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Message: <blockquote data-quote="FireLance" data-source="post: 4429107" data-attributes="member: 3424">Okay, let's try it again. Apparently, my earlier example using a d10 brutal 2 weapon wasn't clear enough. Consider all the possible outcomes of 1,728 (12^3) damage rolls with a d12 brutal 2 weapon. After the first roll:144 (1/12) of them will be 12. 144 (1/12) of them will be 11. 144 (1/12) of them will be 10. 144 (1/12) of them will be 9. 144 (1/12) of them will be 8. 144 (1/12) of them will be 7. 144 (1/12) of them will be 6. 144 (1/12) of them will be 5. 144 (1/12) of them will be 4. 144 (1/12) of them will be 3. 288 (1/6) of them will be re-rolled and still undetermined.Now, consider all the possible outcomes of the 288 re-rolls:24 (1/12) of them will be 12. 24 (1/12) of them will be 11. 24 (1/12) of them will be 10. 24 (1/12) of them will be 9. 24 (1/12) of them will be 8. 24 (1/12) of them will be 7. 24 (1/12) of them will be 6. 24 (1/12) of them will be 5. 24 (1/12) of them will be 4. 24 (1/12) of them will be 3. 48 (1/6) of them will be re-rolled and still undetermined.So, by the end of the first re-roll, all the possible outcomes are:168 (14/144 or 9.72%) of them will be 12. 168 (14/144 or 9.72%) of them will be 11. 168 (14/144 or 9.72%) of them will be 10. 168 (14/144 or 9.72%) of them will be 9. 168 (14/144 or 9.72%) of them will be 8. 168 (14/144 or 9.72%) of them will be 7. 168 (14/144 or 9.72%) of them will be 6. 168 (14/144 or 9.72%) of them will be 5. 168 (14/144 or 9.72%) of them will be 4. 168 (14/144 or 9.72%) of them will be 3. 48 (1/36 or 2.77%) of them will be re-rolled and still undetermined.Now, consider all the possible outcomes of the 48 second re-rolls:4 (1/12) of them will be 12. 4 (1/12) of them will be 11. 4 (1/12) of them will be 10. 4 (1/12) of them will be 9. 4 (1/12) of them will be 8. 4 (1/12) of them will be 7. 4 (1/12) of them will be 6. 4 (1/12) of them will be 5. 4 (1/12) of them will be 4. 4 (1/12) of them will be 3. 8 (1/6) of them will be re-rolled and still undetermined.So, by the end of the second re-roll, all the possible outcomes are:172 (43/432 or 9.95%) of them will be 12. 172 (43/432 or 9.95%) of them will be 11. 172 (43/432 or 9.95%) of them will be 10. 172 (43/432 or 9.95%) of them will be 9. 172 (43/432 or 9.95%) of them will be 8. 172 (43/432 or 9.95%) of them will be 7. 172 (43/432 or 9.95%) of them will be 6. 172 (43/432 or 9.95%) of them will be 5. 172 (43/432 or 9.95%) of them will be 4. 172 (43/432 or 9.95%) of them will be 3. 8 (1/216 or 0.46%) of them will be re-rolled and still undetermined.As mentioned, there is an equal chance of getting each number from 3 to 12, which is essentially d10+2. 8 possibilities are still undetermined (you would have to do a third re-roll; this represents the 1/6 x 1/6 x 1/6 chance that you get a 1 or 2 on the initial roll, the first re-roll and the second re-roll), but hopefully it is obvious by now that those 8 re-rolls would still be distributed evenly between 3-12, with a small chance of a fourth re-roll.</blockquote>

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