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Thread: Serum Powder math

  1. #1
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    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?

  2. #2

    Re: Serum Powder math

    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
    Last edited by rufus; 04-08-2016 at 04:27 PM.

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