I persecuted this ...

Added later:

1 throw with 1 astragalus : 4 possibilities (1, 3, 4, 6)

1 throw with 2 astragali: 10 possibilities (1-1, 3-3, 4-4, 6-6, 1-3, 1-4, 1-6, 3-4, 3-6, 4-6)

1 throw with 3 astragali: 20 possibilities: 1-1-1, 3-3-3, 4-4-4, 6-6-6, 1-1-3, 1-1-4, 1-1-6, 3-3-1, 3-3-4, 3-3-6, 4-4-1, 4-4-3, 4-4-6, 6-6-1, 6-6-3, 6-6-4, 1-3-4, 1-3-6, 1-4-6, 3-4-6

1 throw with 4 astragali:

1111 ... 1 dominates

--

1113 ... 1 dominates

1114

1116

--

11-3-4 ... 1 dominates

11-3-6

11-4-6

---

7x4 = 28

none dominates:

11-33

11-44

11-66

33-44

33-66

44-66

--

1-3-4-6

28+7 = totally 35 possibilities

1 throw with 5 astragali: 56 possibilities as already discussed

1 throw with 6 astragali:

111111 .. 1 dominates

--

111113

111114

111116

---

111133

111144

111166

111134

111136

111146

---

111334

111336

111443

111446

111663

111664

111346

---

4x17 = 68

111333 none dominates

111444

111666

333444

333666

444666

---

113344

113366

114466

334466

--

113346

114436

116634

334416

336614

446613

--

16 possibilities / 68 + 16 = 84

Actually ...

113344 .. is dominated by 6

113366 .. is dominated by 4

114466 .. is dominated by 3

334466 .. is dominated by 1

... so we have 4x18=72 cases of dominance and 12 cases of none dominance

****************

So get

1 astragalus - 4 possibilities

2 astragali - 10 possibilities

3 astragali - 20 possibilities

4 astragali - 35 possibilities

5 astragali - 56 possibilities

6 astragali - 84 possibilities

7 astragali - 120 possibilities

4 x 2.5 = 10

10 x 2 = 20

20 x 1.75 = 35

35 x 1.6 = 56

56 x 1.5 = 84

84 x (10/7) = 120

4 x (5/2) = 10

10 x (6/3) = 20

20 x (7/4) = 35

35 x (8/5) = 56

56 x (9/6) = 84

84 x (10/7) = 120

... so I use my private Pythagoras knowledge to predict that 8 astragali would have ...

120 x 11/8 = 165

... and 9 astragali ...

165 x 12/9 = 220

... and 10 astragali ..

220 x 13/10 = 286

... and 11 astragali ...

286 x 14/11 = 364

... and 12 astragali ...

364 x 15/12 = 455

... which all isn't proved by hand. But I think, that it would work.

Now it starts with the idea, that 1 astragali has 4 possibilities.

If I would use a die with 5 sides (it's possibly difficult to construct one) ...

... I guess, that it starts with 5 possibilities, and 2 dice of that sort would have 5 x 6/2 = 15 possibilities. And 3 dice of that sort would have 15 x 7/3 = 35 possibilities, and 4 dice of that sort would have 35 x 8/4 = 70 possibilities.

Let's see, if I can prove this for the last one: 4 dice with 5 vales a-b-c-d-e.

aaaa ... all dominated by "a"

aaab

aaac

aaad

aaae

aabc

aabd

aabe

aacd

aace

aade

bcde

12x5 = 60

none dominance cases

aabb

aacc

aadd

aaee

bbcc

bbdd

bbee

ccdd

ccee

ddee

------

10 cases

60 + 10 = 70 .... victory.

This seems to be a very nice universal tool to calculate the possibilities of all sorts of dice-systems.

Let's assume a die with 10 faces.

1 die ... 10 possibilities, that's 1 x 10/1 = 10

2 dice ... 10 x 11/2 = 55 possibilities

3 dice ... 55 x 12/3 = 220 possibilities

etc.

I'm too lazy to control that, but it should work.