tag:blogger.com,1999:blog-613387965721174471.post1290168496889372974..comments2024-02-14T18:16:46.226-05:00Comments on the Wood between Worlds: Easier way to replicate funky dice for DCC RPG using regular d6sReecehttp://www.blogger.com/profile/10691425406901685427noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-613387965721174471.post-51350196769334835212021-12-26T21:07:55.180-05:002021-12-26T21:07:55.180-05:00I forgot about the D30:
d30: d3 + d10; if d3 == 2...I forgot about the D30:<br /><br />d30: d3 + d10; if d3 == 2, add 10 to the d10; if d3 == 3 add 20 to the d10formixianhttps://www.blogger.com/profile/15364119484029139355noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-64457997921529069042021-12-26T21:06:18.633-05:002021-12-26T21:06:18.633-05:00This comment has been removed by the author.formixianhttps://www.blogger.com/profile/15364119484029139355noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-43260368631154641052021-12-26T21:04:37.806-05:002021-12-26T21:04:37.806-05:00Mathematically, to have the same chance to hit a n...Mathematically, to have the same chance to hit a number there is a simple way to make the funky dices:<br /><br />d2 : d4 where 1-2 = 1; 3-4 = 2 (divide by 2 round up)<br />d3 : d6 where 1-2 = 1; 2-4 = 2; 5-6 = 3 (divide by 2 round up)<br />d5 : d10 where 1-2 = 1; 3-4 = 2; ...; 9-0 = 5 (divide by 2 round up)<br />d7 : d8 reroll if you obtain 8<br />d14: d2 + d7 if the d2 ==formixianhttps://www.blogger.com/profile/15364119484029139355noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-52245989004400199522020-07-27T15:38:34.758-04:002020-07-27T15:38:34.758-04:00That's true. But it's not nearly as cool ...That's true. But it's not nearly as cool ;)<br /><br />You can get all of the funky ones using tricks like that. I just wanted to show everything except the d7 and d14 can be done only using d6s, and without any "if 1-X add Y" kinds of things. For some people the funky dice is a deal-breaker, and the "fiddly" mechanics are a no-go.Reecehttps://www.blogger.com/profile/10691425406901685427noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-10819996985261859002020-07-26T00:54:07.531-04:002020-07-26T00:54:07.531-04:00There's an easier way to do the d24. You roll ...There's an easier way to do the d24. You roll a d12 and a d6.<br />1-3 = +0<br />4-6 = +12<br />and then add the result of the d12.Blohggehrhttps://www.blogger.com/profile/16661391491230605180noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-5930130243087027902017-08-09T12:26:32.428-04:002017-08-09T12:26:32.428-04:00Check this out: http://rollthedice.online/en/dice/...Check this out: http://rollthedice.online/en/dice/3d2reidhttps://www.blogger.com/profile/06504468664742075947noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-38516498903033094692017-08-09T12:21:52.895-04:002017-08-09T12:21:52.895-04:00I don't think the representation matters as mu...I don't think the representation matters as much as ease or speed, but I just realized it's not as easy/quick as I thought (there is one extra subtraction required). Correction to the formula above: d16 = ((db8 - 1) * 2) + db OR d16 = ((db2 - 1) * 8) + d8<br /><br />Alas, dice aren't 0 based... it would make my suggestion simpler and get rid of the pesky subtraction I erroneously reidhttps://www.blogger.com/profile/06504468664742075947noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-47081011204743787472017-08-08T16:07:28.617-04:002017-08-08T16:07:28.617-04:00That would also work. It'd just be an octal r...That would also work. It'd just be an octal representation as opposed to a binary representation. Octal isn't as common as binary of hexadecimal, though.Reecehttps://www.blogger.com/profile/10691425406901685427noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-76174298848297491042017-08-08T14:56:33.167-04:002017-08-08T14:56:33.167-04:00I meant to start that sentence "I think for r...I meant to start that sentence "I think for rolling d16..."reidhttps://www.blogger.com/profile/06504468664742075947noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-67376748054892590242017-08-08T14:55:37.458-04:002017-08-08T14:55:37.458-04:00I think for d8, it's quicker to roll a d8 and ...I think for d8, it's quicker to roll a d8 and db and compute the answer. But you will have to choose before rolling (I prefer the former): d16 = (db8 * 2) + db OR d16 = (db2 * 8) + d8reidhttps://www.blogger.com/profile/06504468664742075947noreply@blogger.com