The blue shapes have isosceles triangle faces, and so the monohedral bipyramid on top has two planes of reflective symmetry, and all faces are congruent with no reflections needed. The red ones have scalene faces, and so the monohedral bipyramid on top has only one plane of reflective symmetry, and some of the faces are mirror-images of other faces.
(It’s not as obvious as I would have liked that the red faces are scalene, because it turns out that it’s really easy to make the resulting bipyramid non-convex)
How about one bipyramid with vertices (0,0,0), (10,0,0), (5,12,0), and (5, 47/6, \pm5\sqrt{119}/6), and another with vertices (0,0,0), (13,0,0), (99/13,168/13,0), and (70/13, 171/26, \pm3\sqrt{55}/2) ? I tried to keep the numbers more or less simple.
Oh, back to the question of fair (or at least, as fair as possible) dice that aren’t isohedra: The goal isn’t to make each face subtend the same solid angle. I recall a story of a student who had come up with a shape with some peculiar number of faces, where each face did subtend the same solid angle, and declared to their professor that it would thus be a fair die. And in the process, the student set the model down on the teacher’s desk, and it promptly flopped over onto a different side, because the side it was set down on wasn’t even in equilibrium.
To the extent that a fair non-isohedral die can be quantified, what you’d want is for the basin of attraction of each face to subtend the same solid angle. That would at least mean that if it were rolled on a perfectly-dissipative surface, it’d be fair.
Is there a calculator online for this? DPRK threw a five-sided dice 312 times and got the results as given in that post. Is this result something that one can reasonably expect if the dice is fair?
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