Running Probabilities

Let us denote the probability that $x$ wins as $P(x)$ . We want $P(A) + P(B) + P(C) = 1$ , because the total probability is always equal to $1$ . We know that $P(A) = 2P(B)$ , and $P(B) = 2P(C)$ . We can also determine that $P(A) = 4P(C)$ . With this information, we can change the equation to $4P(C) + 2P(C) + P(C) = 1$

$7P(C) = 1$

$P(C) = \frac{1}{7}$

So the probability that $C$ wins is $\frac{1}{7}$ . Likewise, the probability that $B$ wins is $\frac{2}{7}$ , and the probability that $A$ wins is $\frac{4}{7}$ . And indeed, $\frac{1}{7} + \frac{2}{7} + \frac{4}{7} = 1$ . So the answer is $\frac{4}{7} = .57142857...$

Let the probability of $C$ winning the race be $\Pr(C) =p$ . Then the probability of $B$ winning the race be $\Pr(B) =2p$ , and the probability of $A$ winning the race be $\Pr(A) =4p$ . Since there are only $A$ , $B$ , and $C$ in the race, $\Pr(A)+\Pr(B)+\Pr(C) =4p+2p+p =7p=1$ , $\implies p =\frac 17$ and $\Pr(A) =4p =\frac 47 \approx \boxed{0.571}$ .

I have been granted by a boon from the almighty which lets me know the correct answer