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I am horrible at math, lets say on a d20 a four or less is a miss, how accurately could this be translated to represent this roll on a d6 on a d10 and a d12?

So that's 4 in 20 or 1/5

Closest on a d6 is 1/6 (so any single number)

Closest on a d10 would be 2 /10 (so any two numbers, and exactly the same)

Closest on a D12 is either 2 or 3 numbers (giving 1/6 or 1/4)

ok cool now what about a 15 or better on a d20, I think it would be the same as stated previously, correct,

So that's 4 in 20 or 1/5

Closest on a d6 is 1/6 (so any single number)

Closest on a d10 would be 2 /10 (so any two numbers, and exactly the same)

Closest on a D12 is either 2 or 3 numbers (giving 1/6 or 1/4)

Just work out the percentages for success.
It's not hard.
Divide 100 by the number of faces.
To get the chance of success multiply that number by the number of faces that would succeed.
So 15 or better on a d20.
6 chances to succeed on one dice 15,16,17,18,19, 20.
100/20=5
5*6=30
you therefore have a 30% chance to succeed

Every tick on a die is worth 1 divided by the number of faces percent.

So on a d20, 1/20 = 5%.
d6, 1/6 = ~17%

So to equal the 'probabilty' of 1 result on a d6, you would capture between 4 and 5 results on a d20.