popped 8 crystals 3 of those were white mags 3*... the odds of that per crystals drop rate is insane and not sure how its possible. people that understand basic math should understand how it shouldnt be happening this often in game
to me, i see at least 2x 3x dupe pulls in stacks of 10 popped crystals, every other week. in the alliance feed of crystal openings a lot more often. even to the point where 2 separate accounts pull the same champ back to back of the same star lvl. knowing that i wouldnt put it past kabam to probably just being another bug with coding and people pulling duplicates within the same stacks.
That makes so much sense. If birthdays are treated as champion pulls and each person as a crystal, opening just 19 crystals means you have a 50% chance of getting the same champ atleast once. Assuming 250 champion pool
Inputting a pool of 250, and a stack of 10 openings, you get a probability of .1667 of getting a dupe. Basically the same as rolling a die and getting a specific number. You seeing it every other week is probably right on point. You seeing it in your alliance feed much more frequently, when 30 people are opening crystals, makes a ton of sense. What you’re describing is exactly the behavior the math predicts.
The irony of him saying “people that understand basic math should understand how it shouldnt be happening this often in game” and then you proving it should happen often is quite amusing.
Inputting a pool of 250, and a stack of 10 openings, you get a probability of .1667 of getting a dupe. Basically the same as rolling a die and getting a specific number. You seeing it every other week is probably right on point. You seeing it in your alliance feed much more frequently, when 30 people are opening crystals, makes a ton of sense. What you’re describing is exactly the behavior the math predicts.
The irony of him saying “people that understand basic math should understand how it shouldnt be happening this often in game” and then you proving it should happen often is quite amusing.
I think someone said this once on a thread about "coincidences" of bad luck. Given the sheer number of people who open the crystals you should expect it to happen frequently.
I think the only time it worked agaisnt kabam is during the punisher arena crystals issue
The irony of him saying “people that understand basic math should understand how it shouldnt be happening this often in game” and then you proving it should happen often is quite amusing.
I always have to remind myself that the people who say things like that, without being able to reference what math they’re using or walking you through it, are the same people who protested, “when am I going to need this!” from the back of class. They’re relying on how intuitive the result feels because they weren’t excited to learn all the lessons about how unintuitive things like math and physics can be. They’re not the type to write a forum post that says, “The birthday problem is so cool!” because they’re not fascinated by the solution to a mundane question. But still, that doesn’t stop them from “knowing” they’re being scammed because their intuition doesn’t match the results.
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I’m still confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I’m still confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
@SkyLord7000 try like this. 10 doors. You can either choose 1 or choose 9. If you choose 9 you open all doors and get to keep the car if it is there. Else you choose 1 and get to keep the car if it is there.
Now, you choose 1door for you to open and the host chooses 9doors for him to open.
He gives you a chance to swap with him. Ie you get 9 doors to open. Will you switch? Now, whether or not the host has opened the doors(without reward) before asking you to switch doesn't matter because you have 9 doors to check for the car instead of 1
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I still am confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
The part that’s confusing you is that 2 options don’t mean that it’s guaranteed to be 50/50
Let’s go with 100 doors. When you pick your first door, what are the odds that you picked the right one? The host hasn’t done anything yet.
The odds are 1/100 right? Because there are 100 doors and only 1 is right. That means there’s a 99% chance that you didn’t pick right the first time. The host then opens 98 of the doors, and shows you two doors. The odds don’t magically change now, the prize behind your first door hasn’t changed and they are still the original odds of 1% chance.
Let’s go back to a smaller number. Let’s say there’s 5 doors. There are 5 scenarios right? Doors 1-5 are the 5 options you can pick you can pick. Let’s say Door 3 has the prize behind it.
