One of the projects I'm working on in my free time is the adaptation of the Urban Shadows milieu to a live-action environment. (In 2022, this is what one might charitably call a long-term goal.) One of the things I'm playing with is the resolution mechanic. As a PBTA game, Urban Shadows is resolved with 2d6; however, I've always felt that playing cards are a rich and underused system, since they can convey information through rank, suit, color, and even imagery.

How would you map the probabilities of 2d6 onto the drawing of cards from a single, shuffled, unmodified 52 (or 54) card standard deck?

A good answer would be mathematically consistent with the bell curve while also being easy to convey to prospective players — i.e., cards of this rank/suit/color count as this value of roll. I'd consider answers that use a single card draw to be better than those that require multiple draws to work, but either is acceptable. Each draw should be used in determining the outcome. (No "null" cards or "discard and redraw" solutions, in other words.)

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – linksassin
    Jul 6, 2022 at 11:24

11 Answers 11


There are 36 possible rolls and 52/54 cards, so it's easy enough to map one card to one possible roll. Here's one possibility, which has the advantage that you'd mostly be reading the number straight off the card.

Roll Cards
2 2♠️
3 3♠️, 3♥️
4 4♠️, 4♥️, 4♣️
5 5♠️, 5♥️, 5♣️, 5♦️
6 6♠️, 6♥️, 6♣️, 6♦️, A♠️
7 7♠️, 7♥️, 7♣️, 7♦️, Jokers
8 8♠️, 8♥️, 8♣️, 8♦️, K♦️
9 9♠️, 9♥️, 9♣️, 9♦️
10 10♥️, 10♣️, 10♦️
11 J♣️, J♦️
12 Q♦️
Draw again Anything else

You say "unmodified" which I take to mean you don't want to remove cards from the deck, so I've put any of the others down as "draw again". In practice I think it would be simpler to leave them out and use a 36-card deck, which would still be big enough for a satisfying shuffle each time.

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    \$\begingroup\$ Having invalid draws is just modifying the deck after you've drawn instead of before, so I don't think it's a valid solution given the question's constraints. \$\endgroup\$ Jul 5, 2022 at 20:43
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    \$\begingroup\$ This is more or less what we did for Settlers of Catan, except like you said, we just bought two decks of cards, created a deck of 36 cards, and put aside the unused ones in case we need spares. (Never again are we going to roll the same number 5 times in a row.) \$\endgroup\$ Jul 6, 2022 at 12:12
  • \$\begingroup\$ @ThomasMarkov The constraints are arbitrary, and poorly defined. Since the only reason given is aesthetic, and there are no details on it, this seems like a fine answer. \$\endgroup\$
    – fectin
    Jul 6, 2022 at 21:41

With a pre-shuffled deck, including a colour and black-and-white joker, you can draw using the following ranking:

Die score Card rank Card rank Card rank Total cards
1 Aces 7s King of Spades 9
2 2s 8s King of Hearts 9
3 3s 9s King of Clubs 9
4 4s 10s King of Diamonds 9
5 5s Jacks Colour Joker 9
6 6s Queens Black/White Joker 9

Ace-6 represent 1 to 6, 7-Q represent 1 to 6 and the Kings and Jokers represent another 1-6. This gives each die score a 9/54 chance, equal to 1/6.

If you draw twice from the same deck, the "rolls" become dependent and whichever score was "rolled" first becomes slightly less likely in the second drawing (8/54 instead of 9/54). This may be acceptable to your players, but if you truly want 2d6, you can prepare two shuffled decks, or reshuffle after the first draw.

  • 1
    \$\begingroup\$ To be more poker alike you could modify mappings: 1-4 as numbers (2-10) in color order (1 = clubs, 2=diamonds, 3=hearts, 4 = spades), 5 - lower color figures (JQKA clubs+diamonds) & one joker, 6 - upper colors figures + second joker. \$\endgroup\$
    – Marek
    Jul 7, 2022 at 8:13
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    \$\begingroup\$ Alternatively, to be easier to remember you could just remove the Kings and Jokers. The first pull affects the chances of the 2nd slightly more, but should be close enough for a PBTA and less mental gymnastics to figure things out. \$\endgroup\$
    – aslum
    Jul 20, 2023 at 16:27

Pull 3, with oppression.

