Talk:Mana curve
WickedDarkman- The curves I worked out had nothing to do with the spells you would be casting. Nothing at all. The presumtion was that you would want to cast a spell a turn for as many of the first turns as possible with the maximum mana available for that turn.
Sorry, I tend to get carried away :) Also I had some trouble understanding the details of the whole thing. Manacurves are something I am doing a heavy research on using simulations. I would really like to know in more details what it was that you were trying to write about. If we are lucky we could compare the results??? P.S to use a signature hold down ctrl and alt then pound on the button with ~ untill you have four :) Wickeddarkman 04:58, 24 January 2007 (CST)
No worries. To explain a little. My basic curves are all about expected value, i.e.
60 Cards, Draw first
8 at one, 7 at two, 6 at three, 5 at four, 9 at five, 25 Land.
Comes from: turn one 8 draws/60 cards * n successes = 1: for n is approx 8 one drops/land. Total so far 8 + 8 = 16.
turn two 9 draws/60 cards * n successes = 1: for n is approx 7 two drops.
Also turn two 9 draws/60 cards * n successes = 2: for n is approx 13 land. Total so far 8 + 7 + 13(using each land number as a minimum) = 28.
turn three 10 draws/60 cards * n successes = 1: for n = 6 three drops.
... 10 draws/60 cards * n successes = 3: for n = 18 land. Total so far 8 + 7 + 6 + 18 = 39.
turn four 11 draws/60 cards * n successes = 1: for n is approx 5 four drops.
... 11 draws/60 cards * n successes = 4: for n is approx 21 land. Total 8 + 7 + 6 + 5 + 21 = 47.
turn five 12 draws/60 cards * n successes = 1: for n = 5 five drops.
... 12 draws/60 cards * n successes = 5: for n = 25 land. Total 8 + 7 + 6 + 5 + 5 + 25 = 56.
turn six 13 draws/60 cards * n successes = 1: for n is approx 5 six drops
... 13 draws/60 cards * n successes = 6: for n is approx 28 land. Total 8 + 7 + 6 + 5 + 5 + 5 + 28 = 64
Since carrying this algorithim into turn six results in too many cards and turn five doesn't is still four short we replace turn six's calculation with:
turn six 13 draws/60 cards * n successes = 2: for n is approx 9 five drops. New total 8 + 7 + 6 + 5 + 9 + 25 = 60 cards!
Does that clear things up? I'm going to post an example page. BTW, I upgraded your discussion of simulation to it's own section, it deserves it. --Fishysua 08:28, 25 January 2007 (CST)