Yes I have.

The odds of drawing 1 out of 12 if 1 out of 12. It does not matter is I have all 4 suits, or 3 of the 4 suits, or just 1 of the four suits.

The odds of picking 1 out of 12 is 1 out of 12.

We happened to know, AFTER THE FACT, that the other card was the same.

Put another way: Say I draw 4 cards out of the deck, and they are all tens. What are the odds the next card I draw will be a ten?

You are saying 1 out of 48 (52-4) and I am saying 0 chance (of the 12x4 remaining, none are tens.)

The "specificity" is accounted for.

As this is basic combinatorics, I think we can leave it off here.

But just so you know, if you click here I showed how to count all of the Gothic Chess positions before any one piece comes off of the board. That number is 32,099,674,107,692,140,366,789,953,222,888,490,987,180,838,400,000,000 which makes doing "card math" a piece of cake :)