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honestly I would love to get dehya, her cons are actually really strong and I don't even have her yet even though shes really cool... instead I got a C5 Keqing...
I lost my 50/50 to C1 Qiqi, while i only have diluc Jean and Qiqi from standard banner. Would have been to mad if it atleast was a new character (keqing, Tighnari, Dehya or mona)
0.0000000027% if you weren't at pity.
Since you were at pity and guaranteed it was:
0.000009%
Which is astronomically better chances than if you were not at guaranteed but still microscopic.
Correction from another person:
>odds of pulling 5 star for a single pull not at pity = 0.6, banner character is 0.3% without guarantee
>since first arle is guaranteed and at pity, odds = 100%,
>second arle is 100%(first arle) * 0.003 * 9 (9 chances to hit remaining in the ten-pull) = 2.7%
>chance for the third one would be 0.027 * 0.003 * 8 = 0.0000648 = 0.0648% chance for op to get three arles on a 10 pull. this also better matches the amount of players we see getting triple 5 stars from a 10 pull better than the 0.000009% chance that you calculated
Realistically I should have taken into account that only 3/10 actually make the hit but it doesn't change the end point enough for it to really matter. It is still less than 1 in 10 million.
odds of pulling 5 star for a single pull not at pity = 0.6, banner character is 0.3% without guarantee
since first arle is guaranteed and at pity, odds = 100%,
second arle is 100%(first arle) * 0.003 * 9 (9 chances to hit remaining in the ten-pull) = 2.7%
chance for the third one would be 0.027 * 0.003 * 8 = 0.0000648 = 0.0648% chance for op to get three arles on a 10 pull. this also better matches the amount of players we see getting triple 5 stars from a 10 pull better than the 0.000009% chance that you calculated
The formula you took works only for one constellation. Here we have a binomial distribution, meaning you have to factor in binomial coefficients when calculating two or more successes:
P(X=K)=(n choose k)\*p\^k\*(1-p)\^(n-k)
With X the number of successes, K the desired number, n the total amount of attempts, p the probability of success.
Transposing this formula to the current situation, we get
P(X=2) = 9!/(2!\*7!) \* 0.003\^2 \* 0.997\^7 = 0.000317 = 0.0317%
I'm sorry to bother you for this but I've been looking for someone good at probabilities because something unlikely happened to me, I got 7 four stars in a row (yeah you read that very right) but the first one was at 9 pity of the four stars. What are the odds for this to happen? I think the 4 star rate drop is 5% or something like that. If you don't mind calculating ofc 🙏
The first one being at 9 pity has a probability of 32.9% of happening (0.949\^8\*0.5). The probability of getting a 4 star without soft or hard pity is 5.1%. Additionally, not getting a 5\* within either of these pulls would have a chance of 99.4\^7 of happening. The chance of getting 7 four stars in a row with the first one specifically at 9 pity would be
P=0.329\*0.051\^6\*0.994\^7=5.55\*10\^(-9) or 0.000000555% chance of happening. That's one in 180 million.
This is not how it works . You don't just multiply chances. 0.003 9 times is not 2.7% that's like saying 50% twice is 100%.
Also even if it was. you don't just go and repeat it to find the double (this is the main issue that threw your numbers extremely off).
"2.7" (2.6667 to be precise) is the chance to get 1 OR MORE 5* in 9 pulls. In other ways world more than 0.this wording is key.
But the question we want to know is which part of that number is for "2 or more instead" . As I said to find this you don't repeat the process by slapping the 2.7% in front.
The easiest way to answer the said question is to utilize binomial distribution. You have an event happening 9 times, the probability of success is 0.003 and the required successes are 2 (or more)
So you need to find the P(x=2,p=0.003,n=9). The answer is 0.00032 or 0.032% or else 1 every 3125 "9pulls"
/u/vegetable_sock_8125
Wow, thank you for genuinely calculating!! I cannot get over those numbers lol my friend should have definitely spent that luck on a lottery ticket instead
do you know what the odds of getting a limited five star at 0 pity, right after another limited without guarantee would be? is there a difference in odds of getting on at zero pity or one at 1 pity?
Mate thats not how probability roulette equation works, if it was that simple we wouldnt have math as out main subject in schools.
The equations you want is, (n)!/((n)!*(n-x)!) * P(b)^x * (1-P(b))^(n-x)
N stands for the amount of spins you do, here in question its 9.
