Math experts help me here(dual class crystal probability)

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  • SummonerNRSummonerNR Member, Guardian Posts: 12,819 Guardian
    That 0.00045 initial answer, and the ones saying that no matter how many crystals you open that overall you still will never have more than a 2.22% overall chance to pull an AntMan, are why we need our schools re-opened, lol.

    The later ones came in detailing the right answer for you.
    Which can basically be figured in layman’s terms by calculating odds of NOT GETTING HIM at all over 5 rolls.
    44/45 = .977777....
    Then multiply that by itself 4 more times (or to the 5th power on scientific calculators).
    Is .8937 (odds of never getting him), then subtract from 1 to get .1063, or 10.63%

    Pic below was provided by someone earlier, with a note that it was missing Sorceress Supreme, so according to that (if correct) would be 44 Science/Mystic champs instead (so your guess of 45 champs was pretty close).
    So, 10.86% (5 rolls from possible 44 champ pool).
    And from 50 rolls of a 44 champ pool is 68.32% chance to get at least once.
  • SummonerNRSummonerNR Member, Guardian Posts: 12,819 Guardian
    (plus Sorceress Supreme - Mystic)...


  • Morpheus_123Morpheus_123 Member Posts: 792 ★★★
    The chance of getting him is not a stacked probability. You have the same success rate percentage every time you roll the RNG.

    If the pool depth as you say is correct, it's a 1 in 45 chance, which is 2.22%.

    Until you change the depth of the pool, or increase the single possible outcome every time, that percentage will not change.
  • DaddriedaDaddrieda Member Posts: 1,639 ★★★★
    You getting a batch of crystals doesn’t mean you increase your percentage to get the desired champion. The percentage is still the same in all the crystals.

    For example 5 crystals of 2% (I’m being dumb here as it’s just an example) so the percentage doesn’t increase or decrease at all...

    1st crystal - 2%
    2nd crystal - 2%
    3rd crystal - 2%
    4th crystal - 2%
    5th crystal - 2%

    I’m no math geek and I could be off the subject.
  • Roman4544Roman4544 Member Posts: 85

    The chance of getting him is not a stacked probability. You have the same success rate percentage every time you roll the RNG.

    If the pool depth as you say is correct, it's a 1 in 45 chance, which is 2.22%.

    Until you change the depth of the pool, or increase the single possible outcome every time, that percentage will not change.

    Ok, so Cowhale will open 500 featured cavs and he will have only 1.5% of pulling at least 1 6 star...lol
  • DaddriedaDaddrieda Member Posts: 1,639 ★★★★
    Him having 500 crystal doesn’t prove anything. The percentage is still the same. He doing 500 crystals opening is still on luck. You could open 200 and have Ant-man and Cowhale get an Ant-man by opening 500 crystals.
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  • DaddriedaDaddrieda Member Posts: 1,639 ★★★★
    Yeah, but it doesn’t applied here on the game. Kabam have said it many times that the percentage is still the same on each crystal per time
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  • Morpheus_123Morpheus_123 Member Posts: 792 ★★★
    edited October 2020
    Roman4544 said:

    The chance of getting him is not a stacked probability. You have the same success rate percentage every time you roll the RNG.

    If the pool depth as you say is correct, it's a 1 in 45 chance, which is 2.22%.

    Until you change the depth of the pool, or increase the single possible outcome every time, that percentage will not change.

    Ok, so Cowhale will open 500 featured cavs and he will have only 1.5% of pulling at least 1 6 star...lol
    When you say it like that it made you LOL.
    So what you're trying to say is, if he opens 499 of them crystals and has yet to win a 6 star, does that mean he has a 748.5% chance of opening a 6 star on his final crystal? No.
    The probability of crystal openings doesn't change statistically when multiple openings are made, however bizarre or unrealistic it might sound that he won't win a 6 star from 500 cav crystals.

    When it comes to opening crystals the win percentages are singular. It doesn't matter how many crystals you have or how 'unlucky' it may seem, each crystal has statistically the same probability as the one before.
  • OrdalcaOrdalca Member Posts: 543 ★★★

    Roman4544 said:

    The chance of getting him is not a stacked probability. You have the same success rate percentage every time you roll the RNG.

    If the pool depth as you say is correct, it's a 1 in 45 chance, which is 2.22%.