The thing about Urban Shadows is you've got this obvious quadriptych of major city presences -- the factions/circles (depending on edition). It'd be neat if you could lean heavy into that while getting numbers close to 2d6, wouldn't it?

Having each suit match up to one of those and "rolling" 3d4 with a pull of three cards gives you numbers a little bit on the high side. Well, assuming the kings, queens, and knaves of the city have their interests lined up with yours. But they don't, do they?

So here's 3d(card) with oppression. Assign a faction/circle to each of the suits. Also assign a numerical value 1-4 to each of the suits -- this could be absolute for everyone, but it'd be more fun as something variant you printed on each playbook. Your call whether all playbooks of a faction/circle get the same 1-4 spread or if you vary it up a little, but each playbook's "allegiant" suit should always be the one that gets the 4. Pull three cards and add them up, but if you pull a Jack, Queen, or King you've run into the machinations of someone powerful, take -1 for each one you pull.

The numbers line up fairly well with 2d6 (probability breakdown courtesy of anydice), skewing perhaps a bit on the low and complicated side of things, but you also have some interesting hooks for when the nobles show up. Maybe offer a debt so things don't go so badly in the moment. Maybe as you improve faction rating, that means somebody is on your side after all, or at least you can more productively whine to them after one of their discarded spanners lands in your works.

As a practical matter you don't have to reshuffle every time, especially if everybody has a different value system for each suit, but I wouldn't draw the deck to exhaustion, either -- consider cutting in a joker, and replace it in the draw and shuffle when it comes up.


With 2 draws, you can mimic it exactly

Not elegant at all but it works. You draw once from a 54 card-deck (full 52+Jokers), mark what you got, put the card back, shuffle, and draw again.

Split the 54 cards into 6 groups of 9 cards for example:

  • Ace to Nine Clubs = 1
  • Ace to Nine Diamonds = 2
  • Ace to Nine Hearts = 3
  • Ace to Nine Spades = 4
  • Tens and Jacks and Black-and-White Joker = 5
  • Queens and Kings and Color Joker = 6

The order of suites is that of bridge (also, alphabetical).

  • \$\begingroup\$ How much does it really affect the distribution of you draw without replacement? \$\endgroup\$ Jul 4, 2022 at 16:09
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    \$\begingroup\$ @ThomasMarkov I'm not a statistician, probably very little. But if I had not put in the replacement, I sure would be getting comments that without replacement, it is not the same distribution. \$\endgroup\$ Jul 4, 2022 at 16:10
  • 1
    \$\begingroup\$ @thomasMarkov It turns the score that was chosen in the first "roll" from 9/54 to 8/54, and it makes the two "rolls" dependent. In play, I would be happy to draw twice without a reshuffle, but that's a matter for the players and referees to agree on. \$\endgroup\$
    – StuperUser
    Jul 4, 2022 at 16:12
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    \$\begingroup\$ It would skew in favor of odd results, but only slightly, I doubt it would be noticeable in play. \$\endgroup\$ Jul 4, 2022 at 16:16
  • \$\begingroup\$ @ThomasMarkov it certainly reduces the odds of doubles significantly if that’s important \$\endgroup\$
    – Dale M
    Jul 4, 2022 at 23:44

A one draw solution with imperfect mapping.

Obviously with a single draw and no modifications it'll be impossible to get a perfect mapping...

But maybe a "close enough" mapping would be acceptable?

Die roll Probability Ideal (/54) Actual #of Cards Mapping
2 1/36 (2.8%) 1.5 1/54 (1.8%) A (least favorite)
3 2/36 (5.6%) 3 3/54 (5.6%) A(x3)
4 3/36 (8.3%) 4.5 4/54 (7.4%) 2's
5 4/36 (11.1%) 6 6/54 (11.1%) 3's, lower 4's
6 5/36 (13.9%) 7.5 8/54 (14.9%) upper 4's, 5's, lower 6's
7 6/36 (16.7%) 9 10/54 (18.5%) upper 6's, 7's, lower 8's, JOKER's
8 5/36 (13.9%) 7.5 8/54 (14.9%) upper 8's, 9's, lower 10's
9 4/36 (11.1%) 6 6/54 (11.1%) upper 10's, J's
10 3/36 (8.3%) 4.5 4/54 (7.4%) Q's
11 2/36 (5.6%) 3 3/54 (5.6%) K(x3)
12 1/36 (2.8%) 1.5 1/54 (1.8%) K (favorite)

This has the benefit of being no modifications, and reasonably close. Also the 7's line up beautifully.