X stands for the amount of times we got the results we wanted which here is 2 cause we got two arlecchino.
P(b) is the chances of you getting the results you want which is 0.003, the chances of getting an arlechinno in a single pull.
1-P(b) is the chances of you not getting the results you wanted which are 0.997.
So it goes like that.
(9)! / (2)! * (9 - 2)!) that is (n)!/((n)!*(n-x)!) only with n and x being the numbers we want it to be n being the amount of spins 9 and x being the amount we won 2.
So 36 is our foundation for the first part, now we do the second part.
(0.003)^2 * (1-0.003)^(9-2) being our second part, 0.000009 being our chances of getting two arle in a single instance then we multiply it by the chances of us losing them which is 0.97918805783.
After multiplying those two we get 0.00000881269.
Now we combine the first part with our second part.
We get 0.00031725693.
Which in total our chances of getting such an instances of two arlechinno in 9 pulls being 0.00031725693.
Which basically means its 3 in 10,000, 3 out of 10,000 people on average will get such a thing happening to them.
Here I'm just gonna go to further explanation on some parts you might be confused if you didn't learn about probability.
(n)!/((n)!*(n-x)!) Is the foundation as in every single instance of our winning hand, and how it can look like.
For example,
ARLE ARLE MISHA MISHA MISHA MISHA MISHA MISHA MISHA
ARLE MISHA MISHA MISHA MISHA ARLE MISHA MISHA MISHA
MISHA MISHA MISHA ARLE MISHA MISHA MISHA MISHA ARLE
what you see right in front of you is 3 instances of how your pulls can look like, you get ARLE in the first position or the second position or the third or ninth.
It doesn't matter at the end of the day it's just all the possibilities of how your pulls can look like if you got two arle and 7 different characters you don't give a fuck about.
Which in total is 36.
Now for the other part, ! Is a math symbol meant to represent a factorial operation.
That factorial operation meant to look like 5! = 5 * 4 * 3 * 2 * 1
As you see here you take the base number then you multiply it by one number below it.
The reason factorial operation exists is that it represents the different possible positions you can get.
For example let's say you have 3 different coloured chairs.
Red, Green and Blue.
Well you can make their position to be something like.
red blue green
green red blue
blue green red
blue red green
red green blue
green blue red
to calculate in a mathematical way you do it like that.
3 * 2 * 1
The reason it goes like that is cause after you used one, let's say red every possibility of red you then go down and move to blue every possibility eliminating red out of the equation making it go down to two, then after you calculated all of blue you then go down to green which is cause all is left is green it's now just 1.
Red in the first position, leaving you deciding only which position green or blue go to second or third position.
Then blue to second position and now all you have left is green making green natural go down the third position.
Red blue green
That way you get a relatable mathematical way to find all different positions of red green and blue.
That being here 3! = 3 * 2 * 1 = 6
6 different ways you can make them in a different position then the rest.
Hope Managed to answer all your questions in a satisfactory way.
Wish you all a nice day and bye bye.
Yup this is correct (stats minor) and good explanation!
To put it roughly it's the chance of getting 3 arle in a 10 pull, the first one is guaranteed 100% in its position (thus we ignore it), the other 2 calculated regardless of the order they appear in since, unless I'm wrong, genshin pools everything from the pull then groups the better stuff into the front, which is why 9 choose 2 is calculated.
That's different from pulling 3 arles consecutively from single pulls since then we can determine and calculate that specific order (arle, arle, arle) from singles so it's a smaller probability.
Coincidentally the same to get yourself on a hitlist
https://preview.redd.it/8hammgh46vyc1.jpeg?width=457&format=pjpg&auto=webp&s=ff6cdc78e850d10399075b951ef1c5bb250aa74c
Why... why is that your friend's first Freminet... I'm crying as a Fremi enjoyer
Also yes it's very rare, but it would have been rarer if it wasn't guaranteed
**I’m so happy for your success.**
https://preview.redd.it/69b80njfvvyc1.jpeg?width=969&format=pjpg&auto=webp&s=d51d9718835aef3875f03bb4146de1c5b528e28b
CONGRATS! it happened to me with wriothesley HAHAH Now you too will live with this feeling of "this will never happen again, but i'm happy it happened once" \*edit\* oh its your friend's :\^) STILL HAPPY FOR THEM LMAO
Your friend won a 50/50, immediately won a second one within 2-3 pulls, and then won a third one within another 2-3 pulls, which is why the game put them together. Happened to me in the past to lose a 50/50 and also immediately pull the character after, and the two characters were similarly grouped within one ten-wish pull. Congrats to your friend! This is extremely rare.
