How do you put "re-roll x" in dice brackets.

[Mystic Muse]

In a PBP game I'm playing in the DM said "re-roll 1s" However, I don't know how to do that with the dice brackets. I don't want to keep posting until I don't get a 1.

can somebody explain how to do this?

[Mando Knight]

I don't think it's encoded in the dice roller here. [rollv] tags, though, will let you see the raw rolls before adding them together, so you could use those, then roll extras below in a spoiler, just in case.

Myth-Weavers uses XdYrZ for "XdY, reroll Z or lower," but it's not in the giantitp roll tags.

[Djinn_in_Tonic]

Yeah...I do the following:

Roll: Roll here.

Re-Rolls in case of 1: 1+ re-rolls here, depending on the number of dice initially rolled.

[Douglas]

I'd just decrease the die size and add. XdY reroll 1s is equivalent to Xd(Y-1) + X.

[Brewdude]

Yeah, that doesn't help for "reroll and add" if the reroll is the high number.

2d10 reroll and add doubles is one of my favorite die roll systems.

[Zeb The Troll]

I'd just decrease the die size and add. XdY reroll 1s is equivalent to Xd(Y-1) + X.The problem with this method is that it's essentially just turning all 1's into 2's, isn't it? That's not quite the same thing as rerolling 1's.

The Invisible Castle (http://invisiblecastle.com/roller/) site has a very good dice roller with results linkable in your post.

[Fortuna]

The problem with this method is that it's essentially just turning all 1's into 2's, isn't it? That's not quite the same thing as rerolling 1's.

No, actually, that's equivalent to rerolling 1s. Notice that each die is still linear, as opposed to converting each 1 to a 2 which weights it slightly towards 2.

[Zeb The Troll]

Example 1:

I roll 4d6's with the caveat that I will reroll 1's.

I roll 4 x 1's, pick up the dice, and rethrow them. This time I get two 5's, a 4, and a 2. Is this result the same as adding +1 to each die?

Example 2:

I roll 4d6 and get 3 x 6's and a 1. I pick up the 1 and reroll to get a 4. Is this the same result as rolling 3 x 5's and a 1 then adding +4 to the result?

Conclusion:

1d5+1 is not the same thing as 1d6(reroll 1's).

[The Rose Dragon]

It has the same probabilities, though. In either case, you will have a one-fifth probability of coming up with a result from 2 to 6 inclusive. It doesn't get there the same way, but when you roll dice, the probabilities are what matter, not the way you get there.

[Zeb The Troll]

Except that the probabilities don't work out that way. With the +1 option, you have a one in five chance of getting a six. But with the reroll option, you have

(a one in six of getting six) AND (a one in six chance of getting...
(one in six chance) AND (another one in six chance of getting...
...ad infinitum.

I'm not a statistician so I don't know how the numbers actually work out, only that they are not the same.

[The Rose Dragon]

It works out the same.

Basically, you get (1/6) + (1/6)2 + ... (1/6)n, where n is an integer. As n gets arbitrarily large, the sum approaches 1/5.

[Sir Enigma]

But you can apply the same argument to any number - you have a 1-in-6 chance to get a 3, plus a 1-in-6 chance to re-roll and 1-in-6 chance to get a 3 etc.; if you work it out as a series and sum it up, you do in fact get 1-in-5 chance to get any number between 2 and 6.

Mathematically, your chances are 1/6 + (1/6)*(1/6) + (1/6)*(1/6)*(1/6)....

=SUM[n=1...inf](1/6)^n

The formula to work out the value of an infinite geometric series like this is S=a0/(1-r) where a0 is the first term and r is the ratio between two successive terms; for this series, a0 and r are both 1/6, so

S=(1/6)/(1-(1/6))=(1/6)/(5/6)=(1/6)*(6/5)=1/5.

This will work out the same for any N-sided die - 1dNr1 is equivalent to 1d(N-1)+1.

EDIT: Ninja dragons!

[The Rose Dragon]

Gah! Geometric series! That's the term I was looking for!

Sorry, been a while since I took any math-related class.

[Zeb The Troll]

Well, I'll be darned.

I still prefer rerolls because it makes me feel better about my chances, whether they ARE better or not. :smallcool: