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    <p>Python comes with a <a href="http://docs.python.org/2/library/" rel="nofollow">great standard library</a> (as you might already discovered since you are utilizing <em>itertools</em>), in which you'll also find a <a href="http://docs.python.org/2/library/random.html" rel="nofollow">random</a> module.</p> <p>You could use <a href="http://docs.python.org/2/library/random.html#random.randint" rel="nofollow">random.randint</a> to simulate dice rolls. There are multiple ways to solve this. The first code example is somewhat limited, the second is more general.</p> <pre><code>import random # '_' (underscore) is used for values that are generated, but that you do not # care about - here we only want to repeat seven times and do not care about # the actual loop count for _ in range(7): # generate three random numbers between [1 and 6] # and store the values in a, b, c respectively (tuple unpacking) a, b, c = (random.randint(1, 6) for _ in range(3)) # if one of the conditions holds, break out of the loop early if a == b or a == c or b == c or a == b == c: break </code></pre> <p>As @Paulo points out, you can use another <strong>more concise</strong> approach to check, whether the <code>n</code> elements of a list (or tuple) are all different, that is you put all elements in a <a href="http://docs.python.org/2/library/stdtypes.html#set" rel="nofollow">set</a>:</p> <pre><code>for _ in range(7): # instead of three separate variables, we use a list here # the nice thing is, that you can freely vary the number of # 'parallel' dice rolls this way dice = [random.randint(1, 6) for _ in range(3)] # this is more general and will break as soon as there are # duplicate (die) values in the list (meaning, break, if not all elements # are different) if len(dice) &gt; len(set(dice)): break </code></pre> <p>Answering your updated question, just use <a href="http://docs.python.org/2/library/functions.html#sum" rel="nofollow"><code>sum</code></a>:</p> <pre><code> total = sum(dice) </code></pre>
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    1. COIt's `random.randint(1, 6)`.
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    2. CO@PauloAlmeida, thanks, corrected (overgeneralization from slice semantic ;)
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    3. COYeah, I figured :)
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