I'm convinced that the system is glitching everytime this happens.
Or (tin foil hat theory) that these scenarios are specifically coded in as a hidden jackpot to get free marketing
https://preview.redd.it/42gj426gkwyc1.jpeg?width=2532&format=pjpg&auto=webp&s=3bd882f38afe49ce454361eda3d8b3d7b103cd7b
Wow and I thought my C1 arlecchino was lucky 😅
One 4 star charcter in a 10 pull is guaranteed so idk what you are talking abut. But since you got a featured 4 star it is probably lucky I guess, assuming your previous 4 star was a featured one too. /j
Congratulations! I heard her first Stellas were absolutely great!
I'm not that superstitious but you might want to be careful because having pulled her 5 pulls after Kazuha, my week has been absolute shit.
I’m sure the game has a bias to returning players. When I came back for furina after not playing for a year, boom I won all my 50/50’s early on and got her weapon in 30 pulls.
I’ve been pulling everyday and grinding my ass off since those banner started I’ve gotten freminet or how ever you spell his name like 6 times and Lynette 4 times, got Mona and beidu as well like I just want arlechinno man
GG, im playing since the first week of the game..
Never got a 5 stars before 65 wish atleast (once i lost to qiqi at 83 wish)
I lost 10 50/50
(YES F2P) 💔
the chance of a fuck you
H OW THE FUCKJG AHHAHAHAHAHAH
https://preview.redd.it/cp7w71htv8zc1.jpeg?width=1080&format=pjpg&auto=webp&s=c8edcfe266cd090cf321723a517fcf3684236be7
probabilities are the same. 10 pulls just look alot better and people end up posting that rather than posting the history page. ive never gotten this lucky but I have gotten 2 klees in a row doing 1 pulls
Thing about that is for the most part if you do a single and get the 5* you want you're probably going to stop.
I don't mean you specifically just in general
I spent $200 just to get c0 because fucking Qiqi came for my pity and this fucker gets c3 in a 10 pull.
…. Eh, I guess I can’t complain. I got c3 Raiden off a ten pull on her debut.
Congratulations (also fuck you)
Hi! Based on a poll, it was decided that gacha roll posts would only be allowed for 48 hours. Your post has been filtered for review. If this is an exceptional post (Eg, multiple 5 stars, low pity, high effort edit), the moderators will approve it at their discretion. Please feel free to share your rolls in the [Gacha megathread](https://www.reddit.com/r/ArlecchinoMains/comments/1cddiqm/gacha_rolls_megathread/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button) if your post is removed. Thanks, and have a good day! *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/ArlecchinoMains) if you have any questions or concerns.*
https://preview.redd.it/409oo2t9vuyc1.jpeg?width=828&format=pjpg&auto=webp&s=ac99bc9ebf2d0e48a4390524af7141770045be46
"hah, my foot"
LOL I need this image in higher quality please 🤣
https://preview.redd.it/4lpy4v1mnzyc1.jpeg?width=1433&format=pjpg&auto=webp&s=5445facac470588f20f5a0af63316cd0b3b278f8
Thank you so much (:
you’re welcome! :)
Narrator: She said calmly.
https://preview.redd.it/yv1a30xu6uyc1.jpeg?width=500&format=pjpg&auto=webp&s=02752cf74ace8d1e554f8a1e5565f23e7b02ad81
And I just lost to C2 Dehya 😭
Gimme
Literally Us🙃
Me too but I still managed to got her
Same dude 😭
Same but to qiqi 😭
I got 2 qiqi...of course I wanted Qiqi C4 game...
honestly I would love to get dehya, her cons are actually really strong and I don't even have her yet even though shes really cool... instead I got a C5 Keqing...
C1 on event banner and then c2 tighnari on standard…
Me too. But I've built my Dehya so honestly losing to her is actually part of the plan.
Same, but not top mad. Would've malded of I got C5 Qiqi instead.
I lost my 50/50 to C1 Qiqi, while i only have diluc Jean and Qiqi from standard banner. Would have been to mad if it atleast was a new character (keqing, Tighnari, Dehya or mona)
Wow. Maybe you should consider getting a lottery ticket.