Optional Changes

Suggested by @Marek

Shifting the mapping to start at 2 = 2, lets you have the top cards being A (as in Poker).

You can also omit the JOKERS, and use just 52 cards. It'll flatten the curve slightly at 7 (making it as common as 6 and 8), but overall wont really hurt anything.


  1. Rounding Up/Down was done somewhat arbitrarily, in an attempt at simplicity.

  2. You'll have to decide which colors are upper/lower (probably black > red if you think finances), and which suit overall is best/worst.

  3. I don't think this is better than the two-card or deck modifying solutions above, I just wanted to present my best attempt at one-draw no-modifications

  4. I think a perfect one-draw, no-modification mapping can be done with 2 identical decks (a 108 card deck), but it would still require some arbitrary decisions and cutoffs.

  • \$\begingroup\$ This is almost exacly what I was thinking but: 1. I would use 52 deck and on 7 give 8 not 10 cards (it will increase lowest probabilities) 2. I would use difrent card order (poker like) 2<3<...<J<Q<K<A, clubs<diamonds<hearths<spades So 2 clubs = 2, and Ace spade = 12 \$\endgroup\$
    – Marek
    Jul 6, 2022 at 20:29
  • \$\begingroup\$ @Marek I added a note about a 52 card option. The curve flattens a bit but overall no real loss, probably makes it more interesting too (also only half my card decks still have the jokers anyways, so it works out). \$\endgroup\$
    – Lintor
    Jul 6, 2022 at 21:12
  • \$\begingroup\$ I was thinking in post I should have started at 2 = 2, and Ace (of Spades?) being the highest, which works for extremes but makes the middle cards already somewhat confusing, for an idea I really don't think is optimal. \$\endgroup\$
    – Lintor
    Jul 6, 2022 at 21:15

Save the 16 picture cards for the D4 roll or other special purpose instead, and draw just one card

There are 36 cards from 2 through to 10. Mapping them to the outcomes of rolling two D6 is an easy exercise if you care about the order. If you only care about the sum then you could say C<D<H<S and use lexicographic order, so 2 would be 2C, 3 would be 2D-2H, 4 would be 2S-3D, 5 would be 3H-4D, 6 would be 4H-5H, 7 would be 5S-7C, 8 would be 7D-8D, 9 would be 8H-9D, 10 would be 9H-10C, 11 would be 10D-10H and 12 would be 10S.

If you make yourself a diagram this becomes quite clear and the symmetries are quite pretty.


You need two draws - another base 54 solution

It's not possible with one draw from an unmodified deck - your probabilities are multiples of 1/36 and you can't make those from multiples of 1/54. You can however make them from multiples of (1/54)^2.

36 can be divided into factors of 54 in two ways - 2 * 18, 6 * 6. Groody's answer focusses on the 6*6.

Alternatively, you can use the first draw to get the 2 (red/black, say), and the second to get the 18. An example:

First draw Second draw Outcome Probability
Red AKQ of Spades 2 1/2 * 3/54 = 1/36
Red J-6 of Spades 3 1/2 * 6/54 = 2/36
Red 5-2 of Spades, A-10 of Hearts 4 1/2 * 9/54 = 3/36
Red 9-2 of Hearts, A-J of Diamonds 5 1/2 * 12/54 = 4/36
Red Any club, Joker 6 1/2 * 15/54 = 5/36
Red 2-10 of Diamonds 7 1/2 * 9/54 = 3/36
Black 2-10 of Diamonds 7 1/2 * 9/54 = 3/36
Black Any club, Joker 8 1/2 * 15/54 = 5/36
Black 9-2 of Hearts, A-J of Diamonds 9 1/2 * 12/54 = 4/36
Black 5-2 of Spades, A-10 of Hearts 10 1/2 * 9/54 = 3/36
Black J-6 of Spades 11 1/2 * 6/54 = 2/36
Black AKQ of Spades 12 1/2 * 3/54 = 1/36

This is one reasonably symmetric way to get it done, you can find others that may be more aesthetically pleasing for your purposes.

