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Bladesinger - a criticism of its design
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<blockquote data-quote="Ovinomancer" data-source="post: 7260796" data-attributes="member: 16814"><p>Not sure I follow this. What I assume you mean is that the poster generated a LOT of arrays, then pulled the highest roll from each, then averaged the highest rolls and got slightly less than 16? Because the method you presented would return the average roll (lots of elements, average them all), which is between 12 and 13 (12.24). </p><p></p><p>If it's my assumption, then that tracks pretty well with my estimate of the probabilities. I say estimate because working out the math is tediously hard -- the monte carlo simulation is much easier to build.</p><p></p><p>For example, the odds of rolling at least 1 18 is pretty easy to establish. it's the odds I roll 1 18 in 6 rolls, plus the odds I roll 2 18's in six rolls, plus... the odds I roll 6 18's in 6 rolls (turns out you can pretty much ignore everything after 2 18's, which is a 0.37% chance) -- this is 9.34% chance for at least 1 18.</p><p></p><p>But to figure out if your highest roll is 17, you have to figure the odds for 1 17 in 6 rolls minus the odds of rolling an 18 in the other 5 rolls (so 1-5 18's on 5 rolls), then the chance to roll 2 17's minus the chance to roll any 18's on the other 4 rolls, etc. 16's means you're figuring the odds of 1 16 minus the odds of 17 or 18 on the other rolls, and so on. U. G. L. Y. There's not a good shortcut to this kind of probability, at least not that I've been able to find or have ever been aware of. In school, this is when we simulated.</p><p></p><p>Much easier to generate a LOT of sets, pull the highest in each, and then average. If I get the time, I might recreate.</p></blockquote><p></p>
[QUOTE="Ovinomancer, post: 7260796, member: 16814"] Not sure I follow this. What I assume you mean is that the poster generated a LOT of arrays, then pulled the highest roll from each, then averaged the highest rolls and got slightly less than 16? Because the method you presented would return the average roll (lots of elements, average them all), which is between 12 and 13 (12.24). If it's my assumption, then that tracks pretty well with my estimate of the probabilities. I say estimate because working out the math is tediously hard -- the monte carlo simulation is much easier to build. For example, the odds of rolling at least 1 18 is pretty easy to establish. it's the odds I roll 1 18 in 6 rolls, plus the odds I roll 2 18's in six rolls, plus... the odds I roll 6 18's in 6 rolls (turns out you can pretty much ignore everything after 2 18's, which is a 0.37% chance) -- this is 9.34% chance for at least 1 18. But to figure out if your highest roll is 17, you have to figure the odds for 1 17 in 6 rolls minus the odds of rolling an 18 in the other 5 rolls (so 1-5 18's on 5 rolls), then the chance to roll 2 17's minus the chance to roll any 18's on the other 4 rolls, etc. 16's means you're figuring the odds of 1 16 minus the odds of 17 or 18 on the other rolls, and so on. U. G. L. Y. There's not a good shortcut to this kind of probability, at least not that I've been able to find or have ever been aware of. In school, this is when we simulated. Much easier to generate a LOT of sets, pull the highest in each, and then average. If I get the time, I might recreate. [/QUOTE]
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Bladesinger - a criticism of its design
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