Legendary Creatures probability (Don't pay for dupes!)

obsidianpeanutobsidianpeanut Age Unconfirmed Posts: 25 Just Dropped In
edited February 2017 in MtGPQ General Discussion
This is either an appeal to the decision makers to allow micro-transactions for individual cards or an appeal to the players to stop spending money. You choose.
In my frustration of drawing a lot of dupes, I finally decided to do some probability statistics. Of course, commons, uncommons, rares and mythics have different drop rates which complicates things - and I’m not sure that those rates have been published. However, when the powers-that-be decide to sale the “Legendary Creatures” we can do some easy statistics to see if we are throwing away money.

First a brief background. In probability theory there is a very common formula that applies here called the “Birthday problem” – given a set of people, a pair of them will have the same birthday (not considering year). Since there are 366 possible days if you count February 29, then you have a 100% probability with 367. However, would you believe that the probability is 50% at ONLY 23 people?

(read more about that here, https://en.wikipedia.org/wiki/Birthday_problem)

Using the same formula for the Legendary Creatures. There were 15 cards in the set. If you already have 5 of those cards, then you have a 52.5% chance of drawing a dupe for your 1st purchase out of 3 purchases available.

1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15) = 52.5% chance to draw a dupe

If you were lucky and got a new Mythic, now you have 6 from the set, then your second purchase will happen with a 68.3% chance for drawing a dupe.

1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15)*(1-5/15) = 68.3% chance of drawing a dupe

If you were lucky again and got a new Mythic, now you have 7 from the set. Your third and final purchase will happen with a 81% chance for drawing a dupe.

1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15)*(1-5/15)*(1-6/15) = 81% chance of drawing a dupe

You say, well if they would open the entire Mythic set up for Legendary Creatures, we would have better odds, right? Perhaps. It entirely depends on how many Mythics you already have.

For instance, right now there are 8 exclusive Mythics which I will remove. That leave 126 possible Mythics. I currently have 33 Mythics, so there are 93 Mythics in the set that I don’t have. However, with 33 Mythics there is a 98.99% chance that any new Mythic I get will be a dupe.

1-(1-1/126)*(1-2/126)*(1-3/126)*(1-4/126)*(1-5/126)*(1-6/126)*(1-7/126)*(1-8/126)*(1-9/126)*(1-10/126)*(1-11/126)*(1-12/126)*(1-13/126)*(1-14/126)*(1-15/126)*(1-16/134)*(1-17/126)*(1-18/126)*(1-19/126)*(1-20/126)*(1-21/126)*(1-22/126)*(1-23/126)*(1-24/126)*(1-25/126)*(1-26/126)*(1-27/126)*(1-28/126)*(1-29/126)*(1-30/126)*(1-31/126)*(1-32/126) = 98.99% of dupe

I don’t like these odds.

Today the Mythics were sold for 550 crystals. If you are buying crystals, then you may have spent as much as $22.49 for 1 mythic and it was probably a dupe. ($18.33 if you bought 3000 crystals for $99.99)

We should not spend money on duplicate cards. The difference between buying MTGPQ cards and paper Magic cards is that you get a tangible item of value that you can sale. You can also go into any number of stores or online and buy the exact card you want if you have the money.

It is time for D3Go to start doing micro-transactions for individual cards… and at a much more reasonable price. I would propose $0.50 for commons, $1 uncommon, $2.50 rare, $5 Mythic.

Of course, they are not going to do anything as long as people continue to throw away their money. Make them understand you are tired of being ripped-off. Stop spending money for dupes.

Comments

  • VolrakVolrak Posts: 652 Critical Contributor
    Using the same formula for the Legendary Creatures. There were 15 cards in the set. If you already have 5 of those cards, then you have a 52.5% chance of drawing a dupe for your 1st purchase out of 3 purchases available.
    If you randomly get one out of 15 possible mythics, and you already own 5 of them, the chance of the new card being a dupe is simply 5/15, or 33%. Likewise if you own 6 of the mythics, the chance of a dupe is 6/15, and so on.
    You say, well if they would open the entire Mythic set up for Legendary Creatures, we would have better odds, right? Perhaps. It entirely depends on how many Mythics you already have.

