1. #1
    Avatar de mtkyebra Membre de passage
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    escroquerie: 2 carte pack

    Il dit: "2 cartes aléatoires de commune, peu commune ou rare Possibilité d'être doublé".

    Ok, nous allons vérifier la Bibliocarte: 196 cartes différentes -14 épiques +12 événements: 194 cartes possibles (je sais peut-être il ya une certaine carte n'est pas sur la bibliocarte, mais ils sont probablement rares et pas commun), de sorte que nous avons un total de 79 communes , 70 peu commune et 45 rares. 79/194 = 40,72% probabilités qu'une carte est commune. Donc, ce pack sonne bien. Je pensais que sur l'avenir du paquet va augmenter son prix ou de réduire à une seule carte. Mais alors décidé de le tester:

    Résultat après l'ouverture de 22 paquets (44 cartes):
    • 32 communes (72,72% au lieu de 40,72%)
    • 11 peu commun (25% au lieu de 36,08%)
    • 1 rare (2,27% au lieu de 23,20%)

    Alors, quelle est la formule? Parce que cela ne semble pas aléatoire.

    [EN]
    It says: "2 random cards from common, little common or rare. Possibility to be doubled".
    Ok, let's check the Bibliocarte: 196 different cards -14 epic +12 events: 194 cards possible (I know maybe there are some card not on the bibliocarte, but they probably are rare and not common) so we have a total of 79 common, 70 little common and 45 rare. 79/194=40,72% probabilities that one card is common. So this pack sounds great. I thought on future the package will increase it's price or reduce to one card. But then decided to test it:

    Result after opening 22 packages (44 cards):
    • 32 common (72,72% instead of 40,72%)
    • 11 little common (25% instead of 36,08%)
    • 1 rare (2,27% instead of 23,20%)

    So what is the formula? Because this doesn't seem random.
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  2. #2
    No, it's random. Look, for most of the IRL TCG, we say that boosters are random, but it's the cards in their rarity which are random, not the cards in their globality. For example in Magic, you always have 10C/3U/1R (+ foil ; timeshifted : double-sided <- these cards can be C, U or R but replace a common in all cases). Here they won't say a specific xC/yU/zR because these boosters have only 2 cards, but I have understood this from the begining (generally in games, the most exepensive goods are the most interesting)
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  3. #3
    Avatar de zenithale Forumeur fou
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    Well in fact I think you should just imagine instead of a Pack of 12 cards (8 C, 3 U and 1 R) you have 6 Packs of 2 cards amongst the 12 previous. That's why you get more C than U and R. You have not the same probability to get a R or a C even if the number of R and C was the same.
    Of course if you are really lucky you can get 2 R but most of the time you will get 2 C or 1 C and 1 U. As undergroundjin said the safest way is always the Pack of 12 cards.
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  4. #4
    Avatar de mtkyebra Membre de passage
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    Je suis désolé je ne comprends pas votre point.
    Le texte dit aléatoire, c'est une parole concrète, avec un sens très clair.
    Nous ne parlons pas de la pack de 12 cartes, ce qui bien sûr fonctionne comme MagicTG.
    Zenithale, j'ai une certaine 2XU et mon sac 25ème a été 2XU. Mais presque tous les packs sont 1xU 2xc ou 1xC. Vous essayez de dire qu'ils "réduit le pack de 12 à 2"???
    Quoi qu'il en soit, ce n'est pas ce qui est annoncé avant de cliquer sur acheter. Cela signifie donc que tant que mon premier message: L'information est fausse!!

    BTW mon expérience a pris fin à 25 packs:
    • 36 communes (72% au lieu de 40,72%)
    • 13 peu commun (26% au lieu de 36,08%)
    • 1 rare (2% au lieu de 23,20%)

    [EN]
    I'm sorry I can't understand your point.
    The text says random, that is a concrete word with a very clear meaning.
    We are not speaking of the 12 cards pack, which of course works as MagicTG.
    Zenithale, I've got some 2xU and my 25th pack has been 2xU. But nearly all the packs are 2xC or 1xC 1xU. You're trying to mean that they "reduced the 12 pack to 2"????
    Anyway, that's not what is advertised before you click buy. So that means as my first message: The info is false!!!!!!

