Okay I see I've got a lot of reading to do to catch up on all the posts. Meanwhile, yes, let the dealer be a computer. Here's another way to do it:
Scenario A
1. The computer shuffles a deck.
2. It turns over the top card. If it's not an ace, put it on the bottom of the deck.
3. Repeat step 2 until an ace is found. Deal it.
4. Reshuffle the whole deck.
5. Deal the next card.
Scenario B
Same as A except skip step 4.
Scenario C
1. The computer shuffles a deck.
2. It turns over the top two cards. If neither is an ace, put both on the bottom of the deck.
3. Repeat step 2 until at least one ace is found. Deal both.
Compute your odds of getting two aces for all 3 scenarios. I think A is 1/17, C is 1/33, and B is better than 1/17 [perhaps 3 out of (52-7.5) since the computer had to go through on average 6.5 cards to find an ace?]. I hope this version of the problem is value-neutral and clear. -Carl