Rolling multiple d2-1

Hello,

I like to roll multiple d2-1, e.g. 5d2-1 = 5(d2-1), but with this macro he calculates (5d2)-1.
So what is wrong?

&{template:default} [[ [[1d20]] + [[?{Dice}d2-1]] + [[6]] ]] {{name=My Attack}} {{$[[0]] + $[[1]] + $[[2]]==$[[3]]}}

Does this roll element work for you? Also added a default input box value which you can change or remove

[[?{Dice|3}d2 - ?{Dice}]]

Daniel N. said:

Hello,

I like to roll multiple d2-1, e.g. 5d2-1 = 5(d2-1), but with this macro he calculates (5d2)-1.
So what is wrong?

&{template:default} [[ [[1d20]] + [[?{Dice}d2-1]] + [[6]] ]] {{name=My Attack}} {{$[[0]] + $[[1]] + $[[2]]==$[[3]]}}

I'm confused about what you want the -1 to subtract from.

Is it subtracting from the number of dice rolled? 

If ?{Dice} = 5
  [?{Dice}-1]d2 
        [5-1]d2
          [4]d2
    [1,2,2,1]
            =6

Or the number of sides of the dice?

If ?{Dice} = 5
      ?{Dice}[d2-1] 
          [5]d1
    [1,1,1,1]
            =4

Or the final result? 

If ?{Dice} = 5
  ?{Dice}d2-1 
      [5]d2-1
[1,2,2,1,2]-1
          8-1
           =7

It is a rolling of a die with Symbols. 6 sided, with 3 Symbols "Head" and 3 Symbols "blank. (HeXXen 1733 roleplaying game)
3 of 6 sides + 3 of 6 sides = 6 of 6 sides.
So I shortened it into a d2: If I roll d2-1, I will get the result 1 (=Head) or 0 (=blank) in the same probability.

Now I want to count the "Heads" of x dice-rollings. So e.g. for x=5: 5*(d2-1)

The inline roll I posted should give you Xd2 - X, which should be the same as X*(d2-1), right?

Daniel N. said:

It is a rolling of a die with Symbols. 6 sided, with 3 Symbols "Head" and 3 Symbols "blank. (HeXXen 1733 roleplaying game)
3 of 6 sides + 3 of 6 sides = 6 of 6 sides.
So I shortened it into a d2: If I roll d2-1, I will get the result 1 (=Head) or 0 (=blank) in the same probability.

Now I want to count the "Heads" of x dice-rollings. So e.g. for x=5: 5*(d2-1)

Ah, that's not going to work the way you think.  What you are looking for is a 'success' roll: [[?{Dice?}d2>2]]

The full template:

&{template:default} [[ [[1d20]] + [[?{Dice?}d2>2]] + [[6]] ]] {{name=My Attack}} {{$[[0]] + $[[1]] + $[[2]]==$[[3]]}}

David M.'s approach will also work:

&{template:default} [[ [[1d20]] + [[?{Dice|3}d2 - ?{Dice}]] + [[6]] ]] {{name=My Attack}} {{$[[0]] + $[[1]] + $[[2]]==$[[3]]}}

Ah good call, Jarren. While the same result, the success roll would be a cleaner solution.

For this particular usage, it might be worth creating a rollable table. 

There are only 2 entries, one of which would be 0 with the other being 1, and both would have the same weight.  Since it is a 50/50 thing, it doesn't matter if you leave them both with the default weight of 1 or set them both to 3.  You could name the table as something like "hexxen-coin" and incorporate that into your macro with a query for how many times to roll it. If you went about it that way, then [[?{Dice}d2-1]] would become something like:

[[?{Dice?|0}t[hexxen-coin]]]

The dice query is then determining how many times to roll on the hexxen-coin table which will output 0 or 1 for each roll.

Edit: And Jarren came up with an alternative while I was typing that. 

I have rollable tables for all dice of this game. It works well and displays the symbols and also counts the successes.

But I have tried my case with it and it does'nt work. I thougt it can't calculate with the tables so I want to try it with a dice like in my question.

My first attemp was:

&{template:default} {{name=Testkraft}} {{Wurf=[[ ?{Wie viele Hexxenwürfel?|1}t[HeXXenwürfel] ]]}} {{Janus=[[ ?{Wie viele Janswürfel?|0}t[Januswürfel] ]]}} {{Be schrei bung=Der Schaden wird verdoppelt, wenn blöde Sprüche am Tisch gedropped werden. Wer lacht, bekommt Extraschaden}} {{Grund scha den=[[10]]}} {{Summe=$[[0]] + $[[1]] + $[[2]]==$[[3]]}}

But it reaults in an error while calculating:

If you only are interested in the number of successes for X coin tosses, then 

David M. said:

The inline roll I posted should give you Xd2 - X, which should be the same as X*(d2-1), right?

is absolutely sufficient. You can make it more complicated of course, or "nicer", but the solution of David will always be the most simple (and the one with 100% probability of working as expected - a thing I really came to appreciate here).

Okay, I've got it. :)

But there is one problem left. My macro is:

&{template:default} {{name=Power}} {{HexxDice=[[ ?{HexxDice|1}d2>2]]}} + {{JanusDice=[[ ?{JanusDice|0}d2>2]]}} + [[10]] {{Text=lorem ipsum dolor sit amet}} {{Damage=[[10]]}} {{Summe=$[[0]] + $[[1]] + $[[2]]==$[[3]]}}

But the sum of the calculation is wrong. It should be 15, not 10.

Daniel N. said:

Okay, I've got it. :)

But there is one problem left. My macro is:

&{template:default} {{name=Power}} {{HexxDice=[[ ?{HexxDice|1}d2>2]]}} + {{JanusDice=[[ ?{JanusDice|0}d2>2]]}} + [[10]] {{Text=lorem ipsum dolor sit amet}} {{Damage=[[10]]}} {{Summe=$[[0]] + $[[1]] + $[[2]]==$[[3]]}}

But the sum of the calculation is wrong. It should be 15, not 10.

Your last roll for damage isn't actually a roll, it's just a static number: {{Damage=[[10]]}} and that's what's being called with $[[3]].

You need to create a roll that has everything in it, then work backwards to get all of the individual rolls:

Full Roll = [[ [[ ?{HexxDice|1}d2>2]]}} + {{JanusDice=[[ ?{JanusDice|0}d2>2]]}} + [[10]] ]]
HexxDice = $[[0]]
JanusDice = $[[1]]
Damage = $[[2]]
Summe = $[[3]]