OK, so, if you don't know anything about combinatorics at all, I recommend that you learn some, it's probably the most useful thing to know about poker ever. The 2nd most useful thing if you're interested in problems marginally more complicated than this is basic programming. I recommend something like python, ruby, etc. Razz is very easy to write programs for because there are no suits, it makes things remarkably less complicated.
OK, so, let's start with, how many combinations of hole cards ARE there? Well, there are 52 cards and 3 slots, which means that there are C(52,3) distinct combos. Note that we consider A23 and 32A to be the same here. This makes sense if you are considering combos prior to seeing door cards. Google can calculate this for you by the way, "search" for "52 choose 3" and you get 22,100
OK, so, let's consider a particular unpaired hand, such as A23. Any given unpaired hand is going to have the same number of combinations, i.e. there are just as many A23 as 25K or 379 or whatever. Well, you have 4 choices for the first slot, 4 choices for the 2nd and 4 choices for the 3rd. That means there are 4*4*4 combos of A23, which is 64.
So the chance of getting any particular combo is 64/22,100.
So you can count how many 3-card wheels there are. These would be A23 A24 A25 A34 A35A45 235 235 245 345. So total there would be 10*64, i.e. the chance is 640/22100 or about 2.89%
3-card sixes add the following hands: A26 A36 A46 A56 236 246 256 346 356 456, so 10 more sets for a total of 20 sets, which comes out to 1280 combos. About 5.79% of hands dealt.
You can continue on in this way. Counting which combos you're adding is tedious but there's a formula to it. If you want to know how many combos there are of 3 unpaired cards, that are "X or under" (such as 9 or under) then think of it this way:
The first card can be any card X or under, there are X*4 of these.
The second card can be any card X or under, except the first one you chose, there are (X-1)*4 of these
The third card can be any card X or under, except the 1st or 2nd you chose, there are (X-2)*4 of these.
You multiple these 3 numbers together and then divide by 6. Why divide by 6? Because the method above will produce both hands like A23 and 32A and we have already noted these are the same hand. Each logical combination will be produced 6 times, i.e.
So for our 3-card-6 example this would be (6*4)*(5*4)*(4*4)/6 = 1280
You can re-order this as (6*5*4)*(4*4*4) and then you can easily see the relationship to finding out how many combos under, say, 7 there are compared to 6. Consider
(7*6*5)*(4*4*4) = (6*5*4)*(4*4*4)*N
where N is the factor of how much larger 3 card 7s or better are than 3 card 6s or better. So, clearly this is the same as
(7*6*5) = (6*5*4)*N
7 = 4N
N = 7/4 = 1.75
Whew, so there's some tl;dr for you. Now a little chart. Divide the combos by 22,100 to get the probability of getting dealt one.