HigherPOV said:

Okay I can understand now if using the example P30. But isn't that a special case because you can only draw from a pool of the same 16 face cards to achieve a sum of 30. But how would you write that formula for say P20. With P20 couldn't any card can be the 1st card and still have the total ending up being 20. Would you need a different equation to explain P20. Thats the part that I cant get. It looks like you'd need advanced math to explain right?

Yes, of course, each P value has a different equation. Let's look at the other end to get a flavor for what is required.

In a similar fashion to P30, P3 is given only by A,A,A so P3 = (4/52)*(3/51)*(2/50) = 0.018%

Looking at P4 we see we have combinations AA2, A2A, and 2AA that add to 4. The denominator D is always 52*51*50= 132600 so let's just refer to the numerator, always dividing by 132600 (D) to get P4. The numerator is 4*3*4 + 4*4*3+ 4*4*3 = 3*48=144. P4 = 0.109%

For P5, we have the cases A22 (2A2 and 22A) as well as AA3 (A3A and 3AA). Numerators in this case are all 4*4*3 because we have one singleton and a doubleton so overall = 6*48=288. P5=0.217%

Slightly more complicated is P6's numerator, made up of AA4, A23, 222 which are 4*3*4 (3 combos), 4*4*4 (6 combos), and 4*3*2 (1 combo) or 3*48+6*64+1*24 = 552. P6 =0.416%

Now let's see what P20 would look like. We know the first few terms look like A,9,10; A,10,9; 9,10,A; etc. and 2,8,10; 2,10,8; 10,8,2; etc. and 2,9,9; 9,2,9; etc. Let's write the numerator as W(A,T(19))+W(2,T(18))+...+W(10,T(10)), where the W function indicates combinations of the first card with the T function indicating the total of last two cards.

**P20 = (summation [W(I,T(20-i)] from I = 1 to 10) / D**

we can see that there are a lot of terms involved in computing P20. If we had plenty of time, we could tediously write them all down. This is why I wrote a program ( a bit tricky because after the first and second cards are counted they need to be remembered) to automate the computation.

Hopefully this gives a flavor of how to attack problems of this type.