  • \$\begingroup\$ To clarify, are you talking about drawing 2 cards from the deck, or drawing 1 card, shuffling it back in, then drawing a 2nd card from the full 54? \$\endgroup\$
    – Oblivious Sage
    Jul 6, 2022 at 13:14
  • \$\begingroup\$ Drawing 2 cards with replacement - easier if you just have 2 decks. I previously claimed this was robust to drawing without replacement - it isn't, sorry. \$\endgroup\$
    – Cong Chen
    Jul 6, 2022 at 14:04

Since there are 36 outcomes of rolling two 6-sided dice and either 52 or 54 cards in a deck it is impossible to reproduce the distribution of the 2d6 roll with a single draw of a card (without modifying the deck). It seems to me that the best we can aim for is to devise a system where we draw a card and get an answer in most cases, but in some cases we may need to draw additional cards without reshuffling the deck or reintroducing the cards already drawn into the deck.

First, let me describe a simplified system with a modified deck which will be a basis for the actual system explained below. Let's take a deck of 36 cards consisting only of ranks A and 2 through 9. Assign value 1 to A and assign values 1 through 4 to all suits, e.g., ♣️ is 1, ♠️ is 2, ♦️ is 3, ♥️ is 4. If you draw one card from this deck, add the numbers assigned to the rank and the suit and then subtract 1 if the result is 8 or higher, you get a distribution that is very close to 2d6. It differs only in that 7 occurs 8 times while 6 and 8 occur 4 times each. What we want is for 7 to occur 6 times while 6 and 8 to occur 5 times each. We can achieve this by modifying the subtraction rule by saying that one the cards that has the total of 8 is not modified by subtraction while one of the cards with total of 7 is. For example let's say that 3♥️ is 6 (we subtract even though the sum is less than 8) and 4♥️ is 8 (we don't subtract even though the sum is 8 or higher).

Let's say that we have the full deck of 54 cards including the jokers. We extend the values described above by setting the remaining ranks as follows: 10 is 1, J is 2, Q is 3, K is 4. Again, the value of the card is determined by the sum of the values assigned to the rank and the suit. In order to keep things uniform with the previous system, let's subtract one whenever we get 8 or more. (This happens exactly once for K♥️.) We also set the value of both jokers to 6. Now imagine that we draw a card from the deck of the 18 cards listed in this paragraph. The distribution is essentially "the lower half" of the 2d6 distribution. In fact, if we had one extra bit of information, e.g., a coin flip, then we could use it to obtain the missing upper half. For example, if we get tails, then we preserve the value of the drawn card, otherwise, we take 14 minus the value of the card. This operation mirrors the lower half of the distribution onto the upper half. We will simulate the coin toss by drawing a second card. Unfortunately, if we don't want to reshuffle we have to deal with the fact the there is an odd number of cards remaining in the deck. Here is how we can do it: we draw a second card from the remaining 53. If its color (red or black) agrees with the color of the first card, we take the original value, if it disagrees we flip it by subtracting it from 14. This rule needs to be modified slightly for the jokers. If we draw a joker as a second card and it has the same color as the first card, we keep its value. If we draw a joker of opposite color, we discard it, draw again and apply the rule to the resulting card.

Here is the system summarized.

  1. The ranks are given the following values: 2 to 9 their rank values, 10 and A are 1, J is 2, Q is 3, K is 4.
  2. The suits are given the following values: ♣️ is 1, ♠️ is 2, ♦️ is 3, ♥️ is 4.
  3. Each card is given the value of the sum of the values of its rank and suits, lowered by 1 if the total is 8 or higher, with the following exceptions.
    1. 3♥️ is 6.
    2. 4♥️ is 8.
    3. A joker is 6.

To simulate a 2d6 roll we proceed as follows.

