Serum Powder math

I couldn't find any useful sources by googling, so here's my question:

You run 4 copies of Leyline of the Void. I want to increase the chance to get it in my opening hand, but don't have the SB space to run a full playset of Serum Powder.

How much would the chance to lay down a Leyline on T0 increase, if you run a) 2 copies and b) 3 copies of Serum Powder, assuming you don't want to mull down lower than 5 (or 4 cards) total?

Suppose that you are willing to mull to 5 for a leyline. So the scenarios are:

Code:

---

Leyline in first 7
Serum Powder, but no Leyline in first 7
  Leyline in next 7
  Serum Powder but no Leyline in next 7
      Leyline in next 7
      Serum Powder but no Leyline in the next 7
....
(Mulligan)
Leyline in the first 6
Serum Powder, but no Leyyline in the first 7
...

---

The probabilities change cards get exiled, so it's a real chore to make an exact calculation.

When I do lazy estimation, where a serum powder is just a reroll at the current hand size, I get cumulative probabilities of:

EDIT: There was an error here earlier.

Code:

---

4 0.50163757811486        0.71825642761601        0.82167362594705
3 0.48055196939506        0.6968977984162        0.80364752865955
2 0.45677458083869        0.67212521490137        0.78232580423309
1 0.43002501871277        0.64337820484125        0.75704573098066
0 0.4        0.61        0.727

---

Where the columns are having at least one leyline by 7,6,5 cards in hand respectively, and the rows are 4 to 0 powders.

For example:
You have a 0.4 chance to hit a 4-of in the first 7. There's a 0.25 chance to get a 4-of powder and no leyline, so the estimate of the combination hitting a leyline in 7 is 0.4 + 0.25*0.4 = 0.50