I tried to calculate this probability like this:

1. Number of flops that have a flush draw:

C(13,2)*C(39,1) = 3042

2. Number of flops that have a double gutshot straight draw and don't have a flush draw:

This part is the tricky one and I wrote a small script to generate all possible patterns of flops where a double gutshot is possible and they look like this:

x : board (flop) cards
? : outs that give us a straight
o : our hole cards

("x?xxo?o" "x?xox?o" "x?xoo?x" "x?oxx?o" "x?oxo?x" "o?xxx?o")

(BTW this is a brute force script that generates all the permutations of the multiset {x,x,x,o,o,?,?} and identifies the ones that are double gutshot straight draws)

So the pattern "x?xxo?o" which is essentially a flop with one gapper can occur (14-4+1)*C(4,3)=11*C(4,3)=44 times, which is (11 ways to spread the one gapper pattern across all ranks) times (choose 3 suits from the 4 available) so we always have rainbow flops.

Repeating the calculation for all patterns yields:
11*C(4,3)+10*C(4,3)+8*C(4,3)+10*C(4,3)+8*C(4,3)+12 *C(4,3) = 236 flops that have double gutshots.

So the answer is (3042+236)/C(52,3) ~14.8%

Could someone please confirm this calculations are correct?

If this calculation is correct it represents the probability that a given flop is wet since all other types of straight draws are contained in the patterns above (gutshots and open ended straight draws).

Let's fix the flush-draw part first since that is easiest.

If we are looking at xxy flops, there are 13C2 ways to pick the two x cards, and 13 ways to choose the card with suit y. We have four different ways to choose suit x, and for each of those, we have 3 ways to choose suit y. Therefore, our answer is:

13C2 * 13 * 4 * 3 = 12168 flops, or about 55% of the time we have a two-tone flop.

- bachfan

OP's equation is correct for a single suit. Multiply his answer by 4 for any suit and you have a slightly simpler form for getting the result.

2. You are making it way harder than it is, and as a result you came up with a way wrong answer.

There are 11 rank combinations that make a double gutshot flop. Just enumerate them to see this and don't forget the wrap QA3. Thus the total number of double gutshot flops using any suit combination is
11 * 4^3 = 704
704 / C(52,3) = 3.186%

But the number that are rainbow suited is only:
(11*4*3*2) = 264

So

264 / C(52,3) = 1.195%

that are double gutshot rainbow, i.e. no flush draw or flush.

If you wanted to only exclude two-tone flops and count the 44 monotone flops then you would have 308 instead, or 1.394%.

And you now have enough information to answer the title question, if you want to know the combined chance that a flop contains either a gutshot, flush draw, or both. Just deduct the "both" from one of them so you don't double count it.

After posting this I looked at your post again and thought about what you might be asking for when you ask about double gutshot flops. The other interpretation is that you want to know how many flops make it possible that some player has a double gutshot straight draw. But that isn't a useful question because EVERY unpaired flop fits that description. It isn't possible for an unpaired flop to not have a double gutshot draw for some set of hole cards.

Unpaired flops happen 82.824% of the time.

K72 is an unpaired flop and has no gutshot draw.
http://www.spadebidder.com/flop-analysis/part7/
Here you put AQ8 under the "Other 1 gaps" category while it should be under "Double gutshot" instead (you can make a straight on the turn with either a K or a 9 holding JT).
Also AQ3 can't be considered a dbl gutshot because you can't hold KJ and 24 at the same time unless it's Omaha.
I guess this makes some of your calculations wrong, specifically the dbl gutshot and other 1 gaps categories.

Can someone shed some light on this?
I know this is a very old thread but I really need to figure this one out.
All of spadebidder's calcs are right except for this one.
I'm working on this but I need some help.

For his analysis Spadebidder was trying to classify each possible flop. He called the following flops "double gutshots" because they have two one-card gaps between the three different ranks on the flop. Not really directly having anything to do with flopping a gutshot in the normal poker parlance:

A35
246
357
468
579
68T
79J
8TQ
9JK
TQA
QA3

These categories may not be perfect and his labels may not be perfect, but it is pretty clear what he did, why he did it, and what results he found.

Thanks for the explanation, I guess that explains everything. So his calcs are on point.
I really appreciate his work. I think those frequencies are very useful as they give some insight on what types of flop one should focus his studies (i.e focus more on high frequency flops).

It's sad to see he's not active anymore.