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    <p>There is an elegant algorithm attributed by Knuth to A. J. Walker (Electronics Letters 10, 8 (1974), 127-128; ACM Trans. Math Software 3 (1977), 253-256).</p> <p>The idea is that if you have a total of k * n balls of n different colors, then it is possible to distribute the balls in n containers such that container no. i contains balls of color i and at most one other color. The proof is by induction on n. For the induction step pick the color with the least number of balls.</p> <p>In your example n = 10. Multiply the probabilities with a suitable m such that they are all integers. So, maybe m = 100 and you have 20 balls of color 0, 20 balls of color 1, 10 balls of color 2, 5 balls of color 3, etc. So, k = 10.</p> <p>Now generate a table of dimension n with each entry being a probability (the ration of balls of color i vs the other color) and the other color.</p> <p>To generate a random ball, generate a random floating-point number r in the range [0, n). Let i be the integer part (floor of r) and x the excess (r – i).</p> <pre><code>if (x &lt; table[i].probability) output i else output table[i].other </code></pre> <p>The algorithm has the advantage that for each random ball you only make a single comparison.</p> <p>Let me work out an example (same as Knuth). </p> <p>Consider simulating throwing a pair of dice.</p> <p>So P(2) = 1/36, P(3) = 2/36, P(4) = 3/36, P(5) = 4/36, P(6) = 5/36, P(7) = 6/36, P(8) = 5/36, P(9) = 4/36, P(10) = 3/36, P(11) = 2/36, P(12) = 1/36.</p> <p>Multiply by 36 * 11 to get 393 balls, 11 of color 2, 22 of color 3, 33 of color 4, …, 11 of color 12. We have k = 393 / 11 = 36.</p> <p>Table[2] = (11/36, color 4)</p> <p>Table[12] = (11/36, color 10)</p> <p>Table[3] = (22/36, color 5)</p> <p>Table[11] = (22/36, color 5)</p> <p>Table[4] = (8/36, color 9)</p> <p>Table[10] = (8/36, color 6)</p> <p>Table[5] = (16/36, color 6)</p> <p>Table[9] = (16/36, color 8)</p> <p>Table[6] = (7/36, color 8)</p> <p>Table[8] = (6/36, color 7)</p> <p>Table[7] = (36/36, color 7)</p>
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