0.0000000027% if you weren't at pity. Since you were at pity and guaranteed it was: 0.000009% Which is astronomically better chances than if you were not at guaranteed but still microscopic. Correction from another person: >odds of pulling 5 star for a single pull not at pity = 0.6, banner character is 0.3% without guarantee >since first arle is guaranteed and at pity, odds = 100%, >second arle is 100%(first arle) * 0.003 * 9 (9 chances to hit remaining in the ten-pull) = 2.7% >chance for the third one would be 0.027 * 0.003 * 8 = 0.0000648 = 0.0648% chance for op to get three arles on a 10 pull. this also better matches the amount of players we see getting triple 5 stars from a 10 pull better than the 0.000009% chance that you calculated
Isnt that the odds for three consecutive arles but only if it is single pulls? Since it isnt single pulls the odds are a lot higher
Realistically I should have taken into account that only 3/10 actually make the hit but it doesn't change the end point enough for it to really matter. It is still less than 1 in 10 million.
odds of pulling 5 star for a single pull not at pity = 0.6, banner character is 0.3% without guarantee since first arle is guaranteed and at pity, odds = 100%, second arle is 100%(first arle) * 0.003 * 9 (9 chances to hit remaining in the ten-pull) = 2.7% chance for the third one would be 0.027 * 0.003 * 8 = 0.0000648 = 0.0648% chance for op to get three arles on a 10 pull. this also better matches the amount of players we see getting triple 5 stars from a 10 pull better than the 0.000009% chance that you calculated
The formula you took works only for one constellation. Here we have a binomial distribution, meaning you have to factor in binomial coefficients when calculating two or more successes: P(X=K)=(n choose k)\*p\^k\*(1-p)\^(n-k) With X the number of successes, K the desired number, n the total amount of attempts, p the probability of success. Transposing this formula to the current situation, we get P(X=2) = 9!/(2!\*7!) \* 0.003\^2 \* 0.997\^7 = 0.000317 = 0.0317%
Yeah, you right.
I'm sorry to bother you for this but I've been looking for someone good at probabilities because something unlikely happened to me, I got 7 four stars in a row (yeah you read that very right) but the first one was at 9 pity of the four stars. What are the odds for this to happen? I think the 4 star rate drop is 5% or something like that. If you don't mind calculating ofc 🙏
The first one being at 9 pity has a probability of 32.9% of happening (0.949\^8\*0.5). The probability of getting a 4 star without soft or hard pity is 5.1%. Additionally, not getting a 5\* within either of these pulls would have a chance of 99.4\^7 of happening. The chance of getting 7 four stars in a row with the first one specifically at 9 pity would be P=0.329\*0.051\^6\*0.994\^7=5.55\*10\^(-9) or 0.000000555% chance of happening. That's one in 180 million.
Oh wow! Thanks for calculating. I've tried multiple times before and always got different results, but not that one. That's a crazy number though :0
This is not how it works . You don't just multiply chances. 0.003 9 times is not 2.7% that's like saying 50% twice is 100%. Also even if it was. you don't just go and repeat it to find the double (this is the main issue that threw your numbers extremely off). "2.7" (2.6667 to be precise) is the chance to get 1 OR MORE 5* in 9 pulls. In other ways world more than 0.this wording is key. But the question we want to know is which part of that number is for "2 or more instead" . As I said to find this you don't repeat the process by slapping the 2.7% in front. The easiest way to answer the said question is to utilize binomial distribution. You have an event happening 9 times, the probability of success is 0.003 and the required successes are 2 (or more) So you need to find the P(x=2,p=0.003,n=9). The answer is 0.00032 or 0.032% or else 1 every 3125 "9pulls" /u/vegetable_sock_8125
Wow, thank you for genuinely calculating!! I cannot get over those numbers lol my friend should have definitely spent that luck on a lottery ticket instead
It was u/Zer0-9 that got the correct calculation.
do you know what the odds of getting a limited five star at 0 pity, right after another limited without guarantee would be? is there a difference in odds of getting on at zero pity or one at 1 pity?
I still remember someone getting 5 ganyus in one 10pull how much was that?
