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!!!).

ðŸ™„