This is a great question and I record the Price is Right daily to watch only the big wheel. I looked at the proposed answers above and they are brilliant answers to the wrong question.
The reason the calculation of the probabilities is so difficult is that there are three people involved, each with a choice of whether to spin the wheel once or twice, except in the instance where the first two contestants go over a dollar in which case the third contestant automatically wins, but only gets to spin the wheel once for an initial score. The further complication is that there can be ties which does not appear to be accounted for in the calculations above. If there is a tie and the tie is the highest score without going over, there is a spin-off where each contestant gets one additional spin and the dollar bonus is still in play. Further, if any contestant's spins total one dollar, they are awarded $1,000 and an additional spin. If on an additional spin, they land on 5 or 15 cents, they win an additional $10,000 and if they land on a dollar, they win an additional $25,000. One more thing to consider is that the numbers are in a particular non-sequitur order. I would say maybe that doesn't matter except that a contestant can't just spin the wheel targeting a particular range (other than the 5 cent, 1 dollar, 15 cent range on the additional spin) even though many try to do so.
The winning percentages pointed out in the article seem about right, intuitively, and while the calculations above suggest that the first person to spin is severely disadvantaged, by observation, it has not been the case. (I have been recording the show and watching who wins on the big wheel for nearly a year. That's about 420 actual big wheel contests.) The third person already know the scores of the first two contestants and has no incentive to quit spinning with a losing number. The second contestant knows the score of the first and again has no incentive to quit unless they either beat or tie the first contestant. The first contestant knows that there will be two contestants spinning after them and has an incentive to try to get a higher number
But, each contestant, at least on their first spin has no "feel" for the wheel, so it might be fair to say that the first spin of each contestant is completely random. And, the second spin may only be slightly influenced by the "feel" gained in the first spin. Again, the order of numbers on the wheel is not random, but I don't have it to give. So, at least for the initial two spins, I would say you could ignore "feel," but it is the whole idea of the dollar bonus spin, so don't think contestants are unaware of this factor.
All of that said, on the very first spin of each contestant, the probability for any score is going to be 1/20. To determine if the first contestant should take a second spin, the risks are:
A. What is the probability that the score on the first spin, when added to the score of the second spin will exceed one dollar and cause the contestant to automatically be disqualified?
B. What is the probability that on one spin, the second contestant will a)tie or b) surpass the total of the first spin.
C. What is the probability that on two spins, the second contestant will a)tie or b) surpass the total of the first spin without going over a dollar.
D. What is the probability that on one spin, the third contestant will a) tie or b) surpass the total of the first spin.
E. What is the probability that on two spins, the third contestant will a) tie or b) surpass the total of the first spin.
F. What is the probability that the first contestant on a second spin, would score a total of one dollar or less. (This should have been first or second in my list.)
So, what I get from this list is that if:
A + F >= Ba + Bb + Ca + Cb
the first contestant should not spin.
If the first spin is 50 cents, then A + F would be 1. Ba would be .5. Bb would be .25. Ca would be .5 and Cb would be .25. The total for the two following contestants is 1.5, the first contestant should spin again.
If the first spin is 55 cents, then A + F would be 1, so there is a flaw. A should be the probability that the second spin will cause the total to be less than or equal to a dollar so that the term totaling to spin again number goes down as the score in the first spin goes up. Recalculating that way, A+F equals .45 + .45, .90. I'm missing something but will push through to see if it shakes out. Ba would be .45, Bb would be (Okay, well first, tie would only be 1/20 on the first spin always, so that part is wrong. Wow, I am crumbling, I totally forgot D and E above. I guess I'm too sleepy.) Help!
I broke down most of the elements of the probability problem, although something is telling me I still missed two of them (one I suspect is ties on the first contestant's side of the equation). But, I can't think through how to set up the equation. The rules are there. Holy smoke, its 2:30 am.