**WINTER OF WOE - BONUS OBJECTIVE POINT**
As previously announced, the team will be distributing an additional point toward milestones to anyone who completed the Absorbing Man fight in the first step of the Winter of Woe.
This point will be distributed at a later time as it requires the team to pull and analyze data.
The timeline has not been set, but work has started.
There is currently an issue where some Alliances are are unable to find a match in Alliance Wars, or are receiving Byes without getting the benefits of the Win. We will be adjusting the Season Points of the Alliances that are affected within the coming weeks, and will be working to compensate them for their missed Per War rewards as well.

Additionally, we are working to address an issue where new Members of an Alliance are unable to place Defenders for the next War after joining. We are working to address this, but it will require a future update.

What are the odds of getting the same character in a row?

LaserjungeLaserjunge Posts: 33
edited January 2020 in General Discussion
I've just used my 10k 5* and 6* shards to draw my new heros. I thought after getting two times 5* Spiderman Classic, a 5* Rhino and a 5* Iron Man over the last 8 weeks (I am not a super frequent player anymore), what are the odds to draw another NULL. And here we go:

5* Cable and 6* Cable in a row!!

Since this game turned me down so much over the last year, I need at least half a year to get all 6* shards together. So it is effort, in time, in energy and emotional. And then KABAM's terrible RNG machine ruins you another night. The same thing occurs with 5* awakening gems (I've got 5 or 6 and solely 2x mutant, 3x skill, and 1x tech although I am waiting for mystic and cosmic for a year now). If you say, this is just randomness and probability, I'd say finally get a repaying RNG algorithm KABAM!!

For me this is the final killer. I am done with this game and ready for another one, done by better team that actually acknowledges the time and effort people invest in their games.

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Comments

  • KelvinKageKelvinKage Posts: 372 ★★★
    1/22,500 or about .00004% BUT it is possible and just a stroke of momentary bad luck. It happens to all of us at one point and at another you’ll get that momentary stroke of good luck and those are the ones that keep this game exciting lol.
  • Archdemon_Archdemon_ Posts: 613 ★★
    To simplify this the odds are the exactly the same as pulling said champion once.
    The easy way to explain this is... Roll a dice, you chance of rolling a 2 are 6:1
    roll that same dice again and your chance of rolling a 2 are 6:1
    The odds do not change and are not influenced in any way by your previous or subsequent rolls
    (taking human influence, roll speed and force etc out of the equation that have no bearing in a RNG based calculation)
  • Capn_DanteCapn_Dante Posts: 565 ★★★
    I don't know but my 6 star pulls in December were Nick Fury, Captain Marvel Movie and Captain Marvel Movie so if it's rigged I'm fine with it
  • LLStashLLStash Posts: 132
    The odds r good
    Really really good
    I opened my first 6*
    Emma
    I opened my second 6*
    Emma again
    Now she makes life miserable for people on defence
  • Riggs_97Riggs_97 Posts: 111
    My other half pulled a 4* Invisible Woman today after pulling 5* Invisible Woman yesterday. It's just one of those things I suppose.
  • NOOOOOOOOPEEEEENOOOOOOOOPEEEEE Posts: 2,803 ★★★★★
    I know how you feel I've recently pulled a 5* and 6* miles
  • DrZolaDrZola Posts: 8,479 ★★★★★

    To simplify this the odds are the exactly the same as pulling said champion once.
    The easy way to explain this is... Roll a dice, you chance of rolling a 2 are 6:1
    roll that same dice again and your chance of rolling a 2 are 6:1
    The odds do not change and are not influenced in any way by your previous or subsequent rolls
    (taking human influence, roll speed and force etc out of the equation that have no bearing in a RNG based calculation)

    Exactly this right here. And that's the case for any instance of successive openings in this game.

    It's the case for anything really. The odds for events like "rolling [X] [Y] times in a row" are merely a concept. They're not there to acticely draw direct conclusions about real world events.

