Actually, the cast consists of thrice dividing a heap of stalks in two and reducing each of the two resulting piles mod 4 by removing stalks 4 at a time until only 1, 2, 3, or 4 stalks remain.
One begins with 50 stalks but removes one at the start, which plays no further role.
In the first of the three casts the heap of 49 stalks is divided into two piles "at random" (more on this below). One stalk is taken from the right heap and held in the left hand between fingers 4 and 3. Then the sum of the numbers of stalks in the piles, T1 and T2, is 48, which is divisable by 4.
We reduce T1 mod 4, leaving (by convention) a remainder of 1, 2, 3, or 4: R1. Ditto T2, even though this remainder, R2, may be predicted, as R1 + R2 must be 4 or 8. The traditional rule then is to add the remainders, S = R1 + R2, so the four equiprobable remainders, R1 = 1, 2, 3, or 4, result in the sums, S = 4, 4, 4, and 8, respectively. These sums are then converted to scores, 3, 3, 3, or 2, respectively. Hence the equivalence to casting a tetrahedron.
The second of the three casts is done with the heap of cast-out stalks, which number either 40 or 44, divisible by 4 in either case. This time the one stalk taken from the right heap after the random division into two piles is counted along with the two remainders. So we get 1 + S = 1 + R1 + R2 = 4, 4, 4, or 8, with scores of 3, 3, 3, or 2, again, equiprobably. So this second cast is also equivalent to throwing a perfect tetrahedral die with scores of 3, 3, 3, and 2 on its faces.
The third yarrow cast begins with the heap of cast-out stalks, now numbering 32, 36, or 40. Again, it is arithmetically equivalent to casting the same tetrahedral die.
But the manipulation of the stalks in the full ritual sequence is fundamental to the success of the oracle. So will simulate it entire, making use of chaos theory (the chaotic attractor of a two-dimensional map) to create a screen image of a random heap, which then may be divided by a click of the mouse in the midst of the heap. Which map to use is up to the seeker, and we give a menu of a few of the best known maps from chaos theory.
NB: It is completely uncool, although perhaps legal, to divide a heap so that either pile has 4 or less stalks, as then the outcome is already determined. In fact, an experienced yarrowmancer can divide a heap so that either score (4 or 8) will result, even if 8 or more stalks remain in the smaller dividend pile. Thus to be fair, in case of a heap of 32 stalks (ie, after two casts of 8), the good divides are only at: 10, 11, 12, 13, 14, 15, or 16 stalks in the smaller pile. One must select "at random" a number from the set {10, ..., 22} or so, keeping toward the middle most of the time. Thus "at random" is to be interpreted in terms of some bell-shaped curve, and this is where chaos theory comes into our simulation.