For the people who are good at statistics (I am clearly not one of them), can you explain to me the probability of getting a certain card from the Discover mechanic. For example, let's say we're talking about Netherspite Historian and the probability of getting Deathwing. We know there are 14 dragon cards in Standard and 3 unique choices are presented every time from Discover. Can you show me how to calculate this probability? Thanks :)
While I can't dispute the math on this (I'm not a clever man) I was curious as to whether the discover rate is effected by the rarity of the card as well? Are you less likely to find a [card]Deathwing[/card becasue it's a legendary, therefore throwing this off a little?
I don't have sources for this, but I don't think there is differentiation in rarity for discovering effects.
Err, you can't add probabilities like that. Also, what is the division by 3 at the end for? You can't average probabilities either.
e.g. if pulling out of the pool of 6 dragons only (basically 50/50 to hit it, just for the sake of obviousness), the chances with this kind of calculation would be: (1/6 + 1/5 + 1/4) / 3 = 0.2 = 20%
Thank you all for the responses. I really appreciate it.
I looked up whether class cards get a bonus and apparently they do according to the Hearthstone Wiki, 4 times more than neutral cards! In this case, how would you calculate the probability for, let's say, Drakonid Operative?
I know that you normally shouldn't calculate the average of the average, because it won't be the exact number you are looking for. However the lower the numbers the closer you'll get to what the number is. So i think it's save to say that the number will be around the 7.5 and 8%
No, not even close. The whole logic is flawed. Probabilities just don't work that way, no matter how low or high the numbers. The real number is 21.4% (assuming equal probabilities for all dragons), which I'm sure you'll admit doesn't even remotely approximate to an 8%,
Thank you all for the responses. I really appreciate it.
I looked up whether class cards get a bonus and apparently they do according to the Hearthstone Wiki, 4 times more than neutral cards! In this case, how would you calculate the probability for, let's say, Drakonid Operative?
Assuming Standard format, there seems to be 13 neutral dragons and only 1 class one for Priest. So, now instead of looking for a specific 1 card in 14, we're looking for any one of 4 Drakonids in a pool of size 17 (we count the Drakonid 4 times; I'm assuming that's how Blizz does the pooling if they say a chance is x4).
An alternate way to think about is what I also mentioned above: The chance to pull Deathwing is equal to (1 - the chance to miss Deathwing 3 times in a row) from a shrinking pool: 1 - (13/14 * 12/13 * 11/12) = 0.214
This does account for uniqueness and shrinking pool with every pick. Hell, do a simulation if you don't believe it ;)
Your separate pick chances are ok, it's the way you chain them that is wrong (don't add, and certainly don't average!)
Thank you all for the responses. I really appreciate it.
I looked up whether class cards get a bonus and apparently they do according to the Hearthstone Wiki, 4 times more than neutral cards! In this case, how would you calculate the probability for, let's say, Drakonid Operative?
Assuming Standard format, there seems to be 13 neutral dragons and only 1 class one for Priest. So, now instead of looking for a specific 1 card in 14, we're looking for any one of 4 Drakonids in a pool of size 17 (we count the Drakonid 4 times; I'm assuming that's how Blizz does the pooling if they say a chance is x4).
So, the chance to get a Drakonid is:
1 - (13/17 * 12/16 * 11/15) = 0.579 = 57.9%
That would explain why they get it so often :S
Thank you very much! This makes the most sense. :)
I appreciate everyone else's responses also. <3
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For the people who are good at statistics (I am clearly not one of them), can you explain to me the probability of getting a certain card from the Discover mechanic. For example, let's say we're talking about Netherspite Historian and the probability of getting Deathwing. We know there are 14 dragon cards in Standard and 3 unique choices are presented every time from Discover. Can you show me how to calculate this probability? Thanks :)
If all the choices are equal (not sure if they are though, or an expansion/class is preferred?), should be simply 3/14 = 21.4%
Or you can go about it like this (1 - the chance to not pull Deathwing 3 times in a row):
1 - (13/14 * 12/13 * 11/12) = 0.214
Also, what is the division by 3 at the end for? You can't average probabilities either.
e.g. if pulling out of the pool of 6 dragons only (basically 50/50 to hit it, just for the sake of obviousness), the chances with this kind of calculation would be:
(1/6 + 1/5 + 1/4) / 3 = 0.2 = 20%
which is obviously not correct.
Thank you all for the responses. I really appreciate it.
I looked up whether class cards get a bonus and apparently they do according to the Hearthstone Wiki, 4 times more than neutral cards! In this case, how would you calculate the probability for, let's say, Drakonid Operative?
The real number is 21.4% (assuming equal probabilities for all dragons), which I'm sure you'll admit doesn't even remotely approximate to an 8%,
So, now instead of looking for a specific 1 card in 14, we're looking for any one of 4 Drakonids in a pool of size 17 (we count the Drakonid 4 times; I'm assuming that's how Blizz does the pooling if they say a chance is x4).
Me too. I'm talking about the original Deathwing scenario as well.
The chance to pull one specific card out of 14 if picking 3 randomly is 21.4%, not 8%.
@psver: we're running in circles now :)
No, I didn't forget.
An alternate way to think about is what I also mentioned above:
The chance to pull Deathwing is equal to (1 - the chance to miss Deathwing 3 times in a row) from a shrinking pool:
1 - (13/14 * 12/13 * 11/12) = 0.214
This does account for uniqueness and shrinking pool with every pick.
Hell, do a simulation if you don't believe it ;)
Your separate pick chances are ok, it's the way you chain them that is wrong (don't add, and certainly don't average!)
@siritanusik: exactly :)