How do I randomly choose a number in a smaller range than what my random number generator produces?

Question

My uniform random number generator can only produce integers in the range $0 \leq n \leq 255$. I need to use the output from this random number generator to generate another uniformly random integer in the range $0 \leq m \leq 209$. How can I do this?

Context: I need to randomly select an item from a list of 210 items using a random number supplied to me in the form of 8 random binary digits. I cannot change the range of the random number supplied to me.

EDIT: Some of the helpful comments suggest that I should use "rejection sampling". However, I would prefer a solution that has a guarantee on its maximum running time when implemented in a computer program.

You can keep generating random numbers from 0 to 255 until you get a value between 0 and 209, and take this value. — Minus One-Twelfth, Oct 10 '21 at 09:12
You can use rejection sampling for this , as the above comment says. You should try to prove that the above procedure works. — Sarvesh Ravichandran Iyer, Oct 10 '21 at 09:48
Alternative approach, if feasible is to multiply the generated number by the factor of $\frac{209}{255}$ and then round the product to the nearest integer. The problem with this alternate approach is that (for example) $254$ will round down to $208$. Therefore, your chance of having the computed random number be $209$ is $\frac{1}{256},~$ rather than $\frac{1}{210}$. Personally, given that you insist on working with integers rather than real numbers, if you require that the random integers be unbiased, then I see no alternative to the already suggested rejection sampling. — user2661923, Oct 10 '21 at 10:55
If the priority between likelihood of avoiding a rejection and reducing bias is not so clear cut, there are compromises. For example, the largest mutiple of $21$ that is $\leq 256$ is $252$. Therefore, you could accept all numbers from ${0,1,\cdots, 251}$, and compute from the selected number an unbiased equivalence class $\pmod{21}$, and (simultaneously) a very slightly biased equivalence class $\pmod{10}$. So, by introducing a small amount of bias, you have changed the rejection interval from $[210, 255]$ to $[252,255].$ — user2661923, Oct 10 '21 at 11:05
Mathematically, you cannot produce an exact probability of $1/210$ using only a probability space whose elements have probability $1/2^n$ for some integer $n$, which is usually what you get from pseudorandom number generators on a computer. What most people do is to use a RNG that has a much larger probability space (range $0 \leq n \leq 2^{32}-1,$ typically) and divide it into $210$ almost-equal parts. You could do that too by running your RNG four times to build up a $32$-bit number. — David K, Oct 10 '21 at 15:44
You can also do an exact, "lazy" version of @DavidK's suggestion: interpret your random bits as the binary representation of a real number $x$ in $[0,1]$, and stop sampling when you have enough information to determine which one of the 210 equally-sized subintervals contains $x$. — Karl, May 03 '23 at 05:11
Intuitively I think it can't be done exactly in finite steps; It's the same argument as for getting 1/3 odds out of coin flips. Couldn't find a proof on the web. There is a nice algorithm used in CS all the time which samples a larger random integer (let's say 31 bits) and takes that value modulo 210, rejecting only when value is contained in the very last interval [210k, 210(k+1)-1], which is incomplete. It's used in the Java implementation of Random.nextInt(bound) — Pedro Pedruzzi, May 25 '25 at 15:26
There are succinct arguments why you cannot achieve a perfectly uniform distribution with an algorithm that runs in bounded time. See the answers to Uniform distribution on ${1,\dots,7}$ from rolling a die and Can a set of 6 sided die, simulate any other die?. You can get a distribution that is as near uniform as you want simply by planning in advance to roll a fixed number of times, whatever number you want. — David K, May 25 '25 at 20:02
This question is similar to: How to generate a random number between 1 and N with an M sided die. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — David K, May 25 '25 at 20:23
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whoisit · Answer 1 · 2025-05-25T23:46:01.533

(Summarizing the discussion from the comments with added quantitative insights)

Within the given constraints, achieving both - an exactly uniform distribution as well as deterministically finite runtime, is impossible. We can tradeoff one for the other.

Method 1- Rejection Sampling

This method achieves exactly uniform distribution but does not have a deterministic runtime. The idea is, generate $n \in [0, 255]$. If $n \geq 210$, repeat the process; otherwise return the value.

Let $T$ be the number of trials, then the distribution follows: $$\Pr[T>k]=\left(1−\frac{210}{256}\right)^k $$

This gives, $\mathbb{E}[T] \approx 1.22$, and $\Pr[T > 5] \approx 0.02\% $ which is extremely low.
$\Pr[T > 120]$ is much lesser than $1 \over {\text{# atoms in the universe}}$

Method 2- Multiply-and-floor

Each trial of the rng gives 1 byte. Draw $k$ independent bytes, and concat to get the integer $X$. Then, consider $$ m = \left\lfloor X \times \dfrac{210}{2^{8k}} \right\rfloor \in [0, 209] $$

This runs in exactly $k$ draws, but is slightly biased because $2^{8k}$ is never a multiple of 210.

Let's quantify the bias. Exactly $ R_k = 2^{8k}\bmod 210$ of the 210 output-values get a slightly higher probability. Let's call this set $Q_k$. We have, $$ \Pr[m=i] = \begin{cases} \dfrac{\lfloor2^{8k}/210\rfloor+1}{2^{8k}} & i\in Q_k \\ \dfrac{\lfloor2^{8k}/210\rfloor}{2^{8k}} & \text{otherwise}. \end{cases} $$

Maximum bias, from the expected $1/210$ is then approx $2^{-8k}$.
With just $k=1$, the bias is already less than $0.5\%$.

This is a well-organized and thoughtful answer, but it means we now have $16$ very similar questions with similar answers instead of just $15$. — David K, May 26 '25 at 01:51

How do I randomly choose a number in a smaller range than what my random number generator produces?

1 Answers1

Method 1- Rejection Sampling

Method 2- Multiply-and-floor