Bazooka case: or how parts are not random

Day 1,340, 04:38 • Published in Canada Canada • by lucifer_ultionis

In this article I will prove, using basic math that the admins are lying toyou.
Here's how:
In the info box of the collections it is written: You can find the collection parts in the Battle page. Fight and you will receive RANDOMLY some of them!


Is it really RANDOM? 🙂
You probably think it's not! But can you prove it?

Here's my situation. I've receive😛
Stocks: 3
Barrel: 6
Scope: 3
M6A3 rocket: 0 !!!
Triger kit: 6

Random is when all the possibilities have EQUAL Probabilities of appearing. That is a Stock is as likely to appear as a Barrel and as likely as an M6A3 rocket. The logic is the same as that of tossing a dice (the faces of a dice have equal probabilities of showing up)

So, you might wonder, WHAT would be the Probability of my outcome (we can consider yours if you want in the comments) IF we ASSUME the items were indeed appearing RANDOMLY as admins have said (and made us spend lots of our savings, as in my case)

Since there are 5 items, it must be the case that each one of them has a 0.2 probability of appearing (otherwise it's not random)

One can ask: What is the probability of NOT getting a Rocket if N items are received? (N = m1+m2+m3+m4+m5, where m1, m2, ... = number of items of the 1st type, 2nd type and so on)

Probability(of NOT receiving ANY rocket) = (1-0.2)^N
therefore in my case:
The Probability(of NOT receiving ANY rocket) = (1-0.2)^N = 0.8^18 = 0.0180144
In Percentage terms = 1.8% With a 95% confidence interval we can conclude that the distribution of the ITEMS IS NOT RAndom! And admins HAD you! 🙂 Literally! Do you feel ashamed for being stupid and spending your savings? You shouldn't BUT ADMINS SHOULD! Cause it's not nice to lie to the people that bring you revenue and play your game!


Method 2: (a lil bit more advanced)
You can also use a multinomial to calculate what is the probability of getting the EXACT outcome you have, assuming that they appear randomly.
Probability of getting your outcome 😞N!/(m1!+m2!+m3!+m4!+m5!)* 0.2^m1 * 0.2^m2 * 0.2^m3 * 0.2^m4 * 0.2^m5
Where m1, m2, ... = number of items of the 1st type, 2nd type and so on
N = m1+m2+m3+m4+m5 (total number of items you got)
Then if will involve calculating the cumulative distribution of getting up to m1, m2, m3...

ps
And By the way, vote for me 🙂 *if you want to