TL;DR: very very bad.
Stakataka has three of six IVs that are guaranteed to be perfect. 3 IVs * 6 IVs = 18 different ways the Pokemon can have three randomly-generated perfect IVs. They are listed here, assuming x is a random value:
31/31/31/x/x/x
31/31/x/31/x/x <--- this works if the first x is 15, the second x to last is 31 and the last x is 0
31/31/x/x/31/x <--- this works if the first x is 15 and the last x is 0; lacks perfect SAtk
31/31/x/x/x/31
31/x/31/31/x/x
31/x/x/31/31/x <--- this works if the first x is 31, the second x is 15 and the last x is 0
31/x/x/x/31/31
x/31/31/31/x/x
x/x/31/31/31/x
x/x/x/31/31/31
x/x/31/x/31/31
x/31/x/x/31/31
x/31/x/31/x/31
x/31/31/x/x/31
x/31/31/x/31/x
31/x/31/x/31/x
31/x/x/31/x/31
x/x/31/31/x/31
We'll call the three possibilities that would work possibilities a, b and c. We'll calculate them one-by-one:
For a, 1/18 perfect IV options would suffice, and you'd need a 1/32 chance to go in your favour three times. You'd also need the coin toss for Synchronise to work. So:
Pr(a) = 1/18 * 1/32 * 1/32 * 1/32 * 1/2 = 1/1179648
For b, 1/18 perfect IV options would suffice, and you'd need a 1/32 chance to go in your favour twice. You'd also need the coin toss for Synchronise to work. So:
Pr(b) = 1/18 * 1/32 * 1/32 * 1/2 = 1/36864
If you're really going for the imperfect SAtk IV, you'll need to add an extra 31/32 chance (we'll call this bA):
Pr(bA) = 1/18 * 1/32 * 1/32 * 31/32 * 1/2 = 31/1179648 or roughly 1/38053
For c, 1/18 perfect IV options would suffice, and you'd need a 1/32 chance to go in your favour three times. You'd also need the coin toss for Synchronise to work. So it's the same as a:
Pr(c) = 1/18 * 1/32 * 1/32 * 1/32 * 1/2 = 1/1179648
So your chances will be as follows:
First, if you'll allow the SAtk IV to be of any value:
Pr(a + b + c) = 1/1179648 + 1/36864 + 1/1179648 = 17/589824 or roughly 1/34696
Second, if you'll only allow the SAtk IV to be imperfect. You will only accept option bA. So your chance is:
Pr(bA) = 1/18 * 1/32 * 1/32 * 31/32 * 1/2 = 31/1179648 or roughly 1/38053
If you want these rarities in context, the first option is over eight times rarer than finding a random Shiny, and the second over nine times rarer. Long story short, you'll be sat down soft resetting for a very long time if you want this particular Pokemon. :P