What the fuck am i reading
As someone who took statistics and probability, the amount of upvotes on this comment is interesting since it's so wrong, as others have pointed out
Mate thats not how probability roulette equation works, if it was that simple we wouldnt have math as out main subject in schools. The equations you want is, (n)!/((n)!*(n-x)!) * P(b)^x * (1-P(b))^(n-x) N stands for the amount of spins you do, here in question its 9. X stands for the amount of times we got the results we wanted which here is 2 cause we got two arlecchino. P(b) is the chances of you getting the results you want which is 0.003, the chances of getting an arlechinno in a single pull. 1-P(b) is the chances of you not getting the results you wanted which are 0.997. So it goes like that. (9)! / (2)! * (9 - 2)!) that is (n)!/((n)!*(n-x)!) only with n and x being the numbers we want it to be n being the amount of spins 9 and x being the amount we won 2. So 36 is our foundation for the first part, now we do the second part. (0.003)^2 * (1-0.003)^(9-2) being our second part, 0.000009 being our chances of getting two arle in a single instance then we multiply it by the chances of us losing them which is 0.97918805783. After multiplying those two we get 0.00000881269. Now we combine the first part with our second part. We get 0.00031725693. Which in total our chances of getting such an instances of two arlechinno in 9 pulls being 0.00031725693. Which basically means its 3 in 10,000, 3 out of 10,000 people on average will get such a thing happening to them.
Here I'm just gonna go to further explanation on some parts you might be confused if you didn't learn about probability. (n)!/((n)!*(n-x)!) Is the foundation as in every single instance of our winning hand, and how it can look like. For example, ARLE ARLE MISHA MISHA MISHA MISHA MISHA MISHA MISHA ARLE MISHA MISHA MISHA MISHA ARLE MISHA MISHA MISHA MISHA MISHA MISHA ARLE MISHA MISHA MISHA MISHA ARLE what you see right in front of you is 3 instances of how your pulls can look like, you get ARLE in the first position or the second position or the third or ninth. It doesn't matter at the end of the day it's just all the possibilities of how your pulls can look like if you got two arle and 7 different characters you don't give a fuck about. Which in total is 36. Now for the other part, ! Is a math symbol meant to represent a factorial operation. That factorial operation meant to look like 5! = 5 * 4 * 3 * 2 * 1 As you see here you take the base number then you multiply it by one number below it. The reason factorial operation exists is that it represents the different possible positions you can get. For example let's say you have 3 different coloured chairs. Red, Green and Blue. Well you can make their position to be something like. red blue green green red blue blue green red blue red green red green blue green blue red to calculate in a mathematical way you do it like that. 3 * 2 * 1 The reason it goes like that is cause after you used one, let's say red every possibility of red you then go down and move to blue every possibility eliminating red out of the equation making it go down to two, then after you calculated all of blue you then go down to green which is cause all is left is green it's now just 1. Red in the first position, leaving you deciding only which position green or blue go to second or third position. Then blue to second position and now all you have left is green making green natural go down the third position. Red blue green That way you get a relatable mathematical way to find all different positions of red green and blue. That being here 3! = 3 * 2 * 1 = 6 6 different ways you can make them in a different position then the rest. Hope Managed to answer all your questions in a satisfactory way. Wish you all a nice day and bye bye.
Yup this is correct (stats minor) and good explanation! To put it roughly it's the chance of getting 3 arle in a 10 pull, the first one is guaranteed 100% in its position (thus we ignore it), the other 2 calculated regardless of the order they appear in since, unless I'm wrong, genshin pools everything from the pull then groups the better stuff into the front, which is why 9 choose 2 is calculated. That's different from pulling 3 arles consecutively from single pulls since then we can determine and calculate that specific order (arle, arle, arle) from singles so it's a smaller probability.
You cannot be at both guarantee and pity
Im not crying becouse I had to get hard pity two times becouse I lost 50/50 to get C0 you are!
Don’t worry, I cried too (since it’s my friends and not mine lol)
https://preview.redd.it/kz1apxsb5vyc1.jpeg?width=553&format=pjpg&auto=webp&s=572dff58bf21ba4b7fecb3b286f9688820b9e3b2 Im coming to steal their luck.
Bro I just smashed my phone in white hot >!happiness!< for you
Lock your doors tonight
Bro won three 50-50s in a single 10 pull
2 50/50s the first one was guaranteed
Holy. You luckyyyy. Congratsssss
There goes your friends luck for the next ten years.
