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KemosabeTBC
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Re: Ling Ling chance on COF store

Fri Apr 28, 2017 12:18 pm

natsu wrote:
HunterXHunter wrote:
I refreshed 5 times and didn't see a single SS.

there are 13 heroes in CoF shop right now, and there are 6 SS slots. so there is a 46% (approx, IMHO) chance to see Ling in an Refresh.

This is not how you calculate probability. I understand where you get 46%, 6/13. According to you, if there are 2 heroes and 2 SS slots then there is a 100% chance. No, it's 75% chance to see LingLing. LingLing in slot 1 , in slot 2 , in both or none which is 3/4. It is actually less than 1% , around .38% chance to see LingLing.

I assume you used a binomial calculator to calculate this probability. If yes, you got the answer wrong. "Probability" is a value between 0 and 1. In this case the result is 0.38 which means there should be a 38% chance to get at least 1 Ling Ling per refresh.
 
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Re: Ling Ling chance on COF store

Fri Apr 28, 2017 6:00 pm

KemosabeTBC wrote:
natsu wrote:
HunterXHunter wrote:
I refreshed 5 times and didn't see a single SS.

there are 13 heroes in CoF shop right now, and there are 6 SS slots. so there is a 46% (approx, IMHO) chance to see Ling in an Refresh.

This is not how you calculate probability. I understand where you get 46%, 6/13. According to you, if there are 2 heroes and 2 SS slots then there is a 100% chance. No, it's 75% chance to see LingLing. LingLing in slot 1 , in slot 2 , in both or none which is 3/4. It is actually less than 1% , around .38% chance to see LingLing.

I assume you used a binomial calculator to calculate this probability. If yes, you got the answer wrong. "Probability" is a value between 0 and 1. In this case the result is 0.38 which means there should be a 38% chance to get at least 1 Ling Ling per refresh.

First you need to find all possible combinations of all heroes showing up in those 6 slots then divide it by all possible outcome. I am not going to explain how combination and permutation work but you can always google it.
 
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KemosabeTBC
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Re: Ling Ling chance on COF store

Fri Apr 28, 2017 6:15 pm

natsu wrote:
KemosabeTBC wrote:
natsu wrote:
This is not how you calculate probability. I understand where you get 46%, 6/13. According to you, if there are 2 heroes and 2 SS slots then there is a 100% chance. No, it's 75% chance to see LingLing. LingLing in slot 1 , in slot 2 , in both or none which is 3/4. It is actually less than 1% , around .38% chance to see LingLing.

I assume you used a binomial calculator to calculate this probability. If yes, you got the answer wrong. "Probability" is a value between 0 and 1. In this case the result is 0.38 which means there should be a 38% chance to get at least 1 Ling Ling per refresh.

First you need to find all possible combinations of all heroes showing up in those 6 slots then divide it by all possible outcome. I am not going to explain how combination and permutation work but you can always google it.

Dude, I know how to calculate probabilities... This is a binomial probability and I'm telling you the result is 38%

You don't have to find any combinations/permutations but, since I'm a nice guy I will tell you how to do it, instead of telling you "you can always google it":

. Each slot has a 1/13 probability of any specific hero. So the probability for a specific slot to have Ling Ling is 1/13 = 0.076923 (7.6923%)
. There are 6 slots. The probability of getting Ling Ling in a specific slot is independent from the other slots
. We want to know what is the probability of getting at least one Ling Ling in any of the 6 slots.
. This is a Bernoulli trial with the probability of success (getting Ling Ling) = 0.076923

So, what you have to do is calculate a Binomial Probability with the following parameters:
Probability of success = 0.076923
Number of trials = 6
Number of successes = 1

Now, you can calculate this by hand, or use any of the many available binomial calculators. I used a calculator since it's much easier. Here is the result:
Image

Now, the Binomial Probability is the probability of getting Ling Ling exactly once in 6 trials, this is not what we want to know but, in you case you need it it's 30.93%
What we want to know is the probability of getting Ling Ling at least once in 6 trials. This is the Cumulative Probability: P(X >= 1), which is 38%

You're welcome.
 
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Re: Ling Ling chance on COF store

Fri Apr 28, 2017 8:38 pm

ˆ Ouch.
 
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Re: Ling Ling chance on COF store

Sat Apr 29, 2017 4:47 am

what you are doing or the calculator is:
There is a 1/13 chance of getting LingLing and you refresh 6 times to find the chance of getting at least one Ling Ling in those 6 refresh for 1 slot. You completely ignore other possible events that can happen in other slots, you are assuming each slots have independent event but it is not necessarily true because you don't refresh 1 slot at a time, you refresh all slots. Of course, it also depend on how the developer write their code. You don't know exactly what kind of formula they are using so you have to account for all possibilities.
example:
2 heroes ,1 slot = 50% chance
2 heroes ,2 slots = 75% chance ( This is what you want to find )
what you did was similar to this
2 heroes , 1 slot and set number of trial to 2 (meaning 2 refresh) = 75%
2 heroes , 2 slots and trial to 2 = 93.75%
 
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KemosabeTBC
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Re: Ling Ling chance on COF store

