Quasar Zone, where controversy over prize manipulation erupted

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Event giveaways and holiday event questions
The end of the over is pure
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o 11469
2024.11.062144
I don’t know if it was the work of the person in charge of this event’s draw or if the executives and employees of Taysa Zone worked together to get the job done.
Expensive products are monopolized by certain users.
The employees said they couldn’t participate, but I wonder if they secretly participated and took it, or if they encouraged their acquaintances..
The solicitation conducted via live broadcast does not show any signs of that.
Is it okay to get angry because it’s a month for the person in charge of estimating? ?
If that person claims that he or she is out of luck, I have nothing to say; In case of daily cheering comments, please post only for one day.
There are many cases where it goes out of season, and holiday products always end up in our store.
Among users who have been using TaySa Zone for a long time, I think there are many who have never purchased expensive products.
However, people who have only joined for a few months are getting a few points at a time, and the winning rate is very high compared to participating in this event.
There are quite a few people who can see or see it.
Honestly, while I’m looking into it and writing this, I’m wondering what they’re doing, but I need to correct what’s wrong.
Me? If you are doing this because you want to take care of your employees with expensive products, it would be better not to proceed with this move and instead use the ladder internally.
Please hop on
I feel like I’m treating the people who participate in this event like idiots, which makes me feel really bad.
If it is a month for the person in charge of the fortune telling, it will be over if the executives immediately point out this point.

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hello
Regarding this issue, the world wonders whether our store manager is a person with a very heavy cargo.
First of all, how many events have there been so far and how rarely does one person visit our store?
So I think it has no real meaning.
What we need to know is how much the user participates in the event and how much he or she is not interested in, and whether that becomes normal.
It’s about whether it’s within the range of water’s survival.
According to the profile, Mo Yujeo has 445 comments so far and 13 comments are estimated to have been opened in the main office.
For convenience, we will assume that each comment is an entry for a different draw.
Also, I’ll be generous and calculate the cargo for each bar at about 1/300.
Now, I would like to request this information to calculate the first GpT estimate:
Number of trials (n): 445
Successful Fire (p): 1/300
Number of Successes (X): 13
Now, in this situation, calculate whether the 13th point is a situation where there is enough cargo to come out, or whether it is an extremely rare situation.
I’ll try it
1. Expected value and variance calculation
1.1. Expected value (4) calculation
Expectation is calculated as follows: 4 = n
4833
In other words, on average, about 1.48 points per game are expected.
1.2 Calculating variance and standard deviation
The variance is calculated as follows: 0 2 = nX pX (1-p) = 445
The standard deviation is the root of the variance:
1.2160
2.2 – Calculate values
The Z-value is a multiple of the standard deviation of how much the pipe value varies from the reference camera.
z = (K-p)lo = (13- 1.4833)/1.2160 ~ 9.475
In other words, the expected value of 13, which is the number of successes achieved, is offset by about 9.475 standard letters.

2. Two-value calculation
2-sense is how much the tube value deviates from the expected value, expressed as a multiple of the standard deviation.
Z= (X-po= (13 –
4833)/1.2160
9.475
That is, the observed number of successes, 13, is about 9.475 standard deviations away from the expected value.
3 Hwagdeul Goosan
3.1.2-Checks according to value examples 2-If the value is 9.475, it is a very extreme value in the inon normal distribution.
According to the regular distribution table, pCZ > 9.475), the sense of purpose is almost always hunting.
Generally, 2-values ​​greater than 3 are considered significant synchronous differences, and 2 = 9.475 is an extremely unlikely estimate.
3.2 Approximation using Pois power denominators
If p is very small and the number of attempts is close to n, these distributions can be approximated by the Pois power distribution.
Mean of Poisson distribution (1): 1-1
Since it is a Poisson distribution, we calculate all products over X=13: P (
33 Yeseo
4 Conclusion
Expected number of hits: approximately 1.48
Actual number of visits: 13
2 – value approximately 9.475
Therefore, under 1/300 of the given successful stories, the number of stories that will score 13 times in the 445 trials is extremely rare.
It’s hard
5 Possibilities and Interpretations
1Actual success stories are higher
2 Events are not independent
3 Not a random main line
4 Data error or error
I’m asking you to calculate the exact amount because the things discussed here are approximate to 0.
Approximation using normal distribution port: 0.000000000000000000000738
Approximation using Pois power distribution 0.000000000000000254
To help you understand, let me explain:
If the denominators of the regular districts follow the assurances obtained in the East Sea, even if all state offices and party branches are tried to be compressed into seconds,
You have to try for 23,000 years of the universe’s age to get it to happen once.