I'm used to generating stats based on cards. I'm trying to use the system deck52, but it appears not to be configurable.

The way that the simplest method works is that you draw cards between 4 and 9 and create piles of two each and the sum of each pile equals one of the attributes of the player.

However, it seems like the deck52 system cannot handle this.

How do I do this? I'm at wit's end.

You would have to draw all 52 cards, with the dice roller acting as a randomizer for the full deck, then manually edit out the cards you don't want (so anything not between 4 and 9) from the list, then group the remaining numbers as you need them and add them together.

The card draw option on the roller really isn't super robust, for better or worse, so anything complex requires a lot of actual work after the fact.

Between 4 and 9?  Roll a D6 and add 3 for however many cards you need, it'll give you the same basic results (unless you get a ton of 4's or something like that...if you get more than 4 of the same result, re-roll the extras).

You still have to do a little additional math to get the final results...you could simplify it and roll 2D6+6 for each attribute, but that wouldn't give you the individual 'card' totals to verify that you didn't have too many of the same 'card'...so, assuming that's a concern, you roll each die, add 3, record the result, and repeat...add those two results together for your first 'pile'.  Repeat for however many piles you need, re-rolling anything beyond four of the same result (since there are only four of each number card in the deck).

In reply to facemaker329 (msg # 3):

That's exactly not what I want to accomplish, I'm afriad. The entire point is that by drawing the same 12 cards for everyone, the ability scores always add up to the same amount for everyone in the party and if you get an 18 in one pile, you are garanteed a low score in another pile.

Now I'm not sure of the math...4-9 is six cards, times 4 suits, is 24 cards...not sure how that is supposed to be 'drawing the,same 12 cards'.  Or are you saying that there are 12 attributes, so you only get 12 piles?  Because that math does work out...

But if you throw out and reroll results after the the third repeat (plus the original), you still, numerically, only have 24 numbers generated...exactly the same results as you'd get drawing the cards, unless there's some special significance attached to the suits.

It's actually a rather easy mathematical problem. However I don't believe that it's easy to accomplish with the roller. It would require a lot of book keeping to do. I'd really suggest using a real deck of cards.

If you want to go with the roller then you could roll the dice like facemaker suggested but try this:

1. Roll 1d13-7. (assuming you want 12 cards from 4 to 9)
1a. Note number. Reroll if it's a duplicate.
2. If negative -> remove the negative sign.
3. Add three to number and write total down.
4. Repeat.


Another easy way to do this would be to make a simple program.

You could roll 12d24, setting record each die roll and unique rolls settings.  Take each 'die' and subtract six from each until six or less, add 3 to the final.  Unfortunately if you have to pull all the cards, the roller only allows 20 individually recorded dice.

Alternately, you can tell it to Roll all 52 cards and skip the unused cards yourself mentally, if you need to be able to get all instances of cards you mentioned being valid.

A problem with rolling 12d24 is that you're going to run out of rolls if there are any duplicates (and there will be).

So you have two broad methods:

Roll each card uniquely

Create a table with 24 entries (1: 4 of hearts, 2: 4 of clubs, 3: 4 of diamonds, 4: 4 of spades, 5: 5 of hearts, 6: 5 of clubs, etc.).

Roll 1d24 for your first card and look it up in the table.

Roll 1d23 for your second card and look it up in the table, skipping any card you already got.

Roll 1d22 for your third card and look it up on the table, skipping any card you already got.

Examples: Rolls are 12, 7, 5. That's 6 of spades (12), 5 of diamonds (7) and 5 of hearts (5). No skipping happened because the rolls were always decreasing.

Rolls are 5, 7, 12. That's 5 of hearts (5), 5 of spades (7 + 1 because the 5 of hearts is skipped) and 7 of clubs (12 + 2 because the 5 of hearts and 5 of spades are skipped). Simple skipping because the rolls are far apart.

Rolls are 5, 12, 7. That's 5 of hearts (5), 7 of hearts (12 + 1 because the 5 of hearts is skipped) and 5 of spades (7 + 1 because the 5 of hearts is skipped).

Rolls are 3, 3, 3. That's 4 of diamonds (3), 4 of spades (3 + 1 because 4 of diamonds is skipped) and 5 of hearts (3 + 2 because 4 of spades and 4 of diamonds are skipped). Two adjacent cards are skipped together since there is none between them.

Rolls are 3, 4, 3. That's 4 of diamonds (3), 5 of hearts (4 + 1 because the 4 of diamonds is skipped) and 4 of spades (3 + 1 because the 4 of diamonds is skipped). A roll can end up in the gap between two cards.

Rolls are 3, 4, 4. That's 4 of diamonds and 5 of hearts (as above), then 5 of clubs (4 + 2 because both cards are skipped. If only one were skipped, 4 + 1 would give the 5 of hearts, which was already drawn). Counting down the list can be easier than tracking the skip distances.


