Looking at a recruitment post where stats are best 3 of 4d6 with ones rerolled.
Able to do the 6 rolls for stats, but unable to edit and add rerolls for ones. Second rolling set just shows up as the code to reroll however many d6 but no actual roll.
What, pray tell, am I doing wrong?
Thanks in advance.
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Correct, but don't. Roll 5s, and add 1 for each die. Same result, without need to waste posts or run afoul the double post rule.
A d6 in which 1s don't count as a roll is a d5, with +1 to make it so it produces a number between 2-6. In your particular case, roll 3 sets of 4d5 + 4.
Yours,
Grey Wolf
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Correct, but don't. Roll 5s, and add 1 for each die. Same result, without need to waste posts or run afoul the double post rule.
A d6 in which 1s don't count as a roll is a d5, with +1 to make it so it produces a number between 2-6. In your particular case, roll 3 sets of 4d5 + 4.
Yours,
Grey Wolf
While your advice is good, the first bolded part is incorrect. This is an exception where multiple posts in a row are allowed.
Additionally, the second bolded part is also incorrect. He wants a best three of four (roll 4d6, drop lowest). He would need to roll each d5+1 individually.
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Actually, wouldn't it be easier to [roll]4d5b3[/roll] and then add 3 to the result to get it done all at once?
Correct, but +4 would not work. And he would probably need to roll 6 sets of these (the usual amount for a D&D game). Grey_Wolf_c thought he wanted 3 sets of 4d6 (reroll 1s), when he needs an undisclosed amount of 4d6 (reroll 1s, drop lowest).
Also, noparse tags don't work for dice rolls. They show here because dice rolling is disabled for this forum.
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Correct, but +4 would not work. And he would probably need to roll 6 sets of these (the usual amount for a D&D game). Grey_Wolf_c thought he wanted 3 sets of 4d6 (reroll 1s), when he needs an undisclosed amount of 4d6 (reroll 1s, drop lowest).
Also, noparse tags don't work for dice rolls. They show here because dice rolling is disabled for this forum.
So he should use the [rollv] tag with 4d5, however many times; drop the lowest of those four, and add 1 to each individual number, manually?
So he should use the [rollv] tag with 4d5, however many times; drop the lowest of those four, and add 1 to each individual number, manually?
Going for a single post, the following would work: [roll]4d5b3+3[/roll]
-> Works because the lowest roll is dropped, then 1 is added per die to make 1-5 into 2-6.
[roll]1d5+1[/roll]
[roll]1d5+1[/roll]
[roll]1d5+1[/roll]
[roll]1d5+1[/roll]
Manually drop lowest and add together.
-> Works because 1 is added per die to make 1-5 into 2-6 and you are manually dropping the lowest.
[rollv]4d5[/rollv]
Manually drop lowest, add together, and add 3.
-> Works because you are manually doing everything, including adding 1 per die to make 1-5 into 2-6.
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"My Hobby: Replacing your soap with gravy" by rtg0922, Doll and Clint "Rawhide" Eastwood by Sneak
Going for a single post, the following would work: [roll]4d5b3+3[/roll]
-> Works because the lowest roll is dropped, then 1 is added per die to make 1-5 into 2-6.
So that's four, five-sided sice, keeping the best 3, I believe? But doesn't the plus-three add to each roll?
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