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Hearthstone | Posted by | 5 years, 7 months ago
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Discover Effect: What are the probabilities?

TL;DR - Does Discover give every card an equal probability of being displayed?

 

When the Discover effect was changed to "not favor class cards", I just naively assumed that every available card had an equal probabiltiy of being shown as an option. 

For example, Primordial Explorer gives you the option of 3 dragons and there are currently 20 Neutral Dragons and 2 Hunter class Dragons in Standard (not counting Primoridal Explorer itself given the recent change).  Without thinking about it, I just assumed that the probability of getting say Rotnest Drake was 1/22 for the 1st discover option, 1/21 for the 2nd discover option, and 1/20 for the 3rd discover option (I don't remember ever getting the same Dragon shown twice, but if that's possible then the probability would be 1/22 for each option).  

However, this video was randomly recommended to me on YouTube today (had never heard of them before) and I gave it a listen.  The hyperlink skips to the 43:45 mark where an argument is made that Discover instead works as a 2-stage gamble so-to-speak.  For each of the 3 options there's a 1/2 probability that the card will be drawn from the Neutral Pool or the Class Pool.  After this, another die is rolled where every card in the respective pool has an equal probability of being displayed (without replacement).

So in the example above the probability of Rotnest Drake would be 25% for each of the 3 discover options (50% to be a class card and then 50% to be Rotnest Drake given that it's a class card).  Is this really the case??

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  • TL;DR - Does Discover give every card an equal probability of being displayed?

     

    When the Discover effect was changed to "not favor class cards", I just naively assumed that every available card had an equal probabiltiy of being shown as an option. 

    For example, Primordial Explorer gives you the option of 3 dragons and there are currently 20 Neutral Dragons and 2 Hunter class Dragons in Standard (not counting Primoridal Explorer itself given the recent change).  Without thinking about it, I just assumed that the probability of getting say Rotnest Drake was 1/22 for the 1st discover option, 1/21 for the 2nd discover option, and 1/20 for the 3rd discover option (I don't remember ever getting the same Dragon shown twice, but if that's possible then the probability would be 1/22 for each option).  

    However, this video was randomly recommended to me on YouTube today (had never heard of them before) and I gave it a listen.  The hyperlink skips to the 43:45 mark where an argument is made that Discover instead works as a 2-stage gamble so-to-speak.  For each of the 3 options there's a 1/2 probability that the card will be drawn from the Neutral Pool or the Class Pool.  After this, another die is rolled where every card in the respective pool has an equal probability of being displayed (without replacement).

    So in the example above the probability of Rotnest Drake would be 25% for each of the 3 discover options (50% to be a class card and then 50% to be Rotnest Drake given that it's a class card).  Is this really the case??

  • I cannot give you a real answer to this as I don't know how the program exactly resolves the discovery choices (and to my knowledge Blizzard has never said how they are doing it exactly). I would like to point out a couple of issues though:

    • Your conclusion (or the videos) seems to be flawed; if this would be actually resolved like you describe in the last sentence (which I don't think) the probability of getting Rotnest Drake would actually be even higher (e.g. if the first card option is the other class dragon the second option would not be 25% but already 50% as there are no duplicates offered)
    • Having said this based on my experience of discovery results I am pretty sure that this does not work as you described (can't give you any proof though)

    But hey I'm not a statistics expert, so feel free to proof me wrong. Besides even an expert (outside Blizzard) might not be able to answer this properly without the knowledge how this is exactly resolved by the program. I guess there are multiple options.

    • Quote From Thraxus

      I cannot give you a real answer to this as I don't know how the program exactly resolves the discovery choices (and to my knowledge Blizzard has never said how they are doing it exactly). I would like to point out a couple of issues though:

      • Your conclusion (or the videos) seems to be flawed; if this would be actually resolved like you describe in the last sentence (which I don't think) the probability of getting Rotnest Drake would actually be even higher (e.g. if the first card option is the other class dragon the second option would not be 25% but already 50% as there are no duplicates offered)
      • Having said this based on my experience of discovery results I am pretty sure that this does not work as you described (can't give you any proof though)

      But hey I'm not a statistics expert, so feel free to proof me wrong. Besides even an expert (outside Blizzard) might not be able to answer this properly without the knowledge how this is exactly resolved by the program. I guess there are multiple options.

