Two from List 1 and One from List 2, with Overlap

What is the criteria for an item to appear in both sets? Do you know, at the time the first list is made, which items are carryovers? Must the lists be assembled simultaneously? If not, why not just arbitrarily exclude any selected would-be carryover? Why carryover an item that cannot be chosen?

WHY?! WHY?!

Ok, hopefully I understand this:

1. Six cards: (A-F). Use A,B,C,D face down to determine first two random *things*. Pick two cards.
2. If C & D are chosen, then AUTOMATICALLY choose one from face down E & F.
3. Else, place E and F face down with C or D or both, depending on what was chosen, then chose final card.

Alternate
1. Place four dice size labeled cubes (A thru D) in a bag. Choose two.
2. Empty bag and place E and F cubes in the bag along with either C or D cubes or both as determined by what was chosen in step 1.
3. Choose last cube from bag.

It would help to know the game a bit to come up with a more thematic solution. Hope I grasped what were asking.

Cheers

Edit: I didn't see sedj's solution before I posted; I think we have the same solution.

Please note that Nando did not propose a solution. Nando merely rephrased the problem.

Another idea.
There's only 18 possible combinations.
You could have a deck of 18 cards with all 3 picks on each card. So you just draw a single card.
Or you could use a custom 20-sided die, with 2 re-roll spaces, or some other creative event.

EDIT:
A custom 20-side might be expensive. You could use a normal d20 and consult a chart, though you specifically said you didn't like charts.
Alternately, you could use 3 custom 6-siders. Dunno if that would be cheaper, but the spaces are bigger for the three icons or whatnot. So you'd blindly pick one of the dice from a bucket and roll that.
You could even cleverly distribute the choices among the dice sides and let the player choose one, each die perhaps giving slightly different + or - chance to roll certain yields. e.g. The red die has a very good chance to roll C or D, but can't roll A, while the blue die has a good chance for A and B, but rarely gives C and D and never together. Or something.

Have a track with 6 spaces numbered 1-6.
Put the tiles on the track in order: ABCDEF
Roll a d4. Remove the tile on that number from the track. Slide the remaining tiles back to close the gap (thus leaving the number 6 slot empty).
Roll a d3. Remove the tile on that number from the track. Slide the remaining tiles down.

Remove A and/or B, if they're still there. Slide the remaining tiles down.

Roll either a d3 (if there are 3 tiles left) or a d2 (if there are only 2 tiles left).

For the d3's and d2's you can use d6 and divide the result by 2 or 3 accordingly, or have custom d6's for that purpose.

Actually a d12 will handle all those rolls, but you'd have to divide by the right number every time.
Or, you can use a d4 for everything and reroll on the 4 if you're going for a d3.

Ok, now that I have it written down, it's not as elegant as I thought. Oh, well.

Wouldn't that force a player to stop after only 2 tiles, if they drew both C&D?

You could use 4 dice, roll them all at once.
Die 1 is a d4 with these sides: a green C, a blue D, a red E, and a red F (can use little symbols instead of colors for the color blind, but that's probably more expensive and would still have to correspond to the three d6's)
Die 2 is a red d6 with these sides: AB, AC, AD, BC, BD and CD
Die 3 is a green d6 with these sides: AB, AD, BD (each side appears twice)
Die 4 is a blue d6 with these sides: AB, AC, BC (each side appears twice)

Roll them all at once and look at the d4 and the d6 that matches the color of the letter on the d4. The letters on the d4 plus the ones on the matching d6 are your categories.

Hm, I wonder if the probabilities would skew to certain combinations?