Cascade Rounding

How do you round? Consider the number $3.5$. We wish to round to the nearest integer. Notice that $3.5$ is equidistant between $3.0$ and $4.0$. Do we round up or do we round down? We need a tie breaking function.

We define the "round half up" tie-breaking function: $$\lfloor x + \frac{1}{2} \rfloor$$ We can watch what happens to $3.5$: $$\lfloor 3.5 + 0.5 \rfloor = 4 $$

What happens if we want to round to the nearest tenth, or hundredth, or thousandth? We can define a function for any number of digits $n$: $$R_n(x) = \dfrac{\lfloor 10^n x + \frac{1}{2} \rfloor}{10^n}$$ thus, $$3.549 \xrightarrow{R_1} 3.5$$ or, $$3.549 \xrightarrow{R_2} 3.55 \xrightarrow{R_1} 3.6$$

But what if I wanted to round from right to left? We can define this as cascade rounding. $$C_{m \to n}(x) = R_n(R_{n+1}(\cdots R_m(x) \cdots))$$ where $m$ is the digit we start rounding from and $n$ is the digit we finish rounding on.

How do these two methods disagree? They can only ever differ by $10^{-n}$, but how often does that occur?

Consider $3.539 \xrightarrow{R_1} 3.5$ and $3.539 \xrightarrow{C_{2 \to 1}} 3.5$. These values agree. But replace the $3$ in the hundredths place with a $4$ and notice what happens: $$3.549 \xrightarrow{R_1} 3.5 \quad \text{but} \quad 3.549 \xrightarrow{C_{2 \to 1}} 3.6$$ The key issue is the digit $4$: it is too small to round up on its own, but a carry from the right can push it to $5$. A $3$ would absorb any carry without rounding up; a $5$ would round up regardless. Only a $4$ creates a discrepancy. If we assume the digits are i.i.d. and uniformly distributed ^[1], then $$\Pr[d_2 = 4] = \frac{1}{10}$$

How often does a carry survive? A carry can either live or die. We can define both cases:

Digit $= 5, 6, 7, 8, 9$: carry born or kept alive
Digit $= 0, 1, 2, 3, 4$: carry dies, except that a $4$ passes the carry through to the next digit

Let $p$ be the probability that the carry survives to the end of the cascade ^[2] as $m \to \infty$, then $$p = \frac{5}{10} + \frac{p}{10} \implies p = \frac{5}{9}$$ where $\frac{5}{10}$ is the probability that a carry is born, and $\frac{p}{10}$ is the chance that a carry continues.

Finally we can put together these two independent events: $$\boxed{\Pr[\text{disagreement}] = \frac{1}{10} \cdot \frac{5}{9} = \frac{1}{18} \approx 5.6\%.}$$

The i.i.d. uniform digit assumption holds theoretically but fails for most real data, since financial figures, measurements, and discrete datasets all have non-uniform decimal digit distributions. We can treat 1/18 as a theoretical baseline, not a practical error rate. ↩
This can be proven using a Markov chain. ↩