Okay, quick math...something like 6,500 shows, with two Showdowns apiece, so I'm going to use 13,000 Showdowns as our sample size.

The probability that the first contestant goes over, assuming they spin again on 60 and under, is (1/20)(.05+.1+.15+.2+.25+.3+.35+.4+.45+.5+.55+.6) = ((1+2+3+4+5+6+7+8+9+10+11+12)/400) = 78/400 = .26.

The probability that the second contestant goes over, assuming they spin again on 50 and under, is (1/20)(.05+.1+.15+.2+.25+.3+.35+.4+.45+.5) = 55/400 = .1475.

The probability that the third contestant hits $1 on one spin is obviously .05.

Therefore, the probability of the OP's scenario on any given Showdown is (.26)(.1475)(.05), which comes out to .0019175. Multiplying this by 13,000, I get that the expected number of times this has happened is

**24.9275**, or roughly 25, so I'm going to take the over as well.