Scenario 1 You pick Door 1. The host opens door 2,4 and 5 and you have a choice between 1 and 3. Switching gets you the prize
Scenario 2 You pick door 2. The host opens doors 1, 4 and 5 and you have a choice between 2 and 3. Switching gets you the prize
Scenario 3 You pick door 3. The host opens 3 of the other doors, let’s say 1, 2 and 4. You have a choice between 3 and 5. Switching doesn’t get you the prize here
Scenario 4 You pick door 4. The host opens doors 1, 2 and 5 and you have a choice between 3 and 4. Switching gets you the prize
Scenario 5 You pick door 5. The host opens doors 1, 2 and 4 and you have a choice between 3 and 5. Switching gets you the prize
Can you see how in 4 out of 5 of those scenarios if you swap you will get the prize? Because it’s not a 50/50 choice between one door and another. The odds were decided when you had a 1 in 5 chance to pick the correct door at the start.
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I still am confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
The part that’s confusing you is that 2 options don’t mean that it’s guaranteed to be 50/50
Let’s go with 100 doors. When you pick your first door, what are the odds that you picked the right one? The host hasn’t done anything yet.
The odds are 1/100 right? Because there are 100 doors and only 1 is right. That means there’s a 99% chance that you didn’t pick right the first time. The host then opens 98 of the doors, and shows you two doors. The odds don’t magically change now, the pro3 behind your first door hasn’t changed and they are still the original odds of 1% chance.
Let’s go back to a smaller number. Let’s say there’s 5 doors. There are 5 scenarios right? Doors 1-5 are the 5 options you can pick you can pick. Let’s say Door 3 has the prize behind it.
Scenario 1 You pick Door 1. The host opens door 2,4 and 5 and you have a choice between 1 and 3. Switching gets you the prize
Scenario 2 You pick door 2. The host opens doors 1, 4 and 5 and you have a choice between 2 and 3. Switching gets you the prize
Scenario 3 You pick door 3. The host opens 3 of the other doors, let’s say 1, 2 and 4. You have a choice between 3 and 5. Switching doesn’t get you the prize here
Scenario 4 You pick door 4. The host opens doors 1, 2 and 5 and you have a choice between 3 and 4. Switching gets you the prize
Scenario 5 You pick door 5. The host opens doors 1, 2 and 4 and you have a choice between 3 and 5. Switching gets you the prize
Can you see how in 4 out of 5 of those scenarios if you swap you will get the prize? Because it’s not a 50/50 choice between one door and another. The odds were decided when you had a 1 in 5 chance to pick the correct door at the start.
Woah. 🤦♂️It makes more sense now. Thanks for showing the scenarios.
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I’m still confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
@SkyLord7000 try like this. 10 doors. You can either choose 1 or choose 9. If you choose 9 you open all doors and get to keep the car if it is there. Else you choose 1 and get to keep the car if it is there.
Now, you choose 1door for you to open and the host chooses 9doors for him to open.
He gives you a chance to swap with him. Ie you get 9 doors to open. Will you switch? Now, whether or not the host has opened the doors(without reward) before asking you to switch doesn't matter because you have 9 doors to check for the car instead of 1
Don’t understand this though. Seems like 90% chance if I choose the 9 doors and the car is placed randomly. Swapping seems irrelevant
Let’s say their is another game show. Doors doors 99 contestants. Each contestant opens a door one at a time. Your the 99th contestant and you spot out a door your thinking about choosing. Somehow, no one before you chooses your door or gets the prize. Do you go with your gut or switch?
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I’m still confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
@SkyLord7000 try like this. 10 doors. You can either choose 1 or choose 9. If you choose 9 you open all doors and get to keep the car if it is there. Else you choose 1 and get to keep the car if it is there.
Now, you choose 1door for you to open and the host chooses 9doors for him to open.
He gives you a chance to swap with him. Ie you get 9 doors to open. Will you switch? Now, whether or not the host has opened the doors(without reward) before asking you to switch doesn't matter because you have 9 doors to check for the car instead of 1
Don’t understand this though. Seems like 90% chance if I choose the 9 doors and the car is placed randomly. Swapping seems irrelevant
Yup. 90% chance when you choose the 9 doors after swapping. In other cases they open the 8 doors before asking you to swap . Whether or not they open the doors before asking you to swap, it's a 90% chance since you get to open 9 doors total
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I’m still confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
@SkyLord7000 try like this. 10 doors. You can either choose 1 or choose 9. If you choose 9 you open all doors and get to keep the car if it is there. Else you choose 1 and get to keep the car if it is there.
