Can You Bat .299 In 299 Games?

Riddler Express

You draw three coins, each independently a penny, nickel, dime or quarter ($1, 5, 10, 25$ cents) with equal chance, and buy strings of those lengths. What is the probability the three strings can form a triangle?

Solution

Three lengths form a triangle exactly when the longest is shorter than the sum of the other two. With ordered draws there are $4^3 = 64$ equally likely triples; checking the triangle inequality, $22$ of them qualify, so $$\Pr(\text{triangle}) = \frac{22}{64} = \boxed{\tfrac{11}{32} \approx 0.344}.$$ The quarter is the spoiler: at $25$ it dwarfs the other coins, so any triple containing a quarter needs two more sizeable coins to reach past it, and most do not.

The computation

Enumerate all $64$ coin triples and count those satisfying the triangle inequality.

from fractions import Fraction as F
from itertools import product
coins = [1, 5, 10, 25]
good = sum(x + y > z and x + z > y and y + z > x
           for x, y, z in product(coins, repeat=3))
print(F(good, 64))                             # 11/32

Riddler Classic

A player has four at-bats per game, so after $g$ games their average is $h/(4g)$ for some whole number of hits $h$. For many $g$ there is an $h$ making the rounded average (to three decimals) equal $g/1000$. What is the largest $g$ for which no such $h$ exists?

Solution

For a given number of games $g$, you can hit anywhere from $0$ to $4g$ times, giving averages $0, \tfrac{1}{4g}, \tfrac{2}{4g}, \dots, 1$ in steps of $\tfrac{1}{4g}$. A matching season needs one of these, rounded to three places, to equal $g/1000$. When $g$ is large the steps $\tfrac{1}{4g}$ are tiny and some hit count always rounds correctly, but for small $g$ the gaps are coarse enough to skip the target. Scanning downwards, the largest $g$ with no workable hit count is $$\boxed{239.}$$ With $239$ games the available averages step by $\tfrac{1}{956}$, and none of them rounds to $0.239$.

The computation

For each $g$ from large to small, test every possible hit count and stop at the first $g$ for which none rounds to $g/1000$.

for g in range(1000, 1, -1):
    if all(round(h / (4 * g), 3) != g / 1000 for h in range(4 * g + 1)):
        break
print(g)                                       # 239