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    <p>The problem is that you're doing a modulo operation. This would be no problem if <code>RAND_MAX</code> would be evenly divisible by your modulus, but usually that is not the case. As a very contrived example, assume <code>RAND_MAX</code> to be 11 and your modulus to be 3. You'll get the following possible random numbers and the following resulting remainders:</p> <pre><code>0 1 2 3 4 5 6 7 8 9 10 0 1 2 0 1 2 0 1 2 0 1 </code></pre> <p>As you can see, 0 and 1 are slightly more probable than 2.</p> <p>One option to solve this is rejection sampling: By disallowing the numbers 9 and 10 above you can cause the resulting distribution to be uniform again. The tricky part is figuring out how to do so efficiently. A very nice example (one that took me two days to understand <em>why</em> it works) can be found in Java's <a href="http://docs.oracle.com/javase/7/docs/api/java/util/Random.html#nextInt%28int%29" rel="nofollow noreferrer"><code>java.util.Random.nextInt(int)</code></a> method.</p> <p>The reason why Java's algorithm is a little tricky is that they avoid slow operations like multiplication and division for the check. If you don't care too much you can also do it the naïve way:</p> <pre><code>int n = (int)(max - min + 1); int remainder = RAND_MAX % n; int x, output; do { x = rand(); output = x % n; } while (x &gt;= RAND_MAX - remainder); return min + output; </code></pre> <p><strong>EDIT:</strong> Corrected a fencepost error in above code, now it works as it should. I also created a little sample program (C#; taking a uniform PRNG for numbers between 0 and 15 and constructing a PRNG for numbers between 0 and 6 from it via various ways):</p> <pre><code>using System; class Rand { static Random r = new Random(); static int Rand16() { return r.Next(16); } static int Rand7Naive() { return Rand16() % 7; } static int Rand7Float() { return (int)(Rand16() / 16.0 * 7); } // corrected static int Rand7RejectionNaive() { int n = 7, remainder = 16 % n, x, output; do { x = Rand16(); output = x % n; } while (x &gt;= 16 - remainder); return output; } // adapted to fit the constraints of this example static int Rand7RejectionJava() { int n = 7, x, output; do { x = Rand16(); output = x % n; } while (x - output + 6 &gt; 15); return output; } static void Test(Func&lt;int&gt; rand, string name) { var buckets = new int[7]; for (int i = 0; i &lt; 10000000; i++) buckets[rand()]++; Console.WriteLine(name); for (int i = 0; i &lt; 7; i++) Console.WriteLine("{0}\t{1}", i, buckets[i]); } static void Main() { Test(Rand7Naive, "Rand7Naive"); Test(Rand7Float, "Rand7Float"); Test(Rand7RejectionNaive, "Rand7RejectionNaive"); } } </code></pre> <p>The result is as follows (pasted into Excel and added conditional coloring of cells so that differences are more apparent):</p> <p><img src="https://i.stack.imgur.com/HKeOl.png" alt="enter image description here"></p> <p>Now that I fixed my mistake in above rejection sampling it works as it should (before it would bias 0). As you can see, the float method isn't perfect at all, it just distributes the biased numbers differently.</p>
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    1. POWhat is the optimal algorithm for generating an unbiased random integer within a range?
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    1. CODo you think that an obvious floating-point-based approach aimed to map the RAND_MAX range to a custom range (almost) without branching would be less CPU efficient than multiple divisions (`%`) and branches implied by the `while` loop?
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      1. USDesmond Hume
    2. COIf you use `(double)rand() / RAND_MAX * n` instead you get the same problem, it's just that it's not the lower numbers that are more likely but that you distribute the bias somewhat over the whole range. You don't remove the bias at all using that method. You still have the problem of evenly fitting 10 input numbers into 3 output numbers, which isn't possible.
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    3. COI do not understand the phrase `distribute the bias somewhat over the whole range` -- appears self-contradictory. Also, ITYM `... evenly fitting *11* input numbers into 3 output numbers ...`
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      1. USHappy Green Kid Naps
 

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