    Until you change the depth of the pool, or increase the single possible outcome every time, that percentage will not change.

    Ok, so Cowhale will open 500 featured cavs and he will have only 1.5% of pulling at least 1 6 star...lol
    When you say it like that it made you LOL.
    So what you're trying to say is, if he opens 499 of them crystals and has yet to win a 6 star, does that mean he has a 748.5% chance of opening a 6 star on his final crystal? No.
    The probability of crystal openings doesn't change statistically when multiple openings are made, however bizarre or unrealistic it might sound that he won't win a 6 star from 500 cav crystals.

    When it comes to opening crystals the win percentages are singular. It doesn't matter how many crystals you have or how 'unlucky' it may seem, each crystal has statistically the same probability as the one before.
    That's not the calculation being asked. Yes, the chance of pulling a 6* in 500 crystals GIVEN THAT THE FIRST 499 WERE NOT 6* is 1.5%.

    However, without that condition, it's asking "what is the chance that he will get at least 1 6* in the next 500 pulls?" THAT number is 1-(1-0.015)^500 = 99.95%.

    It says nothing about what any specific crystal will give, but only what the entire collection of 500 crystals will give.
  • Was525Was525 Member Posts: 98
    I opened 32 dual crystals going for HT. I’m 0-32. You have a 2% chance no matter how many crystals you open. Don’t over think it. KISS
  • Thicco_ModeThicco_Mode Member Posts: 8,852 ★★★★★
    @SummonerNR and @Vendemiaire are correct
    Roman4544 said:



    The probability that you will pull one in one try is:

    P = 1/N

    P = 1/45 = 2.22%

    The probability will be the same for every crystal opening but the probability of success that you will pull him after a number of tries can be computed with:

    1-(1-P)^n

    where n is the number of tries:

    1-(1-0.0222~)^5

    10.63%

    Mayis said:

    There are 45 champs and we want 1 of them — Ant-man. You have 44/45 chance of not getting ant-man on single try, so if u multiply 44/45 * 44/45 *... five times (its power 5 in english, i guess) u get the probability of NOT getting him in five tries.

    So if we substract this number from 1, we get the chances to get at least 1 ant-man in 5 tries. Its a number from 0 to 1, and to convert it to percents just multiply it by 100. The answer is near to 10.6 %



    Hope this explanation will help u in future

    I can see you guys know what you saying, but you are saying by that explanation that if in 5 tries I have 10 ish %, so if I open 50 crystals I have more than 100% to get him? That's we know for sure it might not happen.
    it doesn't increase like that. I'm not totally sure how to explain it, but the chances aren't additive like that
  • OrdalcaOrdalca Member Posts: 543 ★★★
    edited October 2020

    Roman4544 said:

    The chance of getting him is not a stacked probability. You have the same success rate percentage every time you roll the RNG.

    If the pool depth as you say is correct, it's a 1 in 45 chance, which is 2.22%.

    Until you change the depth of the pool, or increase the single possible outcome every time, that percentage will not change.

    Ok, so Cowhale will open 500 featured cavs and he will have only 1.5% of pulling at least 1 6 star...lol
    When you say it like that it made you LOL.
    So what you're trying to say is, if he opens 499 of them crystals and has yet to win a 6 star, does that mean he has a 748.5% chance of opening a 6 star on his final crystal? No.
    The probability of crystal openings doesn't change statistically when multiple openings are made, however bizarre or unrealistic it might sound that he won't win a 6 star from 500 cav crystals.

    When it comes to opening crystals the win percentages are singular. It doesn't matter how many crystals you have or how 'unlucky' it may seem, each crystal has statistically the same probability as the one before.
    You are using the wrong calculation for that. What you are saying is "the chance of the 500th crystal giving a 6* is 1.5%." That is completely correct. For the spread of 500 crystals, the phrasing would be "What is the chance the 500th crystal is a 6* given that the first 499 were not?" The answer, again, is 1.5%, since the chances are independent, and that is what Kabam has said is constant.

    However, that's not what the question was. The question was, "what is the likelihood that at least 1 of the next 500 pulls is a 6*?" Nothing about which crystal it is that results in the 6*. Only that some crystal, somewhere in the next 500, gives one. It could be the first, the second, the third... any of them, or multiple. The way to calculate that is 1-P(No 6*, 500 times in a row).

