Ridiculous, yet wonderful idea

[MicroRNA-1507]

@MDK22-1519

@MicroRNA-1507

In three eight players games, you will encounter up to 21 unique players. If each player has a 3% chance to be a cheat/toxic, then the odds of all 21 being non-cheats/non-toxix is 52.7% or ~ 0.97^21.

Here is the math. Since you think you understand it but, can’t explain it real well. You have 3% of 100,000 players = 3,000 toxic players. You then take 97,000 / 8 = 12,125 you have a 1 in 12,125 chance of having a cheater. Which is roughly the same odds of finding a pearl in an oyster. Do math right or don’t do it at all. The more people the even greater the odds you will never see it.

Your explanation is so wrong mathematically.

In your example, the odds that the first person is legitimate is 97,000/100,000 = 97%. The odds that the second player is legitimate is 96,999/99,999 = 96.99997%. The odds that the third person is legitimate is 96,998/99,998 = 96.99994%. So on and so forth for the first 21 distinct players.

To calculate the overall probability that all 21 players are not cheater/toxic, you need to multiple all the percentages together. When you do this, the final probability is 52.7% which approximates the binomial distribution of 0.97^21.

If you still do not believe me, you can check the math online as shown below.

@MicroRNA-1507

If you do not believe me, please find a link to a binomial calculator.

Binomial Probability Calculator (stattrek.com)

Probability of success on a single trial Enter 0.97
Number of trials Enter 21
Number of successes (x) Enter 21

This mathematical result that can be independently verified demonstrate that if 3% cheat, then after encountering 21 players, the odds are roughly half (47.3%) that at least one of those 21 players is a cheater.

If you still won’t believe the mathematical truth, here is someone else’s explanation.

@Skimt-2626

The end result is the same.

Hypergeometric

If the population is 200000, and 6000 (3%) are cheaters. We grab a collection of 21 players, and look for 0 players who is cheating.

with(Statistics);
X := RandomVariable(Hypergeometric(200000, 6000, 21));
evalf(Probability(X = 0));

Result: 52.7%.

Binomial

Ignoring max population, we are looking for 21 legitimate players out of 21 players.

with(Statistics);
X := RandomVariable(Binomial(21, 0.97));
Probability(X = 21);

Result: 52.7%

I disagree with pretty much everything MicroRNA says, always, but whether you like it or not, that’s university statistics for you.