Now, you choose 1door for you to open and the host chooses 9doors for him to open.
He gives you a chance to swap with him. Ie you get 9 doors to open. Will you switch? Now, whether or not the host has opened the doors(without reward) before asking you to switch doesn't matter because you have 9 doors to check for the car instead of 1
Don’t understand this though. Seems like 90% chance if I choose the 9 doors and the car is placed randomly. Swapping seems irrelevant
Yup. 90% chance when you choose the 9 doors after swapping. In other cases they open the 8 doors before asking you to swap . Whether or not they open the doors before asking you to swap, it's a 90% chance since you get to open 9 doors total
Inputting a pool of 250, and a stack of 10 openings, you get a probability of .1667 of getting a dupe. Basically the same as rolling a die and getting a specific number. You seeing it every other week is probably right on point. You seeing it in your alliance feed much more frequently, when 30 people are opening crystals, makes a ton of sense. What you’re describing is exactly the behavior the math predicts.
so the math being implied here is for every individual crystal right. and not only that what are the odds of the SAME champ being pulled back to back by 2 different accounts? in a pool of 100 champs and star champs from 3*-6* the chances are a lot slimer yet the same 3*-5* are being pulled. i dont think you are applying all the correct math distributed across a lot more number pulls. you are simplifying the math to make it make sense, but its not as simple as you are making it to be.
Inputting a pool of 250, and a stack of 10 openings, you get a probability of .1667 of getting a dupe. Basically the same as rolling a die and getting a specific number. You seeing it every other week is probably right on point. You seeing it in your alliance feed much more frequently, when 30 people are opening crystals, makes a ton of sense. What you’re describing is exactly the behavior the math predicts.
so the math being implied here is for every individual crystal right. and not only that what are the odds of the SAME champ being pulled back to back by 2 different accounts? in a pool of 100 champs and star champs from 3*-6* the chances are a lot slimer yet the same 3*-5* are being pulled. i dont think you are applying all the correct math distributed across a lot more number pulls. you are simplifying the math to make it make sense, but its not as simple as you are making it to be.
I'm sure somebody has put together a model for a weighted version of the birthday problem, and I invite you to go find that formula and share it here if you want to make your case.
That said, if you're talking about the event cavs, then I suspect the odds of pulling the same champ over multiple pulls is actually quite a bit higher than in a 250 champ pool crystal. The Metal and Mutant cavs only have 27 champs in them with a 50% chance of getting the same rarity and an 81% chance of getting one of two rarities. The Cast and Crew crystals has the same odds with only a 20 champ pool. These pools being so much smaller is going to make collisions (dupes) much much more common.
Imagine for the sake of this example that the coin flip on 3* rarity vs. non-3* rarity happens first each time you open a crystal. In an average pop of 10 crystals, you'd get 5 crystals that are in 3* in rarity. If those 5 crystals are Metal and Mutant, with 27 champs in the pool, those 5 crystals give you a 32% chance of getting a dupe. Regardless of what the other 5 crystals in the 10-crystal pop do, you're looking at a baseline dupe rate of 1 in 3. Those other 5 crystals will also have some smaller percent chance of duplicating among champ and rarity. If we ignore rarity entirely and just ask what the odds are of a repeat champ in a 10 pop of M&M crystals, it's about 85%. Those numbers become 42% and 93%, respectively, in the Cast and Crew crystal with 20 champs in the pool.
You can speculate all day long, but unless you've got some actual math back up your belief that duplicates in these crystals shouldn't be rare, you're just not going to be convincing. I get that it feels shifty, but it's really not. Duplicates are just way more common than your intuition would lead you to believe.
Edit: I realized that I didn't address your first question. It doesn't matter which accounts are opening the crystals unless you're asking about crystals within a particular set limited by the account. The odds of you opening 10 crystals and getting a dupe are the same as the odds of you and I each opening 5 crystals and there being a dupe somewhere in our shared pool. The odds only go up if you're asking about all crystal pulls by a 30 person alliance, where many more crystals are being opened than any one account could open alone.
popped 8 crystals 3 of those were white mags 3*... the odds of that per crystals drop rate is insane and not sure how its possible. people that understand basic math should understand how it shouldnt be happening this often in game
to me, i see at least 2x 3x dupe pulls in stacks of 10 popped crystals, every other week. in the alliance feed of crystal openings a lot more often. even to the point where 2 separate accounts pull the same champ back to back of the same star lvl. knowing that i wouldnt put it past kabam to probably just being another bug with coding and people pulling duplicates within the same stacks.
