First, thanks for providing the picture so that I could figure out what was wrong⌠Simply put, you mis-entered the value of M. You put in 1,500,000 (=30%) that is why your distribution looks so funny relative to my expectations as shown in your graph image shown below.
The most common outcome if there is only 3% cheater is encountering 0 among 8 players. As your graph shows the most common outcome with M mis-entered among 8 players is 2 of 8 cheaters in your distribution. (This is due to 30% was entered) Moreover, you misread the output numbers provided. In your output from the applet, it is 94% of games that have at least 1 cheater when 30% cheat and you are assessing 8 unique players.
(P.S. I did not count in my calculation the you in you encounter players in an 8 player game, so I only look at 7 players.)
The other common sense way is if you know there is a 3% chance of an event happening and that are 8 trials, the odds of that event never happening in those 8 trials had to be much lower than 94%.
For 5 million players where 3% cheat and there is random assortment. To have a single game where the other 7 players in the game are legit can be calculated as follows:
The first player to be legit has the odds 4,850,000 in 5,000,000
The second player to be legit has the odds of 4,849,999 in 4,999,999
The third player to be legit has the odds of 4,849,998 in 4,999,998
The fourth player to be legit has the odds of 4,849,997 in 4,999,997
The fifth player to be legit has the odds of 4,849,996 in 4,999,996
The sixth player to be legit has the odds of 4,849,995 in 4,999,995
The seventh player to be legit has the odds of 4,849,994 in 4,999,994
Overall the odds that all 7 are legit is 80.798274%. This means that the odds that you encounter one or more cheats in a single 8 player game is 19.2% (or roughly 19% as I said in my prior post).