Allow for error and revise dice pools

The first method guarantees that you'll get things done in 23 rolls (rolling 1d1 is sort of pointless). This method can involve more rolls, but the table is much easier to deal with.

Roll 16d24, and look up the results on the table, ignoring all duplicates. This will get you between 1 (you rolled the same number 16 times) and 16 (no doubles) values.

If you still need more numbers, rebuild the table to exclude the numbers you already rolled, and use a smaller die. Make another 12 rolls.

If you still need more numbers, repeat.

You could just roll lots of d24, but if you want to get all the cards it becomes rather difficult and long as you'll get lots of duplicates.

Example: Rolls are 5, 7, 12, 18, 3, 9. No doubles! You get the cards corresponding to 3, 5, 7, 9, 12, and 18 on the table. More rolls are done with d18 rather than d24, excluding the cards corresponding to the numbers you got.

Rolls are 5, 7, 5, 12, 9, 9. Some doubles. You get the cards corresponding to 5, 7, 9, and 12 on the table. More rolls are done with d20 rather than d24, excluding the cards corresponding to the numbers you got.

Rolls are 5, 5, 5, 5, 5, 5. All duplicates. You get the card corresponding to 5 on the table. More rolls are probably done with d24, because rebuilding the table for just one number is sort of wasteful (you'll have to skip any 5s you roll).


Cheat

Ways to bypass the issue!

Use point totals. All characters have the same total number of points, so let them assign them as they wish, with minimum and maximum values (8 to 18). Not random, but gets what you're after. You may want to restrict players to having at most 2 attributes at 18 or 8 (four 4s or four 9s), if it's an issue.

Complementary attributes. Roll 2d6 for a pair of attributes. One of the attributes is set to result + 6, the other is set to 20 - result. This way, the attributes add up to 26 per pair, ranging between 8 and 18, and you get some randomness. The issue is that you every expert has a major weak point (8 and 18) rather than some experts being below par elsewhere (18, 11, 11, 12 across four attributes, avoiding the 8). Someone may be (un)lucky enough to get three or more 18s (and 8s) this way.

Split pools. Building on complementary attributes and point totals, roll 6d6 for a trio of attributes. Assign the results as you wish (6 + 2 dice), then do the same for three other attributes, using the dice complements (7 - die). 18s become more likely (just need two 6s out of six rolls, rather than two rolls), and you still have chances of more than two 18s.

Example: Rolls are 2, 3, 4, 4, 5, 6. The complementary rolls are 5 (7 - 2), 4 (7 - 3), 3 (7 - 4), 3 (7 - 4), 2 (7 - 5), 1 (7 - 6). So your first three attributes could be 12 (2, 4), 13 (3, 4), 17 (5, 6) or 14 (2, 6), 14 (3, 5), 14 (4, 4). Your other three attributes could be 9 (1, 2), 12 (3, 3), 15 (4, 5) or 11 (1, 4), 12 (3, 3), 13 (2, 5). Or many other combinations.


Using the rolling approaches, if you really want to avoid three or more 18s, just require rerolls if they occur (using 1d4+4 instead), so extra values range from 5 to 8. If that gives too many 16s too... don't use these shortcuts.

You won't have duplicates if you use the unique checkbox.

That makes is much easier. My mistake!

12d24 (or up to 24 d24), unique box, and done. 1-4 is a 4, 5-8 is a 5, 9-12 is a 6, 13-16 is a 7, 17-20 is an 8, and 21-24 is a 9. You're using d24 rather than d6 in order to allow duplicates to happen.

Or a different table scheme - LoreGuard's is a 4 with 1, 7, 13, 19 rolled, a 5 with 2, 8, 14, 20 rolled, etc. Same final result, different ways of looking at the numbers to get there.

I like your 1-4=1 idea better, for calculating the result of the d24.  I.e. Divide by 4 and round any fraction at all up.

There is a technical limitation however that limits you to 20d24 due to the limitation on record all die results to a max of 20.

Well, at that point there are only 4 cards left, and they're being drawn in pairs. You have 4 options for the first card and 3 for the second, and since you're adding them their order does not matter, meaning you have 6 possible outcomes. So it's an extra d6 if you finesse the combinations.

The following defines the 11th pair of cards using a d6:

Roll 1: First and Second Cards (12th pair is Third and Fourth)
Roll 2: First and Third Cards (12th pair is Second and Fourth)
Roll 3: First and Fourth Cards (12th pair is Second and Third)
Roll 4: Second and Third Cards (12th pair is First and Fourth)
Roll 5: Second and Fourth Cards (12th pair is First and Third)
Roll 6: Third and Fourth Cards (12th pair is First and Second)

It all depends on knowing what the last four cards are, so the initial 20d24 are required!

Not nessesarily what you may be looking for, but the Roll20 web site has a standard 52 card draw and a customizable card draw. FYI