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copied!<p>EDIT: I've only just noticed this is LINQ to SQL, not LINQ to Objects. Use Marc's code to get the database to do this for you. I've left this answer here as a potential point of interest for LINQ to Objects.</p> <p>Strangely enough, you don't actually need to get the count. You do, however, need to fetch every element unless you get the count.</p> <p>What you can do is keep the idea of a "current" value and the current count. When you fetch the next value, take a random number and replace the "current" with "new" with a probability of 1/n where n is the count.</p> <p>So when you read the first value, you <em>always</em> make that the "current" value. When you read the second value, you <em>might</em> make that the current value (probability 1/2). When you read the third value, you <em>might</em> make that the current value (probability 1/3) etc. When you've run out of data, the current value is a random one out of all the ones you read, with uniform probability.</p> <p>To apply that with a condition, just ignore anything which doesn't meet the condition. The easiest way to do that is to only consider the "matching" sequence to start with, by applying a Where clause first.</p> <p>Here's a quick implementation. I <em>think</em> it's okay...</p> <pre><code>public static T RandomElement<T>(this IEnumerable<T> source, Random rng) { T current = default(T); int count = 0; foreach (T element in source) { count++; if (rng.Next(count) == 0) { current = element; } } if (count == 0) { throw new InvalidOperationException("Sequence was empty"); } return current; } </code></pre>

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