How Long Is the River?

In the language Fiddlish every word is three or four letters long (each equally likely), and words are separated by single spaces. Deep into a long line of text, what is the probability that a given character is a space?

The Fiddler, Zach Wissner-Gross, May 23, 2025(original post)

Solution

Each word contributes, on average, $3.5$ letters plus its one trailing space: $4.5$ characters per word, exactly one of them a space. By the renewal theorem the long-run fraction of spaces is $$\frac{1}{4.5}=\boxed{\frac29}\approx22.2\%.$$

The computation

Generate the actual stream: draw millions of words of three or four letters, append one space to each, and take the fraction of all characters that are spaces.

import numpy as np
rng = np.random.default_rng(0)
words = rng.integers(3, 5, 5_000_000)        # 3 or 4 letters, equally likely
spaces = len(words); total = int((words + 1).sum())   # +1 space per word
print(round(spaces / total, 4))              # 0.2222  = 2/9

Extra Credit

In a monospace font, suppose the $12$th character of a line is a space. A “river” runs diagonally down and to the right: line $2$ continues it if its $13$th character is a space, line $3$ if its $14$th is a space, and so on. How long is the river on average?

Solution

Different lines are independent, so the river continues to each next line with the probability that that line has a space at the relevant position. Crucially those positions ($13,14,15,\dots$) are near the start of the line, where the space probability has not yet settled to $\tfrac29$. Writing $u_q$ for the renewal probability of a space at position $q$ (steps of $4$ or $5$, each $\tfrac12$), $$\mathbb{E}[\text{length}]=1+\sum_{m\ge1}\prod_{j=13}^{12+m}u_j\approx\boxed{1.535}.$$ (The source’s value is paywalled; this is my own, with $u_{13}=u_{14}=\tfrac38$ well above $\tfrac29$.)

The computation

Build the renewal probability $u_q$ that position $q$ is a space (each space sits $4$ or $5$ characters past the previous, equally likely), then accumulate the river’s expected length as the running product of continuation probabilities from position $13$ on.

from fractions import Fraction as F
u = [F(0)] * 160; u[0] = F(1)
for q in range(1, 160):
    u[q] = (u[q - 4] if q >= 4 else 0) / 2 + (u[q - 5] if q >= 5 else 0) / 2
E = F(1); prod = F(1)
for j in range(13, 160): prod *= u[j]; E += prod
print(float(E))                          # 1.5347