    Look at it this way:

    There is no fundamental physical force or magical energy making the die less likely to land on a 6 after it just landed on a 6.
    @DNA3000 has it correct above. Each individual die roll may have a 1/6 probability, but cumulative rolls are treated differently from a statistical perspective. Back to back sixes have a 1/36 chance of happening, just like back to back anything else on a die. That’s easily searchable anywhere on the Interwebs even if you’ve never had a statistics course.

    Not sure how many times this has been pointed out on the forums, but it amazes me that it has to be pointed out over and over again.

    To @Plinko Think of it this way—you won the lottery with those Twin Janes. Granted, it was a Shirley Jackson kind of lottery, but you’re still a lottery winner.

    Dr. Zola
  • JonnySnowJonnySnow Posts: 117
    I once bought the Splatter crystal bundle and pulled back to back 5* AA.

    So I believe the percentage of even pulling a 5* was like 3-5%. What would be the odds of pulling back to back 5* of the sames champ considering the odds??

    Just curious..
  • MassapealMassapeal Posts: 143
    0.5%
  • DeaconDeacon Posts: 4,034 ★★★★★
    ladies and gents i present a 6* Gambit back to back ... one in Cavalier and then right after a 6* crystal. i wasn't devastated because Gambit has a lot of uses BUT i certainly wasn't thrilled either lol
  • PseudouberPseudouber Posts: 748 ★★★
    edited January 2020
    Dont ever expect to get anything good for your hard work or hard earned cash and you wont be disappointed as much. Just expect **** pulls and just be happy if you pull something decent. Its a numbers game, you just need to open alot of crystals and you will get some good pulls sometimes.
  • Sinedd92Sinedd92 Posts: 94
    Pfff, 4 bad champions from 5* in a row.
    I've had like 40-45 bad ones in a row, and then I got dipped stark spidey. As for 6*, I have a roaster of useless things :D
  • MagrailothosMagrailothos Posts: 5,281 ★★★★★
    No one is treating you as a fool, @UmbertoDelRio.


    that's the case for any instance of successive openings in this game...

    The odds for events like "rolling [X] [Y] times in a row" are merely a concept. They're not there to actively draw direct conclusions about real world events...

    ...it doesn't make sense to calculate the odds for a distinct cumulative event...

    Absolutely. Yes, the draws are completely independent; and in a way it doesn't make sense to calculate the odds; but the thing is, people like the OP keep asking these questions.

    People want to know the odds, whether that desire is rational or not. Don't get angry with other community members for trying to help them.
  • Camby01Camby01 Posts: 572 ★★
    It's the difference between flipping a coin and coming up heads twice. Each individual flip is a 50/50 chance. How many times can you keep betting on heads without losing.
    Would you you bet $100 that you could flip a coin on heads 10 times in row if I gave you 10:1 odds.
  • Colonaut123Colonaut123 Posts: 3,091 ★★★★★
    DNA3000 said:

    1/22,500 or about .00004%

    That's the odds of pulling a specific champion twice. That's not the odds of pulling a champion twice. The odds of pulling, say, Cable out of the 5* basic (assuming approximately 150 champs in the crystal) is one in 150, and then the odds of pulling him again are another one in 150, for a combined probability of one in 22,500 to do those two things in a row. But the odds of pulling a champion twice in a row out of the basic crystal are one in 150, because the first pull can be anything, and then the odds of pulling that champion a second time is one in 150.

