I tried, Zeb, but I saw no really easy way to do it (I'm lazy). I only came so far:
I assume you mean drawing without replacement (usual, but not totally necessary in lotteries) and I assume you mean 'at least two consecutive numbers' (for that makes it much more easy).

Then you only have to multiply the five probabilities to get no consecutive number in the first draw (1, obviously), in the second, in the third, in the fourth, in the fifth. And subtract what you get from 1.
My problem were the too many possibilities (cases) to consider. E.g., the second of the above probabilities is 29/31 (for both the number one larger than the first number and the number one smaller than the first are taboo if you do not want to have consecutive numbers). But that is only true if the first draw was not the 1 or the 32. And it gets more complicated later with more different cases (like, e.g., two numbers already drawn that differ by two) to carry on. I did a rough calculation (leaving out all the more difficult cases) to come to a probability for at least two consecutive numbers of more than 50%. But my rough calculation (1 - 29/31 x 26/30 x 23/29 x 20/28 = .54) necessarily leads to a too large probability. And I don't know how far off I was, though I thought not too much and am therefore not completely surprised by your figure above.
Most times however, there is an easy way to do it and I'm looking forward reading about it.

Wolfgang