Talk:Primordial Glyph

Probabilities
Since the same card cannot show up more than once in a given discover, the chance of a specific card (in standard) is (1 - (30/31 * 29/30 * 28/29)), or 3/31. As a more detailed explanation, if you want Ice Block, then the first has a 30/31 chance of not being Ice Block. If the first is Flamestrike, then the second selects from the remaining 30 spells, of which 29/30 are not Ice Block. If the second is Frost Bolt, then the third selects from the remaining 29 spells, of which 28/29 are not Ice Block. --67.247.179.113 00:33, 13 May 2017 (UTC)
 * You are totally right. Probably in order to prevent similar mistakes in the future, &#32;- jerodast created a template that does the computation automatically; I have included that template in place of the fixed values here. -- BigHugger (talk) 10:41, 13 May 2017 (UTC)
 * You're quick! I don't think I've actually been fully awake since starting work on those templates :) Just wanted to mention, as long as the calculation is correct we probably don't need to replace all such probabilities with the template right away. However, if I can get auto-counting templates working, then when the next set comes out and we need to update card counts for those kinds of notes, it will probably be a good choice to prevent us from having to manually change it again later.&#32;- jerodast (talk) 16:09, 13 May 2017 (UTC)
 * I noticed both your work and the recent change by 67.247.179.113 on the recent changes page, and figured I'd stick it in right away before we can forget. ;)
 * I immediately realized that this should work with a dynamic counter rather than a fixed number in the template, but I had no idea how to do that. Hoping that someone will pick that up soon. -- BigHugger (talk) 16:20, 13 May 2017 (UTC)
 * If you make a template whose only output is a number then you can put that in one of the fields of the other template --67.247.179.113 22:25, 13 May 2017 (UTC)
 * For future users' references, the templates being discussed are: DrawChance and DiscoverChance. Aegonostic (talk) 22:42, 13 May 2017 (UTC)

Probabilities2
The chances stated are not correct since they don't include the chance of discovering primordial glyph and then getting the spell you want. The correct chance for standard would be 10.6% or 5/47. 84.107.86.200 23:43, 12 June 2017 (UTC)
 * Very good point. Because a discovered Primordial Glyph would be discounted to 0 mana, you basicall get infinite retries until you either find what you need, or find neither what you need nor another Glyph.
 * I want to change the source but I need to put in a formula, not a fixed number (so that the number automaticall changes as new eexpansions are relased). I struggle to find the correct formula, in terms of number of qualifying cards. The number you quote, 5/47, does not have any obvious link to the relevant base numbers (31 cards, 3 options per discovery, 1 target spell, and 1 spell that allows a retry).
 * Here's what I have so far (based on the currently available 31 spells in standard):
 * Chance to find the spell you want: 3/31 (as explained above)
 * Chance to find neither the spell you want nor another Glyph: (28*27)/(31*30). Based on logic: First card has to be one of the 29 other cards out of 31, then one of the 28 other out of 30 remaining, etc; so 29/31*28/30*27/29, which can be simplified to (28*27)/(31*30).
 * Chance of a retry (i.e. not find desired spell but find Glyph): Remainder, so 1 - 3/31 - (28*27)/(31*30). Make denominators equal: (31*30)/(31*30) - (3*30)/(31*30) - (28*27)/(31*30). Add numerators: ( (31*30) - (3*30) - (28*27) ) / (31*30). First terms have common multiplier: ( (28*30) - (28*27) ) / (31*30). New common multiplier: (28*3)/(31*30).
 * So, the first attempt has a chance of 3/31 of success, and a chance of (28*3)/(31*30) of retry. The second attempt then has a (3/31) * (28*3)/(31*30) chance of succes, and a chance of (28*3)/(31*30) * (28*3)/(31*30) of a seecond retry. And so on.
 * The final chance of finding the required card would be expressed as a summation: SUM (for n = 0 to infinity) of (3/31) * ( (28*3)/(31*30) ^ n ) (where ^ represents "to the power of").
 * I am sure that there must be some way to simplify this indefinite sum into a simple expression, but this is where my math skills give up on me. (Hey, don't judge me, it's been over 35 years since I left high school).
 * If anyone is able to explain the last step and express the chance in terms on the number of cards currently in standard (or in wild), then I can try and pop that formula into the page source. For now, I will just add a note that the actual chance is "slightly higher".
 * -- BigHugger (talk) 13:20, 13 June 2017 (UTC)


 * The last equation you got is indeed correct except for 1 minor typo. The "^n" should be cover one more set of brackets. So now we have:
 * SUM (for n = 0 to infinity) of (3/31) * ( (28*3)/(31*30) )^n = (3/31) * SUM (for n = 0 to infinity) of( (28*3)/(31*30) )^n
 * This is a form of the geometric series and since |(28*3)/(31*30)| < 1 we can say
 * (3/31) * SUM (for n = 0 to infinity) of( (28*3)/(31*30) )^n = (3/31) * (1/(1-(28*3)/(31*30))) = 5/47
 * 84.107.86.200 23:06, 18 June 2017 (UTC)