Talk:Gold foil: Difference between revisions
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1 mettle run gives no foil at all, which fits Muh's formula. [[User:Edivad|Edivad]] 08:04, 7 October 2010 (PDT) | 1 mettle run gives no foil at all, which fits Muh's formula. [[User:Edivad|Edivad]] 08:04, 7 October 2010 (PDT) | ||
11 mettle gives 3 foil. --[[User:Penguinpyro|Penguinpyro]] 02:07, 25 November 2010 (PST) | 11 mettle gives 3 foil all the time. Which confirms Muh further... I guess. --[[User:Penguinpyro|Penguinpyro]] 02:07, 25 November 2010 (PST) |
Latest revision as of 10:43, 25 November 2010
So yeah, I obtained 3 of these at the start of my next retcon after finishing a 12-mettle Naturalist run (no food or sidekick restrictions, 5 pulls/day, no skills). --Prestige 01:58, 30 September 2010 (PDT)
16 mettle, 4 gold foil. Strlikecrazy 1/10/2010
10 mettle = 2 gold foil jpsteel 2 Oct 2010
From current data my guess is you get number of Gold foils equal to 1/4 of mettles your previous run was worth, with minimum of 1. --Shikao 02:13, 2 October 2010 (PDT)
3 foils from 12 mettle run --Shikao 10:58, 5 October 2010 (PDT)
5 foil from a 20 mettle run, rawr. Cristiona
4 foil from a 15-mettle run. --Prestige 03:04, 6 October 2010 (PDT)
With that last data point, the only formula I can think of that seems to fit is floor((difficulty+1)/4). --Muhandes 08:46, 6 October 2010 (PDT)
I will try to do 7 and 11 mettle runs in nearest future to check it out for sure. --Shikao 09:01, 6 October 2010 (PDT)
1 mettle run gives no foil at all, which fits Muh's formula. Edivad 08:04, 7 October 2010 (PDT)
11 mettle gives 3 foil all the time. Which confirms Muh further... I guess. --Penguinpyro 02:07, 25 November 2010 (PST)