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It's a Wand! It's a Crossbow Bolt! It's a Floor Wax!
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<blockquote data-quote="Anax" data-source="post: 2637414" data-attributes="member: 19868"><p>Spatzimus: It's possible to do this task *much much* faster, if you calculate the raw probabilities instead of the full combinatorial possibilities.</p><p></p><p>Start by making a table of (1..n, say 1..100) your first die. The 1 entry is the chance you stop at one roll (1/d), the 2 entry is the chance to stop at two rolls (1/d)((d-1)/d), the 3 entry is the chance to stop at 3 rolls (1/3)((d-1)/d)^2, and so on. Each entry is (1/d) * ((d-1)/d)^(r-1), where d is the die size, and r is the number of rolls.</p><p></p><p>Once you've made a table for each die like this, you can combine two tables at a time. So take your d20 table and combine it with a d12 table to produce the probability of a given number of rolls using a d20 and a d12 together. This is only 10,000 combinations, if you look at chances for up to 100 rolls only. When you're done, sum the probabilities for each resulting number of rolls, and discard any possibilities over 100 rolls. Now repeat the process with the next die size down.</p><p></p><p>The results for d20-d12-d8-d6-d4 are: (whoops--put the 20-12-10-8 in the first time I did this, sorry. Edited to fix.)</p><p></p><p>[code] 0.00% | 5</p><p> 0.01% | 6</p><p> 0.04% | 7</p><p> 0.09% | 8</p><p> 0.17% | 9</p><p> 0.30% | 10</p><p> 0.50% | 11</p><p> 0.76% | 12</p><p> 1.11% | 13</p><p> 1.55% | 14</p><p> 2.08% | 15</p><p> 2.72% | 16</p><p> 3.46% | 17</p><p> 4.31% | 18</p><p> 5.26% | 19</p><p> 6.33% | 20</p><p> 7.49% | 21</p><p> 8.76% | 22</p><p> 10.12% | 23</p><p> 11.57% | 24</p><p> 13.10% | 25</p><p> 14.70% | 26</p><p> 16.38% | 27</p><p> 18.11% | 28</p><p> 19.90% | 29</p><p> 21.74% | 30</p><p> 23.61% | 31</p><p> 25.51% | 32</p><p> 27.44% | 33</p><p> 29.39% | 34</p><p> 31.35% | 35</p><p> 33.31% | 36</p><p> 35.27% | 37</p><p> 37.23% | 38</p><p> 39.17% | 39</p><p> 41.10% | 40</p><p> 43.02% | 41</p><p> 44.91% | 42</p><p> 46.77% | 43</p><p> 48.60% | 44</p><p> 50.40% | 45</p><p> 52.17% | 46</p><p> 53.90% | 47</p><p> 55.59% | 48</p><p> 57.25% | 49</p><p> 58.86% | 50</p><p> 60.43% | 51</p><p> 61.96% | 52</p><p> 63.45% | 53</p><p> 64.90% | 54</p><p> 66.30% | 55</p><p> 67.66% | 56</p><p> 68.97% | 57</p><p> 70.25% | 58</p><p> 71.48% | 59</p><p> 72.67% | 60</p><p> 73.82% | 61</p><p> 74.93% | 62</p><p> 76.00% | 63</p><p> 77.03% | 64</p><p> 78.02% | 65</p><p> 78.97% | 66</p><p> 79.89% | 67</p><p> 80.78% | 68</p><p> 81.62% | 69</p><p> 82.44% | 70</p><p> 83.22% | 71</p><p> 83.97% | 72</p><p> 84.70% | 73</p><p> 85.39% | 74</p><p> 86.05% | 75</p><p> 86.68% | 76</p><p> 87.29% | 77</p><p> 87.88% | 78</p><p> 88.43% | 79</p><p> 88.97% | 80</p><p> 89.48% | 81</p><p> 89.97% | 82</p><p> 90.43% | 83</p><p> 90.88% | 84</p><p> 91.31% | 85</p><p> 91.72% | 86</p><p> 92.11% | 87</p><p> 92.48% | 88</p><p> 92.83% | 89</p><p> 93.17% | 90</p><p> 93.50% | 91</p><p> 93.81% | 92</p><p> 94.10% | 93</p><p> 94.38% | 94</p><p> 94.65% | 95</p><p> 94.91% | 96</p><p> 95.15% | 97</p><p> 95.38% | 98</p><p> 95.61% | 99</p><p> 95.82% | 100</p><p>[/code]</p><p></p><p>(And, of course, the chance of > 100 charges is 4.18%.)</p><p></p><p>This only took a few seconds of computation time to produce using my technique. (I controlled the table generation by hand, so it took a couple of minutes to type the commands to produce all of the intermediate tables and the result.)</p></blockquote><p></p>
