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.
- 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.
- The suits are given the following values: ♣️ is 1, ♠️ is 2, ♦️ is 3, ♥️ is 4.
- 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.
- 3♥️ is 6.
- 4♥️ is 8.
- A joker is 6.
To simulate a 2d6 roll we proceed as follows.
- Draw a card and compute its value as described above.
- If the rank of the card is A or 2 through 9, the value is the final result.
- Otherwise, draw a second card. If it is a joker of color opposite of the color of the first card, draw a third card.
- 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.