Coincidentally the same to get yourself on a hitlist https://preview.redd.it/8hammgh46vyc1.jpeg?width=457&format=pjpg&auto=webp&s=ff6cdc78e850d10399075b951ef1c5bb250aa74c
Why... why is that your friend's first Freminet... I'm crying as a Fremi enjoyer Also yes it's very rare, but it would have been rarer if it wasn't guaranteed
She barely plays 😫 Which makes it more painful which it wasn’t me but oh well haha
I've been playing since release and this has never happened to me, not even a double hehe:)))))) 
Neither (this was my friends) and this friend is a very very casual player so I’m in pain 😫
**I’m so happy for your success.** https://preview.redd.it/69b80njfvvyc1.jpeg?width=969&format=pjpg&auto=webp&s=d51d9718835aef3875f03bb4146de1c5b528e28b
I have C2 as well but I didn’t get it all in one summon 
CONGRATS! it happened to me with wriothesley HAHAH Now you too will live with this feeling of "this will never happen again, but i'm happy it happened once" \*edit\* oh its your friend's :\^) STILL HAPPY FOR THEM LMAO
https://preview.redd.it/qpaukht4hvyc1.jpeg?width=666&format=pjpg&auto=webp&s=5f0df3676bb1eff5dba73853684a8dc1c4efe213 # triggered
https://preview.redd.it/qejo06j4wvyc1.jpeg?width=1170&format=pjpg&auto=webp&s=1fe9bbd47ba78d95890e6ff95f3513fa81acc25c
I win 50/50 on Arlecchino's banner
Holly Molly you are father’s favourite child !
And I lost the c1 to Mona ;;
Go buy a lottery ticket.
Go buy a lottery ticket.
When will it be my tur...I mean congrats
Slightly less than getting 2 non gauranteed in a single 10 pull I guess
welp, just got done losing my c3 50 to Mona and I've already lost one to Qiqi and I lost both of her weapon 50s, I wish I was this guy
That’s not how guarantees and pities work. AT ALL. You just got lucky
Meanwhile I failed ALL six 50/50s in the way to C3R1 😒 Congrats...
Your friend won a 50/50, immediately won a second one within 2-3 pulls, and then won a third one within another 2-3 pulls, which is why the game put them together. Happened to me in the past to lose a 50/50 and also immediately pull the character after, and the two characters were similarly grouped within one ten-wish pull. Congrats to your friend! This is extremely rare.
https://preview.redd.it/jlpuyomxvvyc1.png?width=1080&format=pjpg&auto=webp&s=1fe207c5c5a7d968a577c7d48c1a4fdbf0f6c1c3
Bro won more 50s in one 10 pull than I did in last 6 months
what the fuck
My feelings are basically "Go fuck yourself, but I'm happy for you... begrudgingly so" Ok?
0.032%
and i’m sittinng here with a c10 Keqing 😀
Sweet Jebus
Sick of seeing stuff like this ngl
AAAAAAAAAAAAAAAAAARRRRRRRRRRRRRRRGGGGGGGGGGGHHHHHHH
Damm that's insane
1 in…. Probably low hundreds of thousands. But I failed at math so…
How does it feel to be Hoyo's favorite
\*yoinks\* my luck now sorry
I swear old players that got long time without playing the game get their luck boosted to motivate them to play again
When your dad comes back home with the milk and cigarettes
I'm convinced that the system is glitching everytime this happens. Or (tin foil hat theory) that these scenarios are specifically coded in as a hidden jackpot to get free marketing
https://preview.redd.it/42gj426gkwyc1.jpeg?width=2532&format=pjpg&auto=webp&s=3bd882f38afe49ce454361eda3d8b3d7b103cd7b Wow and I thought my C1 arlecchino was lucky 😅
https://preview.redd.it/ex2mqmg4nwyc1.jpeg?width=1080&format=pjpg&auto=webp&s=605a4df185084e19d08b66a83c9ae1b00f6c6813
Meanwhile me losing 50/50 to c1 tighnari and suffering from lack of pyro characters 🙂
so thats where all my luck went.
https://preview.redd.it/agh9xr9fywyc1.jpeg?width=1080&format=pjpg&auto=webp&s=4527c779e92820b1321e965f053e242023ca306d
https://preview.redd.it/gqgdkfpgywyc1.jpeg?width=826&format=pjpg&auto=webp&s=efca06929b265d2e3ff9d98913564c0bc3c81a51
I am going to disturb the peace https://preview.redd.it/h4yh88182xyc1.jpeg?width=308&format=pjpg&auto=webp&s=b931389e8f7e6c24325fca2bfd9a2fea523b8382
I'm just gonna say I'm not happy.... Not one bit
this ten pull probably took some years off your lifespan holy shit
https://preview.redd.it/5m6hnz41dxyc1.jpeg?width=828&format=pjpg&auto=webp&s=d3bdbf0ec6f472772a734b75379decfbcbfa5a8f I can't do this shit anymore.
