[ipgyst]

I don't see any flaws in the previous answer. Here's how I did it:

First card dealt not an ace = 48/52 (or 12/13)
Second card dealt not an ace and doesn't match the first card = 44/51
Odds of being dealt a non-ace, non-pair hand are 12/13 * 44/51, or 528/663, or 176/221

If you take 176/221 to the 58th power, you get the odds of getting ZERO aces or pairs in 58 hands. This is .0000018407, or 1 in 543272.

First card ace (1/13) second card non-ace (48/51) multiplied is 16/221
First card non-ace (12/13) second card ace (4/51) multiplied is also 16/221
Odds of being dealt one and only one ace are 32/221

So you get (176/221)^57 multiplied by (32/221)^1 multiplied by 58 (because the one-ace hand could occur in any of the 58 hands). This is .000019411, or 1 in 51517.

So if you sat down to play 58 hands of holdem about half a million times, on average one of those times you would get zero aces or pairs, and on average ten times you would get no pairs and one ace. Not at all likely, but possible.