Sat Apr 29, 2017 4:19 pm

natsu wrote:
what you are doing or the calculator is:
There is a 1/13 chance of getting LingLing and you refresh 6 times to find the chance of getting at least one Ling Ling in those 6 refresh for 1 slot. You completely ignore other possible events that can happen in other slots, you are assuming each slots have independent event but it is not necessarily true because you don't refresh 1 slot at a time, you refresh all slots. Of course, it also depend on how the developer write their code. You don't know exactly what kind of formula they are using so you have to account for all possibilities.
example:
2 heroes ,1 slot = 50% chance
2 heroes ,2 slots = 75% chance ( This is what you want to find )
what you did was similar to this
2 heroes , 1 slot and set number of trial to 2 (meaning 2 refresh) = 75%
2 heroes , 2 slots and trial to 2 = 93.75%

No, that is not what I did. Trust me, I have a lot of experience with probability and statistics. The answer is 38%

You completely ignore other possible events that can happen in other slots, you are assuming each slots have independent event but it is not necessarily true because you don't refresh 1 slot at a time, you refresh all slots.

This doesn't make any sense... The only thing I assumed is the only sensible thing to assume: All heroes are equally likely to get, that is all. The slots are independent, doesn't matter if you refresh 1 at a time or all of them. Each one has an independent 1/13 probability of being a specific hero

2 heroes , 1 slot and set number of trial to 2 (meaning 2 refresh) = 75%
2 heroes , 2 slots and trial to 2 = 93.75%

Not what I did at all.
If you have 2 heroes and 1 slot the number of trials is 1 and the answer is 50%
If you have 2 heroes and 2 slot the number of trials is 2 and the answer is 75%
Each slot is an independent trial. If you have 13 heroes and 6 slots the number of trials is 6
See image below

Image
 
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Re: Ling Ling chance on COF store

Sat Apr 29, 2017 4:29 pm

Anyway, I think you probably did the same calculation by hand, and arrived at 0.38. You just incorrectly misinterpreted it as being 0.38% when it is 38%. Do the math again and tell me what the result is.

It is easy to see that it cannot be 0.38%. If you had only one slot the probability of getting ling ling would be 7.7%. It's obvious that if you have 6 slots the probability cannot be lower than this. Maybe It's easy to see if I list the probabilities for an increasing number of slots, from 1-6:
1 slot: 7.69%
2 slots: 14.79%
3 slots: 21.35%
4 slots: 27.40%
5 slots: 32.98%
6 slots: 38.14%
 
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Re: Ling Ling chance on COF store

Sat Apr 29, 2017 7:43 pm

With all due respect to ( both ) of your probability calculation skills :D (I am overwhelmed)
I don't think any of the calculation is totally accurate because we forgot to take in account the 1 fixed hero of the day. If that hero is lingling you are going to get lingling in the next five refreshes as well on the other hand if that hero is not lingling for the next 5 refreshes you are left with 5 boxes only. And this is going to affect the sample space.
[YOU AND I WILL DEFEAT THIS BUG TOGETHER.....]

http://bit.ly/1ZTctlk
 
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Re: Ling Ling chance on COF store

Sun Apr 30, 2017 12:00 am

HunterXHunter wrote:
With all due respect to ( both ) of your probability calculation skills :D (I am overwhelmed)
I don't think any of the calculation is totally accurate because we forgot to take in account the 1 fixed hero of the day. If that hero is lingling you are going to get lingling in the next five refreshes as well on the other hand if that hero is not lingling for the next 5 refreshes you are left with 5 boxes only. And this is going to affect the sample space.

The calculation is accurate and it is a trivial probability.

For the automatic refresh the probability of getting at least one ling ling is 38.14%
For a manual refresh where the fixed hero is not ling ling the probability of getting at least one ling ling is 32.98%
For a manual refresh where the fixed hero is ling ling, the probability of getting at least one ling ling is 100% (obviously)
 
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natsu
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Re: Ling Ling chance on COF store

Sun Apr 30, 2017 2:01 am

The way you approach the problem is wrong. If you are not convinced then let put that into practice and use your Binomial calculator:
Assuming you are right, There is 32.98% chance to see LingLing where there is a fixed hero in first slot
Probability of success on a single trial = .3298
Number of trials = 1 (1 refresh)
Number of successes (x) = 1
result = .3298

Number of trials = 2
result = .5508

Number of trials = .10
result = .9817

Number of trials = 13 ( OP refresh 13 times)
result = .9944
How unlucky can he be if he had 98.17% chance to see at least ONE LingLing by 10 refresh? obviously chance to see LingLing is no where close to 32.98%.
Again, stop doing it slot by slot. let me give you another simple example. let say you throw 1 dice and 2 dices to find #1.
throw 1 dice = 1 , 2 , 3 ,4 , 5 , 6 which is .167 chance
throw 2 dice you get 1-1, 1-2, 1-3, 1-4 , 1-5 , 1- 6 and so on, you get the idea here.
Totally different chance to get #1.
you can't throw 1 dice to get .167 chance
then throw another, use the calculator to get .306 chance
you have to calculate what happen when you throw both.

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