This is a classic problem in geometric probability.

Let:

\( p = \frac{1}{6} \) be the probability of getting a 1 on any single throw. \( q = \frac{5}{6} \) be the probability of not getting a 1.
We want the probability that the first 1 occurs on an even-numbered throw.

This means the first few throws fail (i.e., no 1), and then a 1 comes on 2nd, 4th, 6th, etc.

This gives us an infinite geometric series: \[ P = \sum_{k=1}^{\infty} \left( q^{2k-1} \cdot p \right) = p \cdot q + p \cdot q^3 + p \cdot q^5 + \dots \] \[ P = p \cdot q \left(1 + q^2 + q^4 + \dots \right) \] This is a geometric series with first term \(1\), ratio \(q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}\), so: \[ P = \frac{1}{6} \cdot \frac{5}{6} \cdot \left( \frac{1}{1 - \frac{25}{36}} \right) = \frac{5}{36} \cdot \left( \frac{1}{\frac{11}{36}} \right) = \frac{5}{36} \cdot \frac{36}{11} = \frac{5}{11} \]