Math experts help me here(dual class crystal probability)

I really want to duplicate my Ant Man, let's say(I don't know the exactly numbers of champions of science and mystic in total, but it doesn't matter) there are 45 champions in total of these classes. What is the probability that in 5(50k shards) tries I'm able to get my Ant boy in one of them?
Ps: sorry if there is some grammar error, latino here.

Comments

Elad17

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

Lucifer1810

11 percent

Roman4544

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 %

The_Sentry06

I think you have a 2.22% chance per crystal to get an Antman.

Roman4544

I think you have a 2.22% chance per crystal to get an Antman.

I know, but if I open 5 of them, my chances should be higher than that. I used to know that stuff, but my math nowadays suck

The_Sentry06

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 %

The chances remain the same for every crystal you open.

UnidentifiedCreature

Since Ant Man is a bad character, the chances of you pulling him are through the roof.

Masterpuff

I think you have a 2.22% chance per crystal to get an Antman.
I know, but if I open 5 of them, my chances should be higher than that. I used to know that stuff, but my math nowadays suck

Each crystals has an individual roll of RNG. So each time you open one, the chance remains the same. 2.22%.

Batman_bruce31

Hey mate, these crystals don't work like that.
Each crystal has individual probabilities.
Lets say you have 2 % chance for 1 crystal.
The next crystal also has only 2 % chance.
The probability doesn't add up like that.
Thats why it's called probability.

Correct me if i am wrong but i this is what i know.

SneakyWarrior

@Elad17 ur math indicates the Probability of pulling consecutive Ant Mans. As for the answer, the probability will remain the same no matter how many crystals you open, 2.22% because opening multiple crystals does not change how likely you are to pull a single champion.

Roman4544

I think you have a 2.22% chance per crystal to get an Antman.
I know, but if I open 5 of them, my chances should be higher than that. I used to know that stuff, but my math nowadays suck
Each crystals has an individual roll of RNG. So each time you open one, the chance remains the same. 2.22%.

But I don't wanna pull him multiple times, my question is what is the odds to pull him ONE TIME out of these 5. Let's take heads or tails for example. I wanna bet on Heads to show up 1 time only in 10 tries, my chances definitely should be higher than 1/2 or 50%. It could pop 9 Tails, I only wanna get 1 Heads..

Batman_bruce31

I think you have a 2.22% chance per crystal to get an Antman.
I know, but if I open 5 of them, my chances should be higher than that. I used to know that stuff, but my math nowadays suck
Each crystals has an individual roll of RNG. So each time you open one, the chance remains the same. 2.22%.
But I don't wanna pull him multiple times, my question is what is the odds to pull him ONE TIME out of these 5. Let's take heads or tails for example. I wanna bet on Heads to show up 1 time only in 10 tries, my chances definitely should be higher than 1/2 or 50%. It could pop 9 Tails, I only wanna get 1 Heads..

But you could also pop 10 tails without a single head..... that's how it is mate.... praying to get what we want doesn't change the chance....we think we are lucky when we get what we want and feel unlucky when we don't.

Just pop open those and don't hope anything.
If u get him, you will feel happy and if you don't get him, just try another time.

Crcrcrc

Why is it the game is rigged to only give the champs we are not after 😂 I hope you get antman

Clarkkent76

It's exactly the same as rolling dice hoping for a 6. If you roll one dice, it's 1/6. If you roll 6 dice and you only want one 6 there are 6 available positive outcomes out of 36 available outcomes, so it's 6/36 which is....1/6.

Meciki

Clarkkent76

Actually, I'm completely wrong above.

With 6 dice there would actually be 46,656 available outcomes of which 31031 have a positive outcome of you only need one 6, so the odds would be 66.51%

So your first instinct is right. Even though the events are independent of each other, having more opportunities to get the outcome you want increases your odds.

