gameshowsteve is probably pretty close. Here's an arrangement with 20 Jokers and 8 of everything else, distributed as evenly as possible:
Wheel 1: 6 Jokers, 3 A's, 2 B's, 3 C's, 3 D's, 3 E's
Wheel 2: 7 J, 2 A, 3 B, 3 C, 2 D, 3 E
Wheel 3: 7 J, 3 A, 3 B, 2 C, 3 D, 2 E
For each category, the probability of a natural triple is (.15)(.1)(.15) = .00225
Total probability of a natural triple: five times that, or .01125 (about 1 in 89).
Probability of three jokers: (.3)(.35)(.35) = .03675, or less than once per show but enough to keep it interesting.
.03675/.01125 is about 3.27. I admit it's convoluted, but it produces the result!