OK, here's a slightly different problem solution, but it's a huge step forward in answering this one. I must admit, I got some help from this website: http://mathproblems.info/group4.html (problem No. 69)
This solution involves balancing risk and reward to find the optimum strategy for a continuous wheel from 0 to 1, then calculating the probablility given the strategy. This will give us a good ballpark for the quantized wheel. I'll break the solution into parts to make it easier to digest.
Part I - Probability of winning with 1 spin
Let's start with a two player game, and say Player 1 (P1) spins A1 as a first spin. If P1 stays with his first spin, the probability of losing to Player 2 (P2) with it is as follows:
P2 can win with his first spin --> P = 1-A1
P2 can win with two spins --> P = [1-(1-A1)]*(1-A1) = A1*(1-A1)
(here, A1 comes from spinning under the first time, and (1-A1) comes from not busting on the second spin)
So the total probability of losing with A1 is (1-A1) + A1*(1-A1) = 1-A1^2. Therefore, the probability of winning with A1 is 1-(1-A1^2) = A1^2.
For a three player game, P1 can lose to P2 or not, and lose to Player 3 (P3) or not, so we get the probability of losing as follows:
P = (1-A1^2) + [1-(1-A1^2)]*(1-A1^2) = 1-A1^4
(here, the first term is losing to P2, and the second term is not losing to P2 times losing to P3)
In general, the probability of losing with the first spin in an "n" player game is 1-A1^(2n-2), so the probability of winning with a first spin is A1^(2n-2).
Part II - Probability of winning with 2 spins
Let's start with the 2-player game again. We already know that the probability of winning with one spin is A1^2. This can be generalized to the probability of winning with a given result being R^2 (0<=R<=1). So the probability of winning with 2 spins is (A1+A2)^2.
Now, after the first spin any numbers between 0 and 1-A1 will be OK, and any numbers above 1-A1 will bust P1. So the probability of winning with 2 spins will be the average of all values of A2 between 0 and 1 times the winning result. Since the winning result of A2 above 1-A1 is 0, we can get rid of it (but we must still average over the range 0 to 1). The average winning result is then the integral of [(A1+A2)^2] dA2 from A2=0 to A2=1-A1. After some algebra we get P = 1/3*(1-A1^3). For the three player game, P = 1/5*(1-A1^4). In general, the probability is 1/(2n-1)*(1-A1^(2n-1)).
Part III - Risk vs. reward to find optimum strategy
Two player game again. OK, so the probability of winning with the first spin is A1^2, and the probability of winning with 2 spins is 1/3*(1-A1^3). Now, at some value of A1, call it C1 for "cutoff", these probabilities will be equal, because the risk of busting will balance the advantage of having a bigger result. So we set A1=C1, and set the probabilities equal:
C1^2 = 1/3*(1-C1^3)
So: C1^3+3*C1^2-1 = 0
In general, the optimum cutoff will be the solution to C^(2n-1)+(2n-1)*C^(2n)-1=0.
Using Excel (my favorite piece of MS crapware), we get C1 = 0.5321 (approx.), so if the first spin is less than 0.5321, P1 would spin again. For a 3 player game, C1 = 0.6487 (approx.).
Part IV - Probability of P1 winning
Two player game again. Now we find the probability of winning using this optimum strategy. To do this, we integrate in 2 pieces. First, if the first spin is under C1, P1 will spin again so we integrate 1/3*(1-A1^3) from 0 to C1, and if it's over C1, P1 will stick so we integrate A1^2 from C1 to 1. The overall expression is:
P = int[1/3*(1-A1^3)] (0,C1) + int[A1^2] (C1,1)
The final answer is P = 45.38% (approx.). For the 3 player game, the answer is 30.43% (approx.).
I checked these results on my spreadsheet from the other day. For C1, I used 65 cents, and for C2 I used 55 cents (closest integers). P1's winning percentage is still around 30%, P2's is about 33% (which is about 33/70 = 47% of what's left), and P3's is about 37%. So, the results are pretty close, which is comforting.
Using this method on integer wheel values will involve lots more counting, but I'll give it a go (after some sleep!!!).