Heattblastt

Almost 11%
(5/45) ×100~ 11%

Vendemiaire

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%

unknown

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Roman4544

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%

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.

unknown

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Vendemiaire

No matter what value you put in the formula you can't get past 100%. The best you can get is 99.9999999999...%

Vendemiaire

50 tries is:

1-(1-0.0222...)^50 = 67.49%

Not even close to 100%

Roman4544

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%

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.
No, in 50 tries u should multiply 44/45 with himself 50 times. Its still in range 0-1

Thanks man!

unknown

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unknown

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The_enlightened

Above I gave u the chances of pulling AT LEAST 1 ant-man. Here I'll give the chances of pulling exact 1 ant-man

If we want the event to occur exactly k times out of n tries, the probability will be:

P(k) = (n, k) * p^k * (1-p)^(n-k)
where ^k is k degree
p is probability of occuring the event in single try
(n, k) = n!/((n-k)!k!) (it's called COMBINATION, u can google it if u want to do other calculations. its total number of potential successes, bc we can get ant-man in first, second, third, fourth and fifth crystals)

Our event is getting ant-man, and it's probability in single try is 1/45
1-p = 44/45
our n is 5, because we have 5 tries
and k = 1, because we want to get him 1 time

So we get

P(1) = (5, 1) * 1/45 * (44/45)^4
(5,1) is 5

I'll spare u the math: we get

P(1) = 0.1015

Again, in probability we work with numbers from 0 to 1, so to convert it to percents we just multiply it by 100: chances are 10.15 %

As u see, the chances of getting EXACTLY 1 ant-man are smaller, than getting AT LEAST one, because latter contains situations with 2, 3, 4 and 5 ant-mans.

So, again: chances of getting 1 ant-man are 10.15 %, and chances of getting AT LEAST 1 ant-man are 10,62 %. If u want to, u can use this formulas with another number of tries (n), another number of potential successes (k) and probability (p). For example, if u want to both pull and dupe someone, k will be equal to 2. For incursions crystal p will be equal to 1/10, because there are 10 possible pools.

I was going to post something similar but this seems waaaaay smarter so yeah, I wasted 15 min typing but on the bright side I finally opened my statistics book from 2 semesters ago

Shafeeq

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%

Wish I studied my probability classes from forums

Roman4544

Above I gave u the chances of pulling AT LEAST 1 ant-man. Here I'll give the chances of pulling exact 1 ant-man

If we want the event to occur exactly k times out of n tries, the probability will be:

P(k) = (n, k) * p^k * (1-p)^(n-k)
where ^k is k degree
p is probability of occuring the event in single try
(n, k) = n!/((n-k)!k!) (it's called COMBINATION, u can google it if u want to do other calculations. its total number of potential successes, bc we can get ant-man in first, second, third, fourth and fifth crystals)

Our event is getting ant-man, and it's probability in single try is 1/45
1-p = 44/45
our n is 5, because we have 5 tries
and k = 1, because we want to get him 1 time

So we get

P(1) = (5, 1) * 1/45 * (44/45)^4
(5,1) is 5

I'll spare u the math: we get

P(1) = 0.1015

Again, in probability we work with numbers from 0 to 1, so to convert it to percents we just multiply it by 100: chances are 10.15 %

As u see, the chances of getting EXACTLY 1 ant-man are smaller, than getting AT LEAST one, because latter contains situations with 2, 3, 4 and 5 ant-mans.

So, again: chances of getting 1 ant-man are 10.15 %, and chances of getting AT LEAST 1 ant-man are 10,62 %. If u want to, u can use this formulas with another number of tries (n), another number of potential successes (k) and probability (p). For example, if u want to both pull and dupe someone, k will be equal to 2. For incursions crystal p will be equal to 1/10, because there are 10 possible pools.

Excellent explanation thanks! I will study this tomorrow, my head is so broken right now, long time since I don't practice this.
ps: the detail you mentioned(about one time only and at least) reminded of my math teacher who always make sure that we pay very close attention to the whole title of the question and their tricky meaning

Cliffordcan

Since Ant Man is a bad character, the chances of you pulling him are through the roof.

☝️Comment of the year right here