Let’s say their is another game show. Doors doors 99 contestants. Each contestant opens a door one at a time. Your the 99th contestant and you spot out a door your thinking about choosing. Somehow, no one before you chooses your door or gets the prize. Do you go with your gut or switch?
I see what you're saying here, but this is a different scenario than the Monty Hall Problem. In your scenario, you've brought the options down to 2 doors by the random choices of the contestants who don't know where the car is. In 98 out of 100 shows, the 99th person doesn't even get to guess. The crux of the Monty Hall Problem is that Monty does know where the car is, and will never show it to you.
In the Monty Hall problem, there are effectively two sets of doors: the set of doors you pick on your initial guess and the set of doors you don't pick on your initial guess. The set of doors you pick on your initial guess is 1 door and has a 33% chance of a car. The set of doors you don't pick on your initial guess has a 2 doors and a combined 67% chance of a car. When you get the opportunity to switch doors, you're really getting the opportunity to switch sets--from a set with 1 door to a set with 2 doors.
Because Monty always knows where the car is, him showing you that there's a goat behind one of the doors in the set you didn't choose doesn't change the starting probability of the car being in that set. You already know that there's at least one goat in that set, and Monty changes nothing by showing you that goat. He's not choosing the door randomly, he's just telling you the information you already know and making it feel like new information. You're still picking those two doors combined by switching.
Let's take a look at these scenarios, but instead of thinking about the final outcome, think about the choice you have to make between swapping or staying:
You choose door 1, and the doors break down like this:
Door 1: Goat Door 2: Car Door 3: Goat
Monty is going to show you door 3, and you get to choose between staying on door 1 and your goat or swapping to door 2 and getting the car.
You choose door 1, and the doors break down like this:
Door 1: Goat Door 2: Goat Door 3: Car
Monty is going to show you door 2, and you get to choose between staying on door 1 and your goat or swapping to door 2 and getting the car
In both scenarios, Monty is forced to show you a particular door from the remaining set, and that necessitates that the car is behind the other door. The existence of the car in the set of doors you didn't choose forces him to pick a particular door to show you the goat. So whether the car is behind door 2 or door 3, you effectively get to pick both doors by swapping, because you know the car comes with a goat behind the other door already.
The last scenario is the one where you lose if you swap. You choose door 1, and the doors break down like this:
Door 1: Car Door 2: Goat Door 3: Goat
Monty has the freedom to show you either door 2 or 3, and when you swap, you swap into the set of two goats. But again, there are 3 scenarios and only 1 of those 3 land you with the two goat doors if you swap.
I pick one, Another door is opened to be nothing I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
your odds are locked in once you choose your first door. It’s 1/3 then, and removing the other option doesn’t suddenly make it 50/50 just because there’s only two options now.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
I’m still confused, with 98 empty doors you would have a 50 / 50 shot if you hadn’t chosen prior. So why would choosing something in an event unrelated change the odds.
Tried to italic the parts that confused me the most.
@SkyLord7000 try like this. 10 doors. You can either choose 1 or choose 9. If you choose 9 you open all doors and get to keep the car if it is there. Else you choose 1 and get to keep the car if it is there.
Now, you choose 1door for you to open and the host chooses 9doors for him to open.
He gives you a chance to swap with him. Ie you get 9 doors to open. Will you switch? Now, whether or not the host has opened the doors(without reward) before asking you to switch doesn't matter because you have 9 doors to check for the car instead of 1
Don’t understand this though. Seems like 90% chance if I choose the 9 doors and the car is placed randomly. Swapping seems irrelevant
Yup. 90% chance when you choose the 9 doors after swapping. In other cases they open the 8 doors before asking you to swap . Whether or not they open the doors before asking you to swap, it's a 90% chance since you get to open 9 doors total
I only experienced once where I pulled 6* AV three times consecutively. I wish this RNG could happen again when I pull 6* Aegon / NF / AA / Namor ……and etc.
