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    <p>Starting from <strong>Java 7</strong> (or Android API Level 21 = 5.0+) you could directly use <code>ThreadLocalRandom.current().nextLong(n)</code> (for 0 ≤ x &lt; n) and <code>ThreadLocalRandom.current().nextLong(m, n)</code> (for m ≤ x &lt; n). See <a href="https://stackoverflow.com/a/31682930/224671">@Alex</a>'s answer for detail.</p> <hr> <p>If you are stuck with <strong>Java 6</strong> (or Android 4.x) you need to use an external library (e.g. <code>org.apache.commons.math3.random.RandomDataGenerator.getRandomGenerator().nextLong(0, n-1)</code>, see <a href="https://stackoverflow.com/a/11747812/224671">@mawaldne</a>'s answer), or implement your own <code>nextLong(n)</code>.</p> <p>According to <a href="http://java.sun.com/j2se/1.5.0/docs/api/java/util/Random.html" rel="noreferrer">http://java.sun.com/j2se/1.5.0/docs/api/java/util/Random.html</a> <code>nextInt</code> is implemented as</p> <pre><code> public int nextInt(int n) { if (n&lt;=0) throw new IllegalArgumentException("n must be positive"); if ((n &amp; -n) == n) // i.e., n is a power of 2 return (int)((n * (long)next(31)) &gt;&gt; 31); int bits, val; do { bits = next(31); val = bits % n; } while(bits - val + (n-1) &lt; 0); return val; } </code></pre> <p>So we may modify this to perform <code>nextLong</code>:</p> <pre><code>long nextLong(Random rng, long n) { // error checking and 2^x checking removed for simplicity. long bits, val; do { bits = (rng.nextLong() &lt;&lt; 1) &gt;&gt;&gt; 1; val = bits % n; } while (bits-val+(n-1) &lt; 0L); return val; } </code></pre>
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    1. POJava: random long number in 0 <= x < n range
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    1. COI'm having some problems with "2^x checking" part. Any ideas?
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      1. USVilius Normantas
    2. CO@Vilius: The 2^x checking just makes the generation faster because directly using `rng.nextLong() % n` will be give uniform values (assume all bits are good). You can ignore that part if you want.
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