    So, how do you calculate P(No 6*, 500 times in a row)?

    Well, the formula for independent results is P(A,B,C, and D all happening)=P(a)P(b)P(c)P(d), so it's P(no 6* in first) * P(no 6* in second) * P(no 6* in third) * ... * P(no 6* in 500th).

    Each of those P(no 6* in the Xth crystal) is 98.5%. Multiplying it out, you get P(no 6*, 500 times in a row) = 0.985^500, or about 0.05%.

    Plugging that into the first equation, 100%-0.05%, you get a 99.95% chance that at least 1 of those 500 crystals beat the odds and popped open showing a 6* champ. It doesn't say which one it was, and that's important, but somewhere in that crowd, there's a very good chance that at least one succeeded.
    Was525 said:

    I opened 32 dual crystals going for HT. I’m 0-32. You have a 2% chance no matter how many crystals you open. Don’t over think it. KISS

    Assuming the 2% was still valid, 1-(0.98)^32 = 1-0.5239 = .4761. There's about a 47.6% chance of not pulling the champ you want in 32 crystals.
  • K00shMaanK00shMaan Member Posts: 1,289 ★★★★
    edited October 2020
    the odds of getting him (or anyone in the crystal) are 1/45 per attempt ~2.2%. If you add an attempt, you will increase your overall chance at getting him but it will be by a smaller amount than 2.2%. This way, every time you add an additional crystal to your attempt, your odds of getting him increase overall, but because they increase by a smaller amount for each additional crystal, it will never equal or exceed 100%.

    odds for 1 crystal = 1-(44/45)^1 = 2.22%
    odds for 10 crystals = 1-(44/45)^10 = 20.12%
    odds for 100 crystals = 1-(44/45)^100 = 89.43%
    odds for 952 crystals = 1-(44/45)^952 = 99.99%
    Anything above 952, Google literally just said 100% lol but it technically isn't

    A key point to consider though is as you start opening crystals and fail to get the champion, your odds of getting him decrease as well, they don't increase or stay the same. They are always independent outcomes. That is the Gambler's fallacy.
  • 16wegnerk16wegnerk Member Posts: 214 ★★
    Just reiterating what others have said, but in a very general way. I’m also going to use the 1/45 chance of pulling an ant-man as estimated earlier in the thread.

    The odds of pulling one ant-man in 1 crystal is 1-(44/45), or 1- (odds of pulling no ant-man)

    Now, the odds of pulling AT LEAST 1 ant-man out of, say, 10 crystals is again 1-(odds of pulling no ant-man), though this time it’s done with the formula: 1- (44/45)^10 = ~ 20%

    Similarly, if you plan to open x crystals where x > 0, then the odds of pulling at least one ant-man would be 1-(44/45)^x

    Note that this function has no memory. This means that if you’ve already opened 100 crystals, it won’t change the odds of pulling one ant-man in your next 10 crystals (still ~20%)

    Note also that this includes the odds of pulling multiple ant-men in those x crystals. These are not the odds of pulling EXACTLY 1 ant-man, but rather AT LEAST 1.
  • Negative_100Negative_100 Member Posts: 1,650 ★★★★
    edited October 2020
    0.0169491525
    😂😂😂
  • Negative_100Negative_100 Member Posts: 1,650 ★★★★
    edited October 2020
    JK
  • RookiieRookiie Member Posts: 4,821 ★★★★★

  • DNA3000DNA3000 Member, Guardian Posts: 19,677 Guardian
    K00shMaan said:

    the odds of getting him (or anyone in the crystal) are 1/45 per attempt ~2.2%. If you add an attempt, you will increase your overall chance at getting him but it will be by a smaller amount than 2.2%. This way, every time you add an additional crystal to your attempt, your odds of getting him increase overall, but because they increase by a smaller amount for each additional crystal, it will never equal or exceed 100%.

    odds for 1 crystal = 1-(44/45)^1 = 2.22%
    odds for 10 crystals = 1-(44/45)^10 = 20.12%
    odds for 100 crystals = 1-(44/45)^100 = 89.43%
    odds for 952 crystals = 1-(44/45)^952 = 99.99%
    Anything above 952, Google literally just said 100% lol but it technically isn't

    A key point to consider though is as you start opening crystals and fail to get the champion, your odds of getting him decrease as well, they don't increase or stay the same. They are always independent outcomes. That is the Gambler's fallacy.