      I agree with you that the video's conclusion doesn't jive with my experience, but admittedly I haven't paid much attention to patterns in Discover cards and wanted to ask y'all.

      As you said, if (my understanding of) the argument in the video was correct then you'd be guaranteed Rotnest if you got Veranus as an option already.  They brought this up as a way of explaining why Astromancer Solarian is so often discovered by Mages.  Again, this didn't sound right to me but I've never looked into it.

  • I've always been skeptical of the "random" in Hearthstone, as we already know they've weighted that random a number of time. . . whether it's with discover, or random minion/spell generation, I've always wondered how it was programmed. The idea of multi-step things like Step One : Rarity, step two, Class, Step Three : Card has occurred to me, considering the frequency of certain cards over time (back in the day, i haven't followed as closely lately). Certainly that ups your chances of pulling Legendaries from certain cards, if you only have to clear a 1 in 4 bar on step one.

  • This is just another question that can only be answered by someone from the dev team. All else is speculation, and a pretty impossible one at that. Its just not possible to somehow gauge a formula at something that can be from equal percentage to discover, to slightly skewed towards class cards.

    Another way of seeing this is to consider from what position we were to have came to the current formula. Class cards were 4x more likely than neutrals in the past. If the change is then to configure the formula in such a way to sneak class cards into the equation by means of a 50-50 coin toss, then team5 might well have just never changed the formula in the first place.

    Best to simply take their word for it. Ultimately even if discover effects were still skewed towards class cards, I dont see how they can profit by keeping that information from us.

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    • Quote From dapperdog

      This is just another question that can only be answered by someone from the dev team. All else is speculation, and a pretty impossible one at that. Its just not possible to somehow gauge a formula at something that can be from equal percentage to discover, to slightly skewed towards class cards.

      ...

      Actually it is perfectly possible, and even easy if you have the patience. All you need to do is discover enough times with a given card pool and count how many times you see each possible card. That will give you a set of probabilities for each card, and it doesn't matter how it gets there - all the matters is what it ends up at.

      Of course I don't have that much patience, so I tested the easiest one: how often does Draconic Lackey present Waxadred as an option? As the only dragon in rogue, and a near endless supply of lackeys, this should be a simple question to answer with enough confidence to guess at the algorithm used.

      I'm doing this against the innkeeper (Wild), so the pool is 1 dargon in rogue and 42 in neutral.

      Hypothesis 1: Every card has equal likelihood of being presented, irrespective of class.

      Here the odds of NOT finding Waxadred are (42/43)*(41/42)*(40/43) = 40/43, so the probability to see ol' Waxy in a discover is 3/43 = 6.97%. Or equivalently, you expect to see him about once every 14 discovers.

      Hypothesis 2: Class vs Neutral is chosen before the card

      In this case, the question reduces to the likelihood of not choosing a class card. If that likelihood is 50% per option, you would have (1/2)^3 = 1/8 as the probability to be presented with 3 neutrals. I.e. you would see Waxy in a whopping 7 out of every 8 discovers in both Standard and Wild!

      Results from testing

      Admittedly I got bored pretty quickly, and I only ended up doing 12 discovers, in which I saw Waxadred exactly once, which is perfectly consistent with Hypothesis 1.

      The sample size is too small for that to be taken as proof, and perhaps Hypothesis 2 should use a lower probability, but I see no reason why it would. For it to give 1/12 to see Waxadred it would need to be at 1 - (1 - 1/12)^(1/3) = 2.86%, which is preposterously low and just as untenable a suggestion as 50% is.

      So I have to conclude it is highly likely every card is weighted equally. The only way for it to be consistent with experience otherwise is for every discover pool to use a different probability of choosing between class and neutral cards, which is too complicated for next to no gain.

  • I will say that discovers have a probability of 100% to screw me over.