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=== User: Captain Blue quotes that Loctarjay said "That is exactly the point😂 === Captain Blue quotes that Loctarjay said "That is exactly the point😂 The AI gave what you asked for without any critical thought to it🤣 Ask the AI this instead: This will be initial information needed: * With a playerbase of 370000 players. * A match containing 20 players in total, but excluding me that makes it 19 enemy players. * With the possibility of running into the same people in different matches. * With a high possibility of not running into any player-enemies at all. * With the average amount of enemies you'll ever see in a match being between 0 and 6 enemies (which a large portion of all matches you mainly see between 0 and 3 enemies). * Take into account that there is a chance to have multiple cheaters in 1 match, so that also leaves less cheaters for the rest of the matches * The distribution of players into a match is random and there is no guarentee how many cheaters will be placed in 1 match. * We are capable of meeting the same legit players over and over again. * It is possible to also meet the same cheaters This is the question I want answered: * How big a portion of the playerbase needs to be cheating before the exact same person encounters at least 1 cheater every single match? The answer is 37% of the playerbase NEEDS to be cheating to have 99% chance of meeting a cheater HOWEVER there is 0 evidence of there being 37% cheaters in Arc Raiders For that matter, AnyBrain have not even come out with a statement of how accurate their anti-cheat is and how many cheaters they catch vs how many false-positives there" And then Captain Blue responds with: "- With a high possibility of not running into any player-enemies at all." Why would there be a high chance of not running into anyone? Sounds like skewering the data. "- With the average amount of enemies you'll ever see in a match being between 0 and 6 enemies (which a large portion of all matches you mainly see between 0 and 3 enemies)." It was already calculated in the previous questions. I asked about not running into all 19 players. "- Take into account that there is a chance to have multiple cheaters in 1 match, so that also leaves less cheaters for the rest of the matches" Redundant question it will still depend on the base percentage of cheaters in the population. The question is about likely hood to just run into one cheater. Trying to push more cheaters into a single game to try and skewer the AI would likely not work but that's what you're trying to do. There is always a chance you could get multiple cheaters but that's not the premise of the question and would just be automatically accounted for as the probability is just for one. What you don't realise is the percentage to get 2 cheaters per game IS MORE OR LESS THE SAME PERCENTAGE! * For ~90% chance: p\approx 15\% . * For ~95% chance: p\approx 18\% . * For ~99% chance: p\approx 25\% . This is to get 2 cheaters per game. "- The distribution of players into a match is random and there is no guarentee how many cheaters will be placed in 1 match." That's why it's percentage chance... You do understand a percentage chance? It factors in random distribution. What the player base needs to be for that random distribution to cause running into a cheater nearly every game. "- We are capable of meeting the same legit players over and over again. * It is possible to also meet the same cheaters" Again this will be factored into random chance. "- With a playerbase of 370000 players." Effect of global player count on cheater encounter odds The total number of players (e.g., 700,000) does not change the per-raid probability of meeting a cheater in a 20‑player lobby. What matters is the cheater prevalence within the matchmaking pool you draw from and how many distinct players you personally encounter . The baseline formulas remain: This also supports what I've said before in videos and on stream. If 30% of the population cheats in video games then it will be 30% of all games relative to that population which certain factors that may increase or decrease that amount.
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