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copied!<p>A little bit of math here.</p> <p>A regular die will give each number 1-6 with equal probability, namely <code>1/6</code>. This is referred to as <a href="http://en.wikipedia.org/wiki/Uniform_distribution" rel="noreferrer">uniform distribution</a> (the discrete version of it, as opposed to the continuous version). Meaning that if <code>X</code> is a random variable describing the result of a single role then <code>X~U[1,6]</code> - meaning <code>X</code> is distributed equally against all possible results of the die roll, 1 through 6.</p> <p>This is equal to choosing a number in <code>[0,1)</code> while dividing it into 6 sections: <code>[0,1/6)</code>, <code>[1/6,2/6)</code>, <code>[2/6,3/6)</code>, <code>[3/6,4/6)</code>, <code>[4/6,5/6)</code>, <code>[5/6,1)</code>.</p> <p>You are requesting a different distribution, which is biased. The easiest way to achieve this is to divide the section <code>[0,1)</code> to 6 parts depending on the bias you want. So in your case you would want to divide it into the following: <code>[0,0.2)</code>, <code>[0.2,0.4)</code>, <code>[0.4,0.55)</code>, <code>0.55,0.7)</code>, <code>[0.7,0.84)</code>, <code>[0.84,1)</code>.</p> <p>If you take a look at the <a href="http://en.wikipedia.org/wiki/Uniform_distribution" rel="noreferrer">wikipedia entry</a>, you will see that in this case, the cumulative probability function will not be composed of 6 equal-length parts but rather of 6 parts which differ in length according to the <em>bias</em> you gave them. Same goes for the mass distribution.</p> <p>Back to the question, depending on the language you are using, just translate this back to your die roll. In Python, here is a very sketchy, albeit working, example:</p> <pre><code>import random sampleMassDist = (0.2, 0.1, 0.15, 0.15, 0.25, 0.15) # assume sum of bias is 1 def roll(massDist): randRoll = random.random() # in [0,1) sum = 0 result = 1 for mass in massDist: sum += mass if randRoll < sum: return result result+=1 print roll(sampleMassDist) </code></pre>

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