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    <p>This isn't an answer to your question (I'm not actually sure what you're asking), but I will point out that your shuffling algorithm is not random.</p> <p>First, using <code>%</code> to limit <code>rand</code> to a range of values is usually a bad idea. (See <a href="http://c-faq.com/lib/randrange.html" rel="nofollow noreferrer">Q13.16 from the comp.lang.c FAQ</a>.)</p> <p>Second, it looks like you might be trying to implement the <a href="http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle" rel="nofollow noreferrer">Fisher-Yates shuffling algorithm</a>, but your implementation is wrong. It's easy to see that your approach of swapping with previously-swapped elements is problematic; consider shuffling a deck of three elements { 0, 1, 2 }:</p> <pre><code>First iteration Second iteration ------------------------------------ { 0, 1, 2 } { 1, 0, 2 } { 0, 1, 2 } { 0, 2, 1 } ------------------------------------ { 1, 0, 2 } { 0, 1, 2 } { 1, 0, 2 } { 1, 2, 0 } ------------------------------------ { 2, 1, 0 } { 1, 2, 0 } { 2, 1, 0 } { 2, 0, 1 } </code></pre> <p>The first column represents the possible states of the deck after the first iteration (where you swap <code>deck[0]</code> with <code>deck[rand() % 3]</code>).</p> <p>The second column represents the possible states of the deck after the second iteration (where you swap <code>deck[1]</code> with <code>deck[rand() % 3]</code>). </p> <p>There are 9 equally probable states---but this is obviously wrong since there should be only 3! = 6 permutations. And indeed, several states are duplicated, which means that they have a higher probability of occurring. (Note that even if I had proceeded to the third iteration, you'd end up with 27 states for only 6 permutations, and since 27 is not divisible by 6, you'd clearly end up with some states occurring more than others.)</p>
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