This kind of crystal pull posts always end up with similar high school probability discussion. I really dunno how many times I saw the same responses from similar group of players. I suggest Mods put one of this kinds of post to the top. And I really hope we can see the algorithm behind the RNG someday. Now it seems like we’re watching two group of ppl arguing whether human is created by God or simply Darwin evolution.
One thing that I think is overlooked is that it is obvious to me that everyone likely has different look-up tables from which the RNG selects. Irregardless of how random the RNG is (which is also an issue), it seems everyone has certain champs that are much less likely to them than they are to someone else. This makes any discussion of probabilities here moot unless everyone has the same evenly distributed look-up table. Granted, having different tables is a guess on my part since Kabam has never explicitly said how they do the champ selection to the best of my knowledge. However, seeing how everyone has certain (different) champs that they never pull suggests this is a good possibility.
I only experienced once where I pulled 6* AV three times consecutively. I wish this RNG could happen again when I pull 6* Aegon / NF / AA / Namor ……and etc. two group of ppl arguing whether human is created by God or simply Darwin evolution.
This analogy is pretty much perfect I'd say. Except it's not darwinian evolution. Just evolution would suffice
Comments
I think the only time it worked agaisnt kabam is during the punisher arena crystals issue
I pick one,
Another door is opened to be nothing
I can stay or switch.
Both options are 50/50 now. Where am I wrong?
If someone replies to this please @SkyLord7000 I don’t wanna miss this.
To demonstrate this, imagine there were 100 doors, 1 with the prize and 99 with nothing. You randomly pick one, the host then opens 98 doors showing you nothing. He then asks you whether you want to swap.
The chances of you randomly picking the one door with the prize out of 100 is so low, that the host revealing 98 doors without a prize means that the final door he didn’t reveal likely has the prize in it.
The odds were locked in at 1% chance when you selected the first door. So the odds now that the other door contains the prize is 99%. Even though there are two options, it’s not 50/50.
Tried to italic the parts that confused me the most.
Now, you choose 1door for you to open and the host chooses 9doors for him to open.
He gives you a chance to swap with him. Ie you get 9 doors to open. Will you switch? Now, whether or not the host has opened the doors(without reward) before asking you to switch doesn't matter because you have 9 doors to check for the car instead of 1
Let’s go with 100 doors. When you pick your first door, what are the odds that you picked the right one? The host hasn’t done anything yet.
The odds are 1/100 right? Because there are 100 doors and only 1 is right. That means there’s a 99% chance that you didn’t pick right the first time. The host then opens 98 of the doors, and shows you two doors. The odds don’t magically change now, the prize behind your first door hasn’t changed and they are still the original odds of 1% chance.
Let’s go back to a smaller number. Let’s say there’s 5 doors. There are 5 scenarios right? Doors 1-5 are the 5 options you can pick you can pick. Let’s say Door 3 has the prize behind it.
Scenario 1
You pick Door 1. The host opens door 2,4 and 5 and you have a choice between 1 and 3. Switching gets you the prize
Scenario 2
You pick door 2. The host opens doors 1, 4 and 5 and you have a choice between 2 and 3. Switching gets you the prize
Scenario 3
You pick door 3. The host opens 3 of the other doors, let’s say 1, 2 and 4. You have a choice between 3 and 5. Switching doesn’t get you the prize here
Scenario 4
You pick door 4. The host opens doors 1, 2 and 5 and you have a choice between 3 and 4. Switching gets you the prize
Scenario 5
You pick door 5. The host opens doors 1, 2
and 4 and you have a choice between 3 and 5. Switching gets you the prize
Can you see how in 4 out of 5 of those scenarios if you swap you will get the prize? Because it’s not a 50/50 choice between one door and another. The odds were decided when you had a 1 in 5 chance to pick the correct door at the start.
