The probability of obtaining a 6 is the same for each throw. Therefore, this is a binomial distribution problem. The probability of exactly $k$ successes in $n$ trials, each of which has probability $p$ of success is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways exactly $k$ successes can occur in $n$ trials.
When tossing a fair die, the probability of obtaining a six is $1/6$ for each trial. Hence, the probability of at least one success is
\begin{align*}
\Pr(X \geq 1) & = \sum_{k = 1}^{3} \binom{3}{k}\left(\frac{1}{6}\right)^k\binom{5}{6}^{3 - k}\\
& = \binom{3}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^2 + \binom{3}{2}\left(\frac{1}{6}\right)^2\binom{5}{6} + \binom{3}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^0\\
& = \frac{1 \cdot 3 \cdot 1 \cdot 25}{6^3} + \frac{3 \cdot 1 \cdot 5}{6^3} + \frac{1 \cdot 1 \cdot 1}{6^3}\\
& = \frac{75 + 15 + 1}{6^3}\\
& = \frac{91}{216}
\end{align*}
M D's answer explains why your intuition was incorrect. What follows supplements M D's answer.
Your intuition led you to count each outcome as many times as a six appears, which is why you obtained
\begin{align*}
\Pr(X \geq 1) & = \sum_{k = 1}^{3} \color{red}{k}\binom{3}{k}\left(\frac{1}{6}\right)^k\binom{5}{6}^{3 - k}\\
& = 1\binom{3}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^2 + \color{red}{2}\binom{3}{2}\left(\frac{1}{6}\right)^2\binom{5}{6} + \color{red}{3}\binom{3}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^0\\
& = \frac{3 \cdot 1 \cdot 25}{6^3} + \frac{\color{red}{2} \cdot 3 \cdot 1 \cdot 5}{6^3} + \frac{\color{red}{3} \cdot 1 \cdot 1 \cdot 1}{6^3}\\
& = \frac{75 + \color{red}{2} \cdot 15 + \color{red}{3} \cdot 1}{6^3}\\
& = \frac{75 +\color{red}{30} + \color{red}{3}}{216}\\
& = \frac{\color{red}{108}}{216}\\
& = \color{red}{\frac{1}{2}}
\end{align*}