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Community
General Tabletop Discussion
*Pathfinder & Starfinder
Can we make "charge die" core, please?
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<blockquote data-quote="Lackhand" data-source="post: 3816260" data-attributes="member: 36160"><p>This is easiest to show on a d4:</p><p>Rolling just once, we get (1), (2), (3), and (4); 1/4 of our wands fail.</p><p></p><p>Rolling twice, we get (1 followed by 1, 2, 3, 4), (2 followed by 1, 2, 3, 4), (3 followed by 1, 2, 3, 4) and (4 followed by 1, 2, 3, 4).</p><p>In this, one sample from each of the series that started with 2, 3, and 4 fails, and anything that rolled a 1 on the first die fails. That's 4+3 = 7 of our 16 possible wands that fail.</p><p></p><p>Thrice, the possibilities are:</p><p>111 121 131 141 211 221 231 241 (and because 2 != 1, just run that last set</p><p>112 122 132 142 212 222 232 242 of sixteen twice more for me, swapping</p><p>113 123 133 143 213 223 233 243 3 and 4 for 2 in the hundreds place)</p><p>114 124 134 144 214 224 234 244</p><p></p><p>In this, 1/4 fail because they start with 1, 1/4 of each of the remaining sets of 16 fail because they have a 1 in the hundreds place, and 1/4 of each of the 3 remaining sets (within those 16) fail for the ones place.</p><p>Thus, 16+3*4+3*3*1 = 16+12+9 = 37 of 64 possible wands have failed, or over half.</p><p></p><p>Why do wands that have failed keep rolling? Well, they don't, actually. Once they've gone kaput, they've used up their charged. </p><p>But they still dominate the probability space: If a wand has a 1/4 chance of failing on the first roll, fully 1/4 wands will fail on that first roll.</p><p>Of those that remain, 1/4 will fail on the second, and so that's 1/4+3/16 that fail in two rolls.</p><p></p><p>And so on. The math works out the same.</p><p></p><p> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f631.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":o" title="Eek! :o" data-smilie="9"data-shortname=":o" /> How do I make a table? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f631.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":o" title="Eek! :o" data-smilie="9"data-shortname=":o" /></p><p></p><p>And, on topic: I use something like this rule myself, but would hate for it to be core!</p></blockquote><p></p>
[QUOTE="Lackhand, post: 3816260, member: 36160"] This is easiest to show on a d4: Rolling just once, we get (1), (2), (3), and (4); 1/4 of our wands fail. Rolling twice, we get (1 followed by 1, 2, 3, 4), (2 followed by 1, 2, 3, 4), (3 followed by 1, 2, 3, 4) and (4 followed by 1, 2, 3, 4). In this, one sample from each of the series that started with 2, 3, and 4 fails, and anything that rolled a 1 on the first die fails. That's 4+3 = 7 of our 16 possible wands that fail. Thrice, the possibilities are: 111 121 131 141 211 221 231 241 (and because 2 != 1, just run that last set 112 122 132 142 212 222 232 242 of sixteen twice more for me, swapping 113 123 133 143 213 223 233 243 3 and 4 for 2 in the hundreds place) 114 124 134 144 214 224 234 244 In this, 1/4 fail because they start with 1, 1/4 of each of the remaining sets of 16 fail because they have a 1 in the hundreds place, and 1/4 of each of the 3 remaining sets (within those 16) fail for the ones place. Thus, 16+3*4+3*3*1 = 16+12+9 = 37 of 64 possible wands have failed, or over half. Why do wands that have failed keep rolling? Well, they don't, actually. Once they've gone kaput, they've used up their charged. But they still dominate the probability space: If a wand has a 1/4 chance of failing on the first roll, fully 1/4 wands will fail on that first roll. Of those that remain, 1/4 will fail on the second, and so that's 1/4+3/16 that fail in two rolls. And so on. The math works out the same. :o How do I make a table? :o And, on topic: I use something like this rule myself, but would hate for it to be core! [/QUOTE]
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Community
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Can we make "charge die" core, please?
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