It turns out that after a moment’s thought it is possible to come up with an easy general solution to this problem (for any number of balls). Just round up the quotient of log[n] and log[3]. For example, the minimum number of measurements for 8 balls is ceil(log(8)/log(3)) = 2, using Matlab syntax. The proof lies in the fact that one can always with certainty eliminate in one round 2/3 of the total remaining balls, so you can always reduce your total number of balls to 1/3 or less in any round. Thus the solution for x of n*(1/3)^x <= 1 is the number of rounds required. That solution is, of course, x = ceil(log(n)/log(3)).