Can You Stick It To The Genie?

You roll three dice in order: a $d4$ (faces $1$–$4$), a $d6$ (faces $1$–$6$), then a $d8$ (faces $1$–$8$), each face equally likely. You win when the three rolls strictly increase, like $2$-$4$-$7$. What is your probability of winning? Extra credit: rolling $d4, d6, d8, d10, d12, d20$ in order, what is the chance the six rolls strictly increase?

Solution

Win when $a < b < c$ with $a$ on the $d4$, $b$ on the $d6$, $c$ on the $d8$. The rolls are independent and uniform, so the probability is the count of strictly increasing triples divided by $4 \cdot 6 \cdot 8 = 192$. Count them by fixing the first two and letting the last run free: for each $a \le 4$ and each $b$ with $a < b \le 6$, the top die can be any of the $8-b$ values above $b$. Summing, $$\sum_{a=1}^{4} \sum_{b=a+1}^{6} (8-b) = 48, \qquad \Pr(\text{win}) = \frac{48}{192} = \boxed{\tfrac14}.$$ The clean answer is a small coincidence of these particular face counts, not a general law. With six dice the same idea (count increasing sextuples, divide by $4\cdot6\cdot8\cdot10\cdot12\cdot20$) has no tidy closed form; the exact value is $$\Pr(\text{six increase}) = \frac{2717}{230400} \approx \boxed{0.0118}.$$

The computation

Enumerate every outcome of the dice in order and count the fraction that strictly increase.

from fractions import Fraction as F
from itertools import product
def prob(faces):
    rolls = product(*[range(1, f + 1) for f in faces])
    wins = sum(all(x < y for x, y in zip(r, r[1:])) for r in rolls)
    total = 1
    for f in faces: total *= f
    return F(wins, total)
print(prob([4, 6, 8]))                       # 1/4
print(prob([4, 6, 8, 10, 12, 20]), float(prob([4, 6, 8, 10, 12, 20])))
# 2717/230400, 0.0118