I'm trying to set up "spin the fruit machine" for our summer fair. There will be $T$ tumblers, each with $N$ options. So the total number of possibilities is $T^N$. Ideally, I want the chances of winning to be reasonably good (it is just the company barbecue). Winning is defined as correctly guessing the result on each tumbler (ie. if there are 3 tumblers with 3 options on each, then ACB is a win but ACC or CBA is not).
The probability of winning is $1-[(T^N-1)/T^N]^Z$ where $Z$ is the number of tries. So all I do is find the $Z$ where this is >= 50%, right?
My initial arrangement of 3 tumblers with 5 options each appears to give the improbably large number of 169 goes required for each expected win. Oops... lots of unhappy punters at that rate! Looks like 2 tumblers with 4 options and about 10 goes required is about perfect.
Is my math right?