Just going to post that this is inaccurate.

1,073,741,824 is just over 1 billion (not trillion).

Second, the calculations you posted for all of your calculations assume that if you are looking for a 2 IV Pokemon there are only 2 stats. There are 15 possible combinations of 2 IVs existing. That means that there are 13,852,815 possible combinations out of the 1,073,741,824

So about 1 in 78 to get a 2 IV Pokemon.

Now you just need to remember that these are the "binary" (maxed or not maxed) possibilities.

1 outcome for zero IVs

6 outcomes for 1 IV

15 outcomes for 2 IVs

20 outcomes for 3 IVs

15 for 4 IVs

6 for 5 IVs

1 for 6 IVs

Aside from the 6 IVs scenario, you just need to use

z=x*31^y

where

x = the number of binary combinations for desired IVs

y = 6 - number of maxed IVs desired.

You're chances are z/32^6

The guy above me, Le Scraf, the chance for a 0 IV Pokemon is only

1 in 1,073,741,824