Serum Powder math

[Barook]

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?

[rufus]

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


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.



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