My math is correct.
If 3% cheat, then 97% are legitimate. In three 8 player games, let’s assume you will encounter 21 distinct players.
The odds that all 21 players are legitimate is 0.97 raised to the power of 21 (or as I wrote 0.97^21 = 52.7% or roughly 50% as I said).
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.