Hearthpwn

Can someone easily explain the chances to discover a certain card in my deck?

for example, if i discover a spell in my deck using Shadow Visions and i have 8 spells in my deck, how high are the chances to discover that 1 spell in my deck, and if there are 2 perfect spells in my deck to discover, how high are the chances to discover 1 or 2 of those?

thank you for explaining... 

If you have 8 single spells left in the deck, the % chance is 3/8 x 100

Duplicates increase or decrease this chance.

Quote from MrBurger >>

Are there duplicates of any spell in your deck?

 sometimes yes, sometimes no...

but I want to know how to calculate rather then have a direct answer to the scenario above... 

so dont tell me the chance is x% but tell me how to calculate it myself so I can calculate it myself...

Quote from Devilseye >>

If you have 8 single spells left in the deck, the % chance is 3/8 x 100

Duplicates increase or decrease this chance.

 is it realy that simple?

Discover is always three different options, so duplicates will not really count as a + to the nominator. I'm actually not sure if discover only looks at unique cards, giving single and duplicate cards the same chance or it generates one of the three options at a time and then excludes duplicates from the pool. so  3/8 or 1/8+1/7+1/6 I can't tell 

Much hangs by the detail of exactly how the Shadow Visions treats duplicates, but let's give some principles.  Perversely, it is often easier to work out the chances of not getting the spell that you want, then your chance of getting the spell that you do want is 100% minus the chance of net getting it.  So if I have 8 spells and want just one of them, the chance of not getting it first draw is 7/8.  Assuming that the 7/8 chance came up, there are now 7 cards left, 6 of which I don't want, so the odds of not getting it with the second draw are 6/7.  If I was 'unlucky' twice, the chances are 5/6.  So my chance of not getting the card I want are (7/8)*(6/7)*(5/6) which comes to 210/336.  So my chance of not getting the card I want is 62.5% , making my chance of getting it 37.5%.  

Quote from RuthnAbi >>

Much hangs by the detail of exactly how the Shadow Visions treats duplicates, but let's give some principles.  Perversely, it is often easier to work out the chances of not getting the spell that you want, then your chance of getting the spell that you do want is 100% minus the chance of net getting it.  So if I have 8 spells and want just one of them, the chance of not getting it first draw is 7/8.  Assuming that the 7/8 chance came up, there are now 7 cards left, 6 of which I don't want, so the odds of not getting it with the second draw are 6/7.  If I was 'unlucky' twice, the chances are 5/6.  So my chance of not getting the card I want are (7/8)*(6/7)*(5/6) which comes to 210/336.  So my chance of not getting the card I want is 62.5% , making my chance of getting it 37.5%.  

 oh wow, ok thanks a lot... now i need to speed up my calculations... :)

It strikes me that this expression (7/8) * (6/7) * (5/6) can be simplified. The 7s and the 6s cancel, and you'd be left with 5/8.

Generalizing, you'd have (n-1)/(n)*(n-2)/(n-1)*(n-3)/(n-2) or (n-3)/(n) being the chance of not getting the card you want, with n representing the number of candidates.

But since we prefer the chance of getting the card we want, we could write this as 1 - (n-3)/n. This also can reduce to 1 - (n/n - 3/n), or just 3/n.

Wow, I guess is that simple. I hadn't initially trusted it, but doing the math out... yeah, checks out.

I didn't read anything other than fig's last post, which is correct based on the applicable assumptions.

However, I do remember a thread similar to this when shadow visions first came out where someone had theorized that, essentially, having two copies of a spell in your deck DID NOT affect the possibility of discovering it.  In other words, shadow visions selects from a list of the different kinds of spells in the deck, but said list only contains one instance of each spell regardless of whether or not multiple copies of said spell appear in the deck.

So, it would take a fairly large sample size to say with any relevant level of certainty whether or not doubles affect the chance of discovery.  You can only be certain that the simpler calculations are correct if there is ONLY ONE copy of the spell in the deck at the time.