    BTW my experiment ended at 25 packs:
    • 36 communes (72% au lieu de 40,72%)
    • 13 peu commun (26% au lieu de 36,08%)
    • 1 rare (2% au lieu de 23,20%)
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  5. #5
    Avatar de mdurandg Fidèle des forums
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    For each card in the booster, the game randomly determines if you will loot a Common (72% chance according to your latest data), an Uncommon (26%) or a Rare (2%).
    Then once the rarity is determined, the game randomly determines which card amongst said rarity will be opened.

    Random does not mean you have a 33/33/33 chance to loot a Common / Uncommon / Rare. Probabilities are not equal, but it's still random.
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  6. #6
    Avatar de zenithale Forumeur fou
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    The text is not wrong, but right, you can misinterpret it. Random does not mean all cards have the same probability at all.
    Pick 80 C, 30 U and 10 R = 120 cards = 60 packs of 2 cards.
    (I'm not sure about the numbers, maybe it's even worse.)
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  7. #7
    Avatar de mtkyebra Membre de passage
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    That's it Zenithale, the number is easy: 79xC, 70xU 45xR = 194 (also there are 14xE which I didn't count)
    Acording to the 196 cards on Bibliocarte -14 epic + 12 (5xC 5xU 2xR) events already counted on the first line.

    no mdurandg, random wouldn't be 33% C - 33% U - 33% R. Random is 40,72% C - 36,08% U - 23,20% R THIS IS RANDOM whatever other thing MUST be specified on the info displayed or it's false advertising. image

    But of course maybe there are one or two cards not on the bibliocarte.

    Precisely in France there are very restrictive laws regarding false advertisment, and THQ is having a lot of problems with that.
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  8. #8
    Avatar de zenithale Forumeur fou
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    You are really stubborn Elder . The game can give you 2 random cards with 80% for Common, 15% for Uncommon and 5% for Rare probabilities or whatever other probabilities, even 95%/4%/1% and that's still random. The only thing you can protest against is the lack of information regarding the probabilities, but anyway the number of cards in each categorize (C/U/R) are completely unrelated.

    As mdurandg said previously, when you buy the 2 cards pack the game will...
    1) randomly determine the rarity of your 2 cards (the probabilities are probably 8/12 for C, 3/12 for U and 1/12 for R as in the 12 cards pack),
    2) randomly determine what cards from the corresponding (C/U/R) pool.
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  9. #9
    Envoyé par zenithale Go to original post
    1) randomly determine the rarity of your 2 cards (the probabilities are probably 8/12 for C, 3/12 for U and 1/12 for R as in the 12 cards pack),
    .

    That's right.
    67% C/ 25% U/ 8% R
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  10. #10
    Avatar de mtkyebra Membre de passage
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    têtue? LOL quand nous sommes plus ou moins dit la même chose, pour résumer:

    - Les cartes obtenues ne sont pas vraiment aléatoire. Vérifiez la description sur un dictionnaire. http://en.wikipedia.org/wiki/Random *
    - Les infos sur la publicité (image) est fausse.
    - Il peut conduire à des problèmes juridiques ont.

    [EN]
    stubborn, LOL when we are more or less saying the same, to summarize:

    - The cards obtained aren't really random. Check the description on a dictionary. http://en.wikipedia.org/wiki/Random *
    - The info on the advertise (image) is false.
    - It can lead to juridic problems.

    *
    [...] In situations where the population consists of items that are all
    distinguishable, a random selection mechanism would require equal
    probabilities for any item to be chosen. That is, if the section process
    is such that each member of a population, of say research subjects, has
    the same probability of being chosen then we can say the selection process is random.[...]
    [FR] Je ne dis pas de changer le système, mais d'en préciser la distribution de probabilité réelle sur la description de l'objet.
    [EN] I'm not saying to change the system, but to specify the real probability distribution on the description of the item.
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