Optimal "Price is Right" strategy

[BadPenguin]

I was sitting in a deli the other day waiting for my sandwich to be made and "The Price is Right" was playing on the TV. It was the part at the end of the first half where they see which one of several people gets to go on to the ... big thing at the end, I forget what it's called. Anyway, the game they play at this stage is the following:

There's a big wheel marked from $.05 to $1.00 in increments of 5 cents. There are 3 players (maybe sometimes 2 or 4, I'm not sure) who play in some predetermined order. The first player spins the wheel and sees where it stops - he can stop and take the result of that spin as his score or he may immediately spin again, in which case the result of the second spin is added to the first to give his score. The object is to get as high a score as possible without exceeding $1.00 . If you go over $1 you lose.

Then the second player does the same thing, again with the option to spin either once or twice. He needs to score higher than the first player (if the first player didn't go over $1, that is) in order to take the lead. The third player then follows in the same way.

Obviously it's a big disadvantage to be "under the gun" in this game. The question is, what's the optimal strategy to follow if you're spinning first? (Or, more generally, what's the optimal strategy if there are n players yet to spin behind you and everyone in front of you went over $1 and is out.) Clearly if your first spin is .05 you're going to spin again, and if it's $.95 or $1.00 you're not. Where is the crossover point? In other words, what's the lowest first spin that will cause you to stand on that score?

I tried to solve this both mathematically and computationally and got different answers, so either my math or my code is wrong - probably my math, I'm guessing. (It might be easier to approach the problem - at least on the mathematical side - if you assume the wheel can return any real number between 0 and 1.00 rather than only multiples of .05.)