The Bob Barker Problem

Forget Monty Hall, Bob Barker punches out golf pro's. I was watching the Price is Right the other day, and came upon a the following fun probability problem.

Before the showcases, three contestants spin a wheel with the numbers 5 to 100 (in increments of 5), with the goal of getting closest to a dollar in a combination of two spins without going over. Here is the scenario, the first contestant spins a 60 and stays. The second contestant spins a 60, and is asked if he desires to spin again. A third contestant remains that may defeat him in a combination of two spins. Should he stay and the third contestant fail to defeat his 60, he would have a spin off with the first contestant. He thought for four seconds and spun. He 'went over' and lost. In my gut I felt he should have stayed. What about you? What is the 'right' course of action?

Don't forget that the third guy can tie him when calculating your probability