Difference between revisions of "Talk:3015: D&D Combinatorics"
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Rolling 22 or lower on percentile dice (or, equivalently, 79 or higher) is close enough, and easier to come up with. (Give or take whether 00 is treated as 100 or zero.) Or directly represent the action: roll a d10. If it's 1-5, you lose. If it's 6-10, roll again; if it's 1-5 you lose, 6-9 you win, 10 roll again. (Modify slightly if you want to distinguish the case of grabbing *two* cursed arrows.) [[User:Jordan Brown|Jordan Brown]] ([[User talk:Jordan Brown|talk]]) 03:26, 23 November 2024 (UTC) | Rolling 22 or lower on percentile dice (or, equivalently, 79 or higher) is close enough, and easier to come up with. (Give or take whether 00 is treated as 100 or zero.) Or directly represent the action: roll a d10. If it's 1-5, you lose. If it's 6-10, roll again; if it's 1-5 you lose, 6-9 you win, 10 roll again. (Modify slightly if you want to distinguish the case of grabbing *two* cursed arrows.) [[User:Jordan Brown|Jordan Brown]] ([[User talk:Jordan Brown|talk]]) 03:26, 23 November 2024 (UTC) | ||
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| + | == Alternative exact solution == | ||
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| + | Roll: 1d8, 2d6, 1d4 succeed on 19 or higher. | ||
Revision as of 03:54, 23 November 2024
The bot originally created this page as “D Combinatorics”. I renamed it to the correct title and tried to get as many of the references as possible (including a few redirects). JBYoshi (talk) 00:54, 23 November 2024 (UTC)
- The title in the Atom feed (which I'm assuming the bot consumes) is "D Combinatorics". I'm guessing something in Randall's pipeline didn't like the ampersand. --162.158.154.160 01:41, 23 November 2024 (UTC)
- By raw combinatorics: 71 + 52 + 34 + 20 + 10 + 4 + 1 ways to get each of 16 - 22 respectively, for a total of 192, out of 4(6^3) = 864 total. 192/864 simplifies to exactly 2/9. I have no idea how Randall found this; if anyone has an idea, please let me know. Kaisheng21 (talk) 01:33, 23 November 2024 (UTC)
It seems like we edited the transcript at the same time. The odds of rolling 16 or higher in this situation seem to be 2/9? Darkmatterisntsquirrels (talk) 01:29, 23 November 2024 (UTC)
- There are 864 possible rolls (6 * 6 * 6 * 4). If you enumerate all of the rolls you will find that 192 are 16 or higher. 192/864 = 2/9, the value from the explanation. 172.68.54.139 01:41, 23 November 2024 (UTC)
A much simpler approach: Roll two six sided dice and sum the result. You are successful if the result is 5 or 9. That happens 8 times out of 36. 8/36 = 2/9. (Or successful if the sum is 4 or 6, or 2 or 7, or 2,3,4 or 11, or several other combinations.) 172.68.54.139 01:41, 23 November 2024 (UTC)
- Clever, but dice rolls in D&D involving summing all the dice, applying modifiers, if any, and then comparing to one or more threshold values. Your method makes it very difficult to apply modifiers. 162.158.41.8 02:49, 23 November 2024 (UTC)
Minor quibble, arrows aren't fired (unless they're flaming or self-propelled, perhaps), they are shot. (Shotguns are fired of course.) 162.158.41.73 02:52, 23 November 2024 (UTC)
Rolling 22 or lower on percentile dice (or, equivalently, 79 or higher) is close enough, and easier to come up with. (Give or take whether 00 is treated as 100 or zero.) Or directly represent the action: roll a d10. If it's 1-5, you lose. If it's 6-10, roll again; if it's 1-5 you lose, 6-9 you win, 10 roll again. (Modify slightly if you want to distinguish the case of grabbing *two* cursed arrows.) Jordan Brown (talk) 03:26, 23 November 2024 (UTC)
Alternative exact solution
Roll: 1d8, 2d6, 1d4 succeed on 19 or higher.
