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    <p><strong>Option 1: zip, shuffle, unzip</strong></p> <p>To expand on Marcelo's comment, and assuming you don't mind creating <em>new</em> lists instead of shuffling the <em>existing</em> ones:</p> <pre><code>var zipped = listA.Zip(listB, (a, b) =&gt; new { a, b } ).ToList(); Shuffle(zipped); var newListA = zipped.Select(pair =&gt; pair.a).ToList(); var newListB = zipped.Select(pair =&gt; pair.b).ToList(); </code></pre> <p><strong>Option 2: shuffle the indexes</strong></p> <p>To expand on MAK's answer with code:</p> <pre><code>var indexes = Enumerable.Range(0, listA.Count).ToList(); Shuffle(indexes); var newListA = indexes.Select(index =&gt; listA[index]).ToList(); var newListB = indexes.Select(index =&gt; listB[index]).ToList(); </code></pre> <p>Of course both approaches <em>could</em> mutate the original lists, with a bit more work.</p> <p><strong>Option 3: shuffle both lists with the same random seed</strong></p> <p>I personally like to pass a <code>Random</code> (or whatever) into methods/classes which need them instead of creating new ones. So I'd give my <code>Shuffle</code> a <code>Random</code> parameter. It avoids various problems, and expresses the dependency nicely. You can use this to your advantage, by creating two <code>Random</code> instances with the same seed:</p> <pre><code>int seed = existingRandom.Next(); Shuffle(listA, new Random(seed)); Shuffle(listB, new Random(seed)); </code></pre> <p>Assuming <code>Shuffle</code> does the same thing when given the same sequence of random numbers, this will shuffle both lists in the same way.</p>
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    1. CO+1, I was hoping someone would be less lazy than I.
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    2. CO@MarceloCantos: While I was hoping you wouldn't mind me using your idea :)
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    3. COdon't you mind Shuffle(indexes);? ;)
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