Are You Sure this is a Fair Spinner?

Let $p_i$ be the probability player $i$ wins on a single spin (ie, the proportion of the spinner assigned to them) and let $P_i$ be their probability of winning overall.

Let $L=(1-p_1)(1-p_2)(1-p_3)$ be the probability that in any given round, no-one wins.

We have $P_1=p_1+Lp_1+L^2 p_1 \ldots=\frac{p_1}{1-L}$

$P_2=(1-p_1)p_2+L(1-p_1)p_2+L^2 (1-p_1)p_2 \ldots=\frac{(1-p_1)p_2}{1-L}$

$P_3=(1-p_1)(1-p_2)p_3+L(1-p_1)(1-p_2)p_3+L^2 (1-p_1)(1-p_2)p_3 \ldots=\frac{(1-p_1)(1-p_2)p_3}{1-L}$

Since we want the $P_i$ to be equal, this just becomes

$p_1=(1-p_1)p_2=(1-p_1)(1-p_2)p_3$

The largest portion of the spinner will be assigned to Player $3$ ; so $x=p_3$ .

Cancelling and rearranging the above, we find

$p_1 = \frac{x}{1+2x}$ and $p_2 = \frac{x}{1+x}$

Since $p_1+p_2+p_3=1$ , we have

$\frac{x}{1+2x}+\frac{x}{1+x}+x=1$

Clearing denominators, this is $2x^3+4x^2=1$ , so $(A,B,C)=(2,4,0)$ and the answer is $\boxed6$ .

Let $p_1$ be the percentage of Player 1's spinner, $p_2$ be the percentage of Player 2's spinner, and $p_3$ be the percentage of Player 3's spinner. Player 3 has the disadvantage of going last, so to have an equal probability of winning the game, his percentage of the spinner will be the biggest, therefore we are trying to find $x = p_3$ .

All three percentages must add up to the whole spinner, so

$p_1 + p_2 + p_3 = 1$

Let $q_1$ be the overall probability of Player 1 winning. Player 1 has a $p_1$ chance of winning on the first spin and has a $(1 - p_1)(1 - p_2)(1 - p_3)$ chance of everyone else not spinning their correct color so that the game can start over again. Therefore, $q_1 = p_1 + (1 - p_1)(1 - p_2)(1 - p_3)q_1$ . Since there are $3$ people playing, each player must have a $\frac{1}{3}$ probability of winning, so $q_1 = \frac{1}{3}$ . Therefore, $\frac{1}{3} = p_1 + (1 - p_1)(1 - p_2)(1 - p_3)\frac{1}{3}$ or

$\frac{1}{3}(1 - (1 - p_1)(1 - p_2)(1 - p_3)) = p_1$

By a similar argument, let $q_2$ be the overall probability of Player 2 winning. Player 2 has a $1 - p_1$ chance of Player 1 not spinning his correct color and a $p_2$ chance of winning on his first spin, and then a $(1 - p_1)(1 - p_2)(1 - p_3)$ chance of everyone else not spinning their correct color so that the game can start over again. Therefore, $q_2 = (1 - p_1)p_2 + (1 - p_1)(1 - p_2)(1 - p_3)q_2$ . Since there are $3$ people playing, each player must have a $\frac{1}{3}$ probability of winning, so $q_2 = \frac{1}{3}$ . Therefore, $\frac{1}{3} = (1 - p_1)p_2 + (1 - p_1)(1 - p_2)(1 - p_3)\frac{1}{3}$ or

$\frac{1}{3}(1 - (1 - p_1)(1 - p_2)(1 - p_3)) = (1 - p_1)p_2$

From the last two equations, $\frac{1}{3}(1 - (1 - p_1)(1 - p_2)(1 - p_3)) = p_1 = (1 - p_1)p_2$ , so $p_2 = \frac{p_1}{1 - p_1}$ . Substituting this into the first equation we have $p_1 + \frac{p_1}{1 - p_1} + p_3 = 1$ which solves to $p_1 = \frac{1}{2}(3 - p_3 - \sqrt{p_3^2 - 2p_3 + 5})$ .

Then after substituting $p_2 = \frac{p_1}{1 - p_1}$ and then $p_1 = \frac{1}{2}(3 - p_3 - \sqrt{p_3^2 - 2p_3 + 5})$ into $\frac{1}{3}(1 - (1 - p_1)(1 - p_2)(1 - p_3)) = p_1$ , this solves to the cubic equation $2p_3^3 + 4p_3^2 = 1$ , so $A = 2$ , $B = 4$ , $C = 0$ , and $A + B + C = \boxed{6}$ .

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Note : the numerical approximations for $p_1$ , $p_2$ , and $p_3$ are $p_1 \approx 0.237286$ , $p_2 \approx 0.311108$ , and $p_3 \approx 0.451606$ .