Ankit, Box, and Clark are taking the tiebreakers for the geometry round, consisting of three problems. Problem $k$ takes each $k$ minutes to solve. If for any given problem there is a $\frac13$ chance for each contestant to solve that problem first, what is the probability that Ankit solves a problem first?
1. Let's denote the probability that Ankit solves a given problem first as \( P(A) \). Given that each contestant has an equal chance of solving any problem first, we have:
   \[
   P(A) = \frac{1}{3}
   \]
   Similarly, the probability that Ankit does not solve a given problem first is:
   \[
   P(\text{not } A) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}
   \]

2. We need to find the probability that Ankit solves at least one of the three problems first. To do this, we first calculate the probability that Ankit does not solve any of the three problems first. Since the events are independent, the probability that Ankit does not solve any of the three problems first is:
   \[
   \left( \frac{2}{3} \right)^3 = \frac{8}{27}
   \]

3. The probability that Ankit solves at least one of the three problems first is the complement of the probability that Ankit does not solve any of the three problems first. Therefore, we have:
   \[
   P(\text{Ankit solves at least one problem first}) = 1 - \left( \frac{2}{3} \right)^3 = 1 - \frac{8}{27} = \frac{27}{27} - \frac{8}{27} = \frac{19}{27}
   \]

The final answer is \(\boxed{\frac{19}{27}}\).