    The odds of pulling a champion from the 5* basic and then the same champ again from the 6* basic is also about one in 150. Since order doesn't matter the odds of pulling a 5* champ and then an identical 6* champ are the same as pulling a 6* champ and then an identical 5* champ, which means the odds of pulling identical champs in a row are the same as above: one in 150. It doesn't even matter how many champs are in the 6* crystal: if there was only one champ in the 6* crystal then the odds of pulling the same champ twice in a row would still be one in 150. The 6* champ would be fixed (it could only be one thing) and then the odds of pulling that champ out of the 5* basic would be one in 150.
    Incorrect math. As both pulls are independent, you should multiply the odds thus 1/22,500. Still, with the millions of crystal openings, that is bound to happen once in a while. I think I opened a 4* Domino from a basic 4* crystal and not much later a 4* Domino from a PHC.
  • DNA3000DNA3000 Posts: 18,554 Guardian

    In an rng-system that is supposedly strictly not cumulative it doesn't make sense to calculate the odds for a distinct cumulative event.

    I'm not sure why so many people believe this, but the entire study of statistics is essentially the study of calculating the cumulative odds of sequences of independent events. In fact, the fundamental principle of statistics is that the odds of two independent events happening together are the product of the chance for each to occur, and is a logical consequence of the counting principle of statistics.

    We presume each crystal opening is an independent event that has nothing to do with each other, and from this assumption we calculate the odds of two crystal openings happening in a set. This is not only a fundamental calculation of statistics that is specifically intended to determine the odds of real world sequences of events, it isn't even a particularly interesting or controversial calculation. This is a completely trivial calculation.

    I honestly have no idea where this misconception about "cumulative events" comes from. I can only assume it is a mental glitch where if you believe the first crystal and the second crystal are independent and thus do not affect each other, it is somehow "invalid" to treat the two rolls as a single event whose odds can be calculated. But that is precisely what the counting principle of statistics mathematically describes.

    In a throw of two dice, the dice are independent, so the first throw can be any one of six results, each having the same chance to happen. Regardless of the first roll there are six possibilities for the second roll, all equally likely, because the second roll is independent of the first. That means there are thirty-six possible net outcomes: 1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 2/1, 2/2, 2/3, 2/4, 2/5, 2/6, 3/1, 3/2, 3/3, 3/4, 3/5, 3/6, 4/1, 4/2, 4/3, 4/4, 4/5, 4/6, 5/1, 5/2, /53, /54, 5/5, 5/6, 6/1, 6/2, 6/3, 6/4, 6/5, 6/6. Out of the thirty six possible outcomes, six are identical throws: 1/1, 2/2, 3/3, 4/4, 5/5, 6/6. So there are six out of thirty six possibilities in which the dice are identical throws, which means the odds are six in thirty six or one in six.

    All of statistical calculations are math that simplifies this basic fundamental counting of possible outcomes. We aren't somehow claiming that the two throws are a "cumulative event" we count all possibilities of two throws as independent events to determine the odds of a sequence of independent events to occur.

    This is something you have to accept, or fail statistics on day one. In fact, I would like to see the face of a statistics teacher when you tell them "The odds for events like 'rolling [X] [Y] times in a row' are merely a concept." Yeah: that concept is called "statistics."
  • DrZolaDrZola Posts: 8,479 ★★★★★
    edited January 2020
    The odds of having to revisit this explanation again within the next 90 days are almost definitely 100%.

    To @Kabam Vydious — ouch. But at least we have hope you’re on the inside putting in the good word for an IP buff.

    Dr. Zola
  • DNA3000DNA3000 Posts: 18,554 Guardian

    DNA3000 said:

    1/22,500 or about .00004%

    That's the odds of pulling a specific champion twice. That's not the odds of pulling a champion twice. The odds of pulling, say, Cable out of the 5* basic (assuming approximately 150 champs in the crystal) is one in 150, and then the odds of pulling him again are another one in 150, for a combined probability of one in 22,500 to do those two things in a row. But the odds of pulling a champion twice in a row out of the basic crystal are one in 150, because the first pull can be anything, and then the odds of pulling that champion a second time is one in 150.

