Distributed voting is a fundamental topic in distributed computing. In pull voting, in each step every vertex chooses a neighbour uniformly at random, and adopts its opinion. The voting is completed when all vertices hold the same opinion. On many graph classes including regular graphs, pull voting requires $Θ(n)$ expected steps to complete, even if initially there are only two distinct opinions. In this paper we consider a related process which we call two-sample voting: every vertex chooses two random neighbours in each step. If the opinions of these neighbours coincide, then the vertex revises its opinion according to the chosen sample. Otherwise, it keeps its own opinion. We consider the performance of this process in the case where two different opinions reside on vertices of some (arbitrary) sets $A$ and $B$, respectively. Here, $|A| + |B| = n$ is the number of vertices of the graph. We show that there is a constant $K$ such that if the initial imbalance between the two opinions is ?$ν_0 = (|A| - |B|)/n \geq K \sqrt{(1/d) + (d/n)}$, then with high probability two sample voting completes in a random $d$ regular graph in $O(\log n)$ steps and the initial majority opinion wins. We also show the same performance for any regular graph, if $ν_0 \geq K λ_2$ where $λ_2$ is the second largest eigenvalue of the transition matrix. In the graphs we consider, standard pull voting requires $Ω(n)$ steps, and the minority can still win with probability $|B|/n$.