That said, if you're talking about the event cavs, then I suspect the odds of pulling the same champ over multiple pulls is actually quite a bit higher than in a 250 champ pool crystal. The Metal and Mutant cavs only have 27 champs in them with a 50% chance of getting the same rarity and an 81% chance of getting one of two rarities. The Cast and Crew crystals has the same odds with only a 20 champ pool. These pools being so much smaller is going to make collisions (dupes) much much more common.
Imagine for the sake of this example that the coin flip on 3* rarity vs. non-3* rarity happens first each time you open a crystal. In an average pop of 10 crystals, you'd get 5 crystals that are in 3* in rarity. If those 5 crystals are Metal and Mutant, with 27 champs in the pool, those 5 crystals give you a 32% chance of getting a dupe. Regardless of what the other 5 crystals in the 10-crystal pop do, you're looking at a baseline dupe rate of 1 in 3. Those other 5 crystals will also have some smaller percent chance of duplicating among champ and rarity. If we ignore rarity entirely and just ask what the odds are of a repeat champ in a 10 pop of M&M crystals, it's about 85%. Those numbers become 42% and 93%, respectively, in the Cast and Crew crystal with 20 champs in the pool.
You can speculate all day long, but unless you've got some actual math back up your belief that duplicates in these crystals shouldn't be rare, you're just not going to be convincing. I get that it feels shifty, but it's really not. Duplicates are just way more common than your intuition would lead you to believe.
Edit: I realized that I didn't address your first question. It doesn't matter which accounts are opening the crystals unless you're asking about crystals within a particular set limited by the account. The odds of you opening 10 crystals and getting a dupe are the same as the odds of you and I each opening 5 crystals and there being a dupe somewhere in our shared pool. The odds only go up if you're asking about all crystal pulls by a 30 person alliance, where many more crystals are being opened than any one account could open alone.
In the Monty Hall problem, there are effectively two sets of doors: the set of doors you pick on your initial guess and the set of doors you don't pick on your initial guess. The set of doors you pick on your initial guess is 1 door and has a 33% chance of a car. The set of doors you don't pick on your initial guess has a 2 doors and a combined 67% chance of a car. When you get the opportunity to switch doors, you're really getting the opportunity to switch sets--from a set with 1 door to a set with 2 doors.
Because Monty always knows where the car is, him showing you that there's a goat behind one of the doors in the set you didn't choose doesn't change the starting probability of the car being in that set. You already know that there's at least one goat in that set, and Monty changes nothing by showing you that goat. He's not choosing the door randomly, he's just telling you the information you already know and making it feel like new information. You're still picking those two doors combined by switching.
Let's take a look at these scenarios, but instead of thinking about the final outcome, think about the choice you have to make between swapping or staying:
You choose door 1, and the doors break down like this:
Door 1: Goat
Door 2: Car
Door 3: Goat
Monty is going to show you door 3, and you get to choose between staying on door 1 and your goat or swapping to door 2 and getting the car.
You choose door 1, and the doors break down like this:
Door 1: Goat
Door 2: Goat
Door 3: Car
Monty is going to show you door 2, and you get to choose between staying on door 1 and your goat or swapping to door 2 and getting the car
In both scenarios, Monty is forced to show you a particular door from the remaining set, and that necessitates that the car is behind the other door. The existence of the car in the set of doors you didn't choose forces him to pick a particular door to show you the goat. So whether the car is behind door 2 or door 3, you effectively get to pick both doors by swapping, because you know the car comes with a goat behind the other door already.
The last scenario is the one where you lose if you swap. You choose door 1, and the doors break down like this:
Door 1: Car
Door 2: Goat
Door 3: Goat
Monty has the freedom to show you either door 2 or 3, and when you swap, you swap into the set of two goats. But again, there are 3 scenarios and only 1 of those 3 land you with the two goat doors if you swap.
Edit: @SkyLord7000 sorry, forgot to tag
This kind of crystal pull posts always end up with similar high school probability discussion. I really dunno how many times I saw the same responses from similar group of players. I suggest Mods put one of this kinds of post to the top. And I really hope we can see the algorithm behind the RNG someday. Now it seems like we’re watching two group of ppl arguing whether human is created by God or simply Darwin evolution.
👀👀👀🤦🤦🤦
...I knew this would happen when DNA mentioned the Monty Hall problem...