  1. Draw a card and compute its value as described above.
  2. If the rank of the card is A or 2 through 9, the value is the final result.
  3. Otherwise, draw a second card. If it is a joker of color opposite of the color of the first card, draw a third card.
  4. If the last card drawn has the same color as the first card, the result is the value of the first card. Otherwise, the result is 14 minus the value of the first card.

You will draw one card ⅔ of the time, two cards around ⅓ of the time and three cards ⅔% of the time. I think that this is a fairly satisfying solution to the puzzle, but I don't know that I would recommend to use it in an actual game.

One final note: one can modify the value computing rule by making K♥️ an 8 and one of the jokers a 7. This gives the same distribution, but I'm conflicted on whether this is easier to use or not.

Another final note: if you are OK with reshuffling, you can avoid the potential third draw by modifying the procedure as follows. If the first car drawn is 10, J, Q, K or a joker, return it to the deck, reshuffle and draw a second card. Use the color of the second card to modify the value of the first one as above.


A bit of a frame challenge, but simple and keeps the gambling motif: dominos.

Each domino originally represented one of the 21 results of throwing two six-sided dice (2d6).

Modern sets add a blank or zero to their faces and contain a single permutation i.e. a single domino of (2,3) rather than a (2,3) and a (3,2). 2 sets with the blanks and spare doubles removed gives you a 1:1 mapping to every result of 2d6.


Frame Challange, the mechanic isn't as important as the bell curve. Let's look at the most common situations and aproximate % chances of each result

Bonus 6- 7-9 10+
2d6-1 58 33 8
2d6 42 42 17
2d6+1 28 44 28
2d6+2 17 42 42

You're asking how to simulate this kind nuanced chance with a deck of cards, any kind of mapping of cards to specific results is going to run afoul of the fact that you've got a 1 in 52 chance of any particular card and ANY system you make is likely going to have to be pretty dang complicated and is going to probably require some sort of lookup table which will definitely slow the game down.

The best I can think is to try and mimic the probabilities.

Bonus 6- 7-9 10+
-1 Number (61) J-K (31) A (8)
0 2-6 (38) 7-Q (46) K,A (15)
1 2-5 (30) 6-10 (38) J,Q,K,A (30)
2 2-3 (15) 4-9 (46) 10,J,Q,K,A (38)

This still has the problem of a look-up chart, but it's a LOT easier to remember, the probabilities are close to what you'd get with a pair of 2d6 and you don't have to pull then do math. Ace is Always a full success, and 2 is Always a full failure. Most of the time (+1 or +0) 7-10 will be a mixed success.

As a bonus, it only actually requires 1 suit, so one deck of cards could be split for four players.

Alternatively, if you really wanted to lean into the "cards and suits" theme, you could have abilities tied to a specific suit, so when "diamonds" is drawn it counts as the next higher (or lower) result, though that would certainly muck with probabilities. Without actually doing the math saying "diamonds count as a better success" would reduce chance of failure by 25% of what it would have been for a particular pull. If you want to try this, I'd spend some time with excel figuring out how it modifies chances.


Just use black cards to represent one die, and red cards to represent the other.

[Just noticed the strange requirement that no draws be discarded, so if strictly respecting the question, this answer is invalid. However, it's easy to remember and it's a mathematically identical result to dice rolls]

Ignore Kings. A23456 represent 123456 rolled, 78910JQ also represent 123456. Once you draw a valid card (ie anything except a King) a subsequent draw of the same colour is ignored. Just keep going until you get a valid card of the other colour. There's your 2d6.

For 3d6 you'd count the first spade, the first heart and the first diamond drawn, and ignore clubs. For 4d6 you'd use each suit to represent one die. For d4 rolls you'd assign blocks of 4 (so A,5,9 all count as 1 rolled).

You could remove all the high-value cards and use a 24-card deck with A-5 of each suit. But it's slightly harder to shuffle just 24 cards.

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    \$\begingroup\$ The "strange requirement" that no draws be discarded is because having unused or invalid cards is no different than just modifying the deck beforehand, except that it's just more steps. It's what OP said you couldn't do, but worse. \$\endgroup\$ Jul 6, 2022 at 12:22

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