[QUOTE="Anax, post: 2637414, member: 19868"] Spatzimus: It's possible to do this task *much much* faster, if you calculate the raw probabilities instead of the full combinatorial possibilities. Start by making a table of (1..n, say 1..100) your first die. The 1 entry is the chance you stop at one roll (1/d), the 2 entry is the chance to stop at two rolls (1/d)((d-1)/d), the 3 entry is the chance to stop at 3 rolls (1/3)((d-1)/d)^2, and so on. Each entry is (1/d) * ((d-1)/d)^(r-1), where d is the die size, and r is the number of rolls. Once you've made a table for each die like this, you can combine two tables at a time. So take your d20 table and combine it with a d12 table to produce the probability of a given number of rolls using a d20 and a d12 together. This is only 10,000 combinations, if you look at chances for up to 100 rolls only. When you're done, sum the probabilities for each resulting number of rolls, and discard any possibilities over 100 rolls. Now repeat the process with the next die size down. The results for d20-d12-d8-d6-d4 are: (whoops--put the 20-12-10-8 in the first time I did this, sorry. Edited to fix.) [code] 0.00% | 5 0.01% | 6 0.04% | 7 0.09% | 8 0.17% | 9 0.30% | 10 0.50% | 11 0.76% | 12 1.11% | 13 1.55% | 14 2.08% | 15 2.72% | 16 3.46% | 17 4.31% | 18 5.26% | 19 6.33% | 20 7.49% | 21 8.76% | 22 10.12% | 23 11.57% | 24 13.10% | 25 14.70% | 26 16.38% | 27 18.11% | 28 19.90% | 29 21.74% | 30 23.61% | 31 25.51% | 32 27.44% | 33 29.39% | 34 31.35% | 35 33.31% | 36 35.27% | 37 37.23% | 38 39.17% | 39 41.10% | 40 43.02% | 41 44.91% | 42 46.77% | 43 48.60% | 44 50.40% | 45 52.17% | 46 53.90% | 47 55.59% | 48 57.25% | 49 58.86% | 50 60.43% | 51 61.96% | 52 63.45% | 53 64.90% | 54 66.30% | 55 67.66% | 56 68.97% | 57 70.25% | 58 71.48% | 59 72.67% | 60 73.82% | 61 74.93% | 62 76.00% | 63 77.03% | 64 78.02% | 65 78.97% | 66 79.89% | 67 80.78% | 68 81.62% | 69 82.44% | 70 83.22% | 71 83.97% | 72 84.70% | 73 85.39% | 74 86.05% | 75 86.68% | 76 87.29% | 77 87.88% | 78 88.43% | 79 88.97% | 80 89.48% | 81 89.97% | 82 90.43% | 83 90.88% | 84 91.31% | 85 91.72% | 86 92.11% | 87 92.48% | 88 92.83% | 89 93.17% | 90 93.50% | 91 93.81% | 92 94.10% | 93 94.38% | 94 94.65% | 95 94.91% | 96 95.15% | 97 95.38% | 98 95.61% | 99 95.82% | 100 [/code] (And, of course, the chance of > 100 charges is 4.18%.) This only took a few seconds of computation time to produce using my technique. (I controlled the table generation by hand, so it took a couple of minutes to type the commands to produce all of the intermediate tables and the result.) [/QUOTE]
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