dihedral symmetry)
Wikipedia also handily lists the symmetry group of each polyhedron, which is what we are interested in. This post on Stackexchange lists more known examples; in your own example, you are not counting reflections, though? You said it’s just a tetrahedron glued to its own mirror image.
My example (which doesn’t seem to be in that StackExchange post) has some faces which are mirror-images of each other, but still congruent. Though if you don’t want mirror-image faces, you could use an isosceles triangle as the face, and still avoid isohedrality.
Okay, for equalising the angles, H/B=1 is too low. But again, I don’t actually recall what it was I was trying to equalise. Still, my empirical result of 19% (with a cardboard shell rather than a solid) gives me some confidence that I had a reasonable solution.
If I understood your gluing, the remaining figure has only the one reflection symmetry remaining?
I don’t believe the Stackexchange post was meant to be any kind of comprehensive catalogue. Though there is probably one online somewhere, if we can find it.
Seems like more testing is in store. Cardboard, you say?