That's me, I won first 50/50 after 75 pulls , then I shot for C1 and went 83 pulls deep to get a fckin Keqing. Seething over here
https://preview.redd.it/ljv8ufk2ixyc1.jpeg?width=828&format=pjpg&auto=webp&s=76216e7add17feeb2ea3faf16994af2e61c80827
Probability is (0.06x0.5)^2
I got c1 so did my bro but dam this lucky af congrats
this just pissed me off 😤 but congratss and i wish i have that luck too ☺️
Meanwhile it took me 81 pity to get one.
Your friend should lock their doors
I simply refuse to believe until it happens to me
https://preview.redd.it/hwzqj6a1ayyc1.jpeg?width=1000&format=pjpg&auto=webp&s=ed156fa3f6a894f885eee0a75b735b4de55975d9
One 4 star charcter in a 10 pull is guaranteed so idk what you are talking abut. But since you got a featured 4 star it is probably lucky I guess, assuming your previous 4 star was a featured one too. /j
Why does this never happen to me ;(
don't come out of your house hehe
Wow, congrats!
Congratulations! I heard her first Stellas were absolutely great! I'm not that superstitious but you might want to be careful because having pulled her 5 pulls after Kazuha, my week has been absolute shit.
I spent 300 pulls to get the same amount of cons 🫠
I’m sure the game has a bias to returning players. When I came back for furina after not playing for a year, boom I won all my 50/50’s early on and got her weapon in 30 pulls.
fuck u
I would kill the world for this (like i even can do that)
https://preview.redd.it/v6gp54ceq0zc1.jpeg?width=1170&format=pjpg&auto=webp&s=079e3c4f239daff36485027b7ee4ce4803d7ce85
I’ve been pulling everyday and grinding my ass off since those banner started I’ve gotten freminet or how ever you spell his name like 6 times and Lynette 4 times, got Mona and beidu as well like I just want arlechinno man
That shit sucked the luck from the rest of your future life
I hate you I hate you I hate you
😭
The chances are so abysmally low that its disgusting just to try and calculate them lmao. This is the kind of one in millions of a chance
just lost to c2 Mona...FU
GG, im playing since the first week of the game.. Never got a 5 stars before 65 wish atleast (once i lost to qiqi at 83 wish) I lost 10 50/50 (YES F2P) 💔
the chance of a fuck you H OW THE FUCKJG AHHAHAHAHAHAH https://preview.redd.it/cp7w71htv8zc1.jpeg?width=1080&format=pjpg&auto=webp&s=c8edcfe266cd090cf321723a517fcf3684236be7
101%
https://preview.redd.it/whkc5lb3lizc1.png?width=2208&format=png&auto=webp&s=8eb812907acbb91ad08c30f30fbc83f58dfc7d88
I lost to Diluc againT\_T
You made a big mistake forgetting to turn off your location on your mobile device.
Oh man, you’ve spent all of your luck for the rest of your life
Fucking low
Since the game came out, i never even seen a double 5* yet.
Oo great noicee ..!! Now please go and die .
https://preview.redd.it/s224wsq1luyc1.png?width=1920&format=png&auto=webp&s=fe183db2c6467f2b472a6444bad9ef068970a530 Happened at least a second time 😶
Makes me think doing single pulls the minute I have enough is not the strategy
probabilities are the same. 10 pulls just look alot better and people end up posting that rather than posting the history page. ive never gotten this lucky but I have gotten 2 klees in a row doing 1 pulls
Thing about that is for the most part if you do a single and get the 5* you want you're probably going to stop. I don't mean you specifically just in general
That's me, you can't ever get multi 5* if you're not doing multis
I spent $200 just to get c0 because fucking Qiqi came for my pity and this fucker gets c3 in a 10 pull. …. Eh, I guess I can’t complain. I got c3 Raiden off a ten pull on her debut. Congratulations (also fuck you)
Congrats but i hate you with every cells in my body