    For instance, right now there are 8 exclusive Mythics which I will remove. That leave 126 possible Mythics. I currently have 33 Mythics, so there are 93 Mythics in the set that I don’t have. However, with 33 Mythics there is a 98.99% chance that any new Mythic I get will be a dupe.
    If you open a random mythic from 126 possible mythics, of which you already own 33, the odds of a dupe is simply 33/126: 26%

    These calculations don't relate to the birthday problem. The birthday problem would apply if you had 5 random mythics in sealed envelopes, which could be dupes; what are the odds that you have at least one pair of dupes from your unknown mythics? In contrast, your starting point is a collection of known, non-dupe mythics.
  • OrionStarOrionStar Posts: 5 Just Dropped In
    This is more like the coupon collector's problem... https://en.wikipedia.org/wiki/Coupon_collector's_problem

    If I wanted all 15 of the available mythics in the Legendary Creatures set, it would take me at least 50 tries to get them all!

    (Since I have 12 out of the 15 available mythics, and only 3 shots at a new one, I won't be spending any crystals on this, because the odds are good that I'll get a dupe.)
  • blacklotusblacklotus Age Unconfirmed Posts: 589 Critical Contributor
    thanks for the interesting topics on probability. i learm something new today. going to read more about these cases. icon_e_smile.gif

    i got lucky. i received 3 new mythics out of all 3 purchases.

    but they are Kozilek, General Tazri and Omnath. Only the last chap is top tier. The other 2 are somewhat average.

    So maybe not so lucky after all. icon_e_sad.gif
  • [Deleted User][Deleted User] Posts: 0 Match Maker
    Having wasted £40 this morning in getting 3 duplicate mythics in a row I've decided to delete the app.

    That and I've somehow spent over £250 since Christmas and I can't believe what appalling value I'm getting for my money.
  • hawkyh1hawkyh1 Posts: 780 Critical Contributor
    1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15) = 52.5% chance to draw a dupe

    I think your probability is wrong.

    by your assumptions fishing for dupes

    1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15)...*(1-11/15) = 52.5+%

    how can can fishing for dupes and fishing for non dupes add up
    to greater than 100%. (it's either a dupe or it's not, regardless of
    what you already own)

    HH
  • obsidianpeanutobsidianpeanut Age Unconfirmed Posts: 25 Just Dropped In
    obsidianpeanut wrote:
    1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15) = 52.5% chance to draw a dupe


    I think your probability is wrong.

    by your assumptions fishing for dupes

    1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15)...*(1-11/15) = 52.5+%

    how can can fishing for dupes and fishing for non dupes add up
    to greater than 100%. (it's either a dupe or it's not, regardless of
    what you already own)

    HH


    Hawkyh1

    As Volrak pointed out, I misapplied the formula here. The Birthday Problem is about a probability of a dupe within a given set of n, not the probability that the next addition to the set will be a dupe.

    I want to be clear here. I misapplied the formula, so we have to step away from thinking of this as I used it above. In which case as you asked fishing for dupes versus fishing for none dupes may not apply. It should really be the probability that a dupe already exists in the set or the probably that all objects in the set are unique.

    To finish that out:
    I gave:
    1-(1*(1-(1/15))*(1-(2/15))*(1-(3/15))*(1-(4/15))) = 0.52545185185

    OR 52.5% that given k=15 and n=5, the probability that at least 2 of the 5 are dupes

    You asked about the inverse and gave this formula….
    1-(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15)...*(1-11/15) = 52.5+%

    But that isn’t the inverse. The inverse would mean that you don’t subtract from the leading 1:

    (1*(1-1/15)*(1-2/15)*(1-3/15)*(1-4/15)) = 0.47454814814

    OR 47.5% that in k=15 and n=5, the probability that all 5 are uniquely different.

    52.5 + 47.5 = 100%

    But again, I don’t think this is worth more discussion with the Birthday problem formula as I misapplied it in the first place.
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