    The odds of pulling a champion from the 5* basic and then the same champ again from the 6* basic is also about one in 150. Since order doesn't matter the odds of pulling a 5* champ and then an identical 6* champ are the same as pulling a 6* champ and then an identical 5* champ, which means the odds of pulling identical champs in a row are the same as above: one in 150. It doesn't even matter how many champs are in the 6* crystal: if there was only one champ in the 6* crystal then the odds of pulling the same champ twice in a row would still be one in 150. The 6* champ would be fixed (it could only be one thing) and then the odds of pulling that champ out of the 5* basic would be one in 150.
    Incorrect math. As both pulls are independent, you should multiply the odds thus 1/22,500. Still, with the millions of crystal openings, that is bound to happen once in a while. I think I opened a 4* Domino from a basic 4* crystal and not much later a 4* Domino from a PHC.
    By your reckoning, the odds of rolling two six sided dice and getting the same number on both dice would be one in thirty six. Which is obviously wrong.

    The flaw here is that while each crystal opening is independent, the desired sequence is not an independent sequence. So the analysis has to look at the requirements of each crystal. The requirement on the second crystal is that it match the first crystal, and whatever the first one was there's only a one in 150 chance of matching. But there's no requirement on the first crystal, it can be anything. So the odds of the first crystal meeting the sequence requirements is 100%.

    Alternatively, you can choose to use the fundamental counting principle, which is always correct, just maddeningly tedious. There are 150 possible drops for crystal 1. For each of those possibilities there's 150 possible drops for crystal 2. There are thus 22500 possibilities for the two crystals. However, there are 150 possibilities in which the two drops match. So the odds are 150 out of 22500, or one in 150.

    Let's try this with the simplest possible case to make life easy. What are the odds of flipping a coin twice and getting the same result. By your calculation, each flip is independent and has two possibilities, so you multiply and get one in four. But that's clearly wrong. The fundamental counting principle says there are four possibilities: Heads/Heads, Heads/Tails, Tails/Heads, Tails/Tails, and two of them are matching: Heads/Heads, and Tails/Tails; so the odds are two out of four, or one in two. You count all possibilities, count all desired ones, and the odds are the latter out of the former. That's actually what saying "X out of Y" means. There are Y possibilities all of which have equal odds, and X of them are the desired ones.

    So the two valid ways of calculating this are: calculate the number of possibilities and the number of desired ones and divide. Or use constraint logic, which is how Markov Chains fundamentally work. There's no constraint on the first flip, but a 50% constraint on the second flip (it must match the first one), for a net odds of 1/1 * 1/2 = 1/2 = 50% on the net result.

    There's a simple double check on logic here that I used to use when teaching statistics. When someone says, for example, that the odds of pulling the same champ twice are one in 22500, because they just multiplied the numbers, I ask what are the odds of pulling Winter Soldier twice in two crystals. Of course that's one in 22500. But if that's the case, how can the odds of pulling *any* champion twice in two crystals be the same? Obviously, it has to be more likely to pull two identical champs than to pull WS specifically twice in a row. You have to realize that asking what the odds of pulling two identical champs in a row is really asking what are the odds of pulling Winter Soldier twice in a row, plus the odds of pulling Storm twice in a row, plus the odds of pulling Aegon twice in a row, etc. There's 150 ways to pull the same champ in a row, whereas there's only one way to pull Winter Soldier twice in a row.
  • DNA3000DNA3000 Posts: 18,554 Guardian
    DrZola said:

    The odds of having to revisit this explanation again within the next 90 days are almost definitely 100%.

    The first time I explained this situation in the context of a game, I was doing it on Fidonet. I'm pretty sure I've done it over a hundred times on USENET. I did it so many times on a couple game forums it was enshrined as an official FAQ once. Statistical calculations are probably the single most misunderstood game mechanical subject in all of gaming.

    Which is weird because I always thought probability and statistics was a required math subject for anyone who has graduated high school much less undergraduate college. Either you had to take it, or you had to take more sophisticated math that presumes it. I don't expect people to have command of the more sophisticated stuff, but I would think most people would at least be able to recognize it again when it is presented.
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