Exercise 1 solution

from itertools import product, combinations_with_replacement
from collections import Counter
from fractions import Fraction

def flower(genotype):
    return {0: "white", 1: "pink", 2: "red"}[genotype.count("R")]

for p1, p2 in combinations_with_replacement(["RR", "Rr", "rr"], 2):
    c = Counter(flower(a + b) for a, b in product(p1, p2))
    total = sum(c.values())
    print(p1, "x", p2, {k: str(Fraction(v, total)) for k, v in sorted(c.items())})

Output:

RR x RR {'red': '1'}
RR x Rr {'pink': '1/2', 'red': '1/2'}
RR x rr {'pink': '1'}
Rr x Rr {'pink': '1/2', 'red': '1/4', 'white': '1/4'}
Rr x rr {'pink': '1/2', 'white': '1/2'}
rr x rr {'white': '1'}

Reading it: combinations_with_replacement gives each unordered pair of parent genotypes once (6 pairs). Because each genotype has its own look, the genotype ratio and the look ratio are the same. Only one pair, Rr x Rr, gives all three looks, in the ratio 1/4, 1/2, 1/4.