    This is a bit of an odd way to describe what's happening, particularly the "odds decrease" part.

    The correct way to think about probability is that it is a prediction of what will happen, before it happens. When the situation changes, the prediction changes. The odds of flipping a coin heads once is 50%. The odds of flipping a coin heads twice in a row before you flip is 25%. But if you flip the coin once and it comes out heads, the odds of the coin now coming up heads twice in a row is 50%, because the first head already happened.

    Mathematically, the odds of getting two heads in a row was the odds of getting the first head which was 50% times the odds of getting the second head which was also 50%, so the odds of getting two heads is 0.5 x 0.5 = 0.25. But once you flip the first head, the odds of getting the first head is now 100%, because it actually happened. So the odds of getting two heads in a row are now 100% x 50% = 50%.

    It isn't the odds of future events that change. It is the odds of the future events that become past events that change. They start off with some probability of happening, and then become 100% when they actually happen or 0% if they didn't happen. In fact, for people who have difficulty with the notion that the odds of past events changes once they occur to either 100% or 0%, consider what are the odds of a coin coming up heads twice in a row if the first flip is *tails*. Intuitively, we know the odds of getting heads twice in a row on the second flip is zero: it can't happen. But why? Isn't the odds still one in four? Of course not, because the first flipped failed, so no matter what happens on the second flip you can't get two in a row. The odds of the first one being heads isn't still 50/50. It is now zero. So the odds of getting two in a row are also zero.

    So the odds of getting a particular result in, say, ten crystals is calculated on the assumption that none of those ten crystals have been opened yet, so each has some odds of something happening. But once they are opened, those opened crystals no longer have a chance for something to happen: they now have a 100% chance for the result they generated to occur, and a 0% chance for anything else to happen, because of course. That changes the calculation, not because the odds of future events change, but rather because there are fewer events in the future. There were ten crystals in the future, and now there are less.

    The gambler's fallacy is the belief that if something happens less frequently than you'd expect statistically, it will happen more frequently in the future to "balance out." But if those future events have independent odds, the past can't affect the future. No matter what has happened in the past, the future odds of things happening in the future that haven't happened yet are only governed by the odds of those future events happening.
  • phillgreenphillgreen Member Posts: 4,127 ★★★★★
    There are 5 champs I really wanted from the science/mystic crystal. Opened 13, got none. I hope your luck is better than mine.

    This is why I dont gamble.
  • GroundedWisdomGroundedWisdom Member Posts: 36,573 ★★★★★
    Roman4544 said:

    Elad17 said:

    1/45 per roll so out of 5 crystals your chances are .00045

    I know the probability to get him in one crystal is 1/45, something around 2%. But what you saying is that I have less chances to pull him, even if I'm opening more crystals??? I should have more chances to get him at least once I think than 2 ish %
    Each Crystal you open is one CHANCE. The odds of that chance is 1/x (x=number of outcomes). Each Crystal is not swayed or affected by the previous outcome.
  • GroundedWisdomGroundedWisdom Member Posts: 36,573 ★★★★★
    Meaning, opening 5 of them doesn't increase your odds of pulling him, just the number of chances at 1/x.
  • DrZolaDrZola Member Posts: 9,125 ★★★★★
    edited October 2020
    https://www.mathsisfun.com/data/probability-events-independent.html

    Just going to leave this here. I would recommend anyone citing statistics in this thread (other than @DNA3000 ) take at least a casual perusal.

    Dr. Zola
  • GroundedWisdomGroundedWisdom Member Posts: 36,573 ★★★★★
    Probability can't answer the OP's question. If the odds are 1/45, the first Crystal is 1/45. If he doesn't pull Ant-Man the first time, the next Crystal is 1/45. If he does pull Ant-Man, the next Crystal will still be 1/45.
  • GroundedWisdomGroundedWisdom Member Posts: 36,573 ★★★★★
    The OP stated that they believed the chance should increase the more they open, but only the number of chances taken increases. The odds will still be 1/45 each time they open the Dual Crystal.
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