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Is there a winning strategy for 3 rounds for Duplicator? I cannot see any.

Strategy for Ehrenfeucht's game.

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    @bof - Who was married with Tarski's daughter ? Ehren or Feucht ? – 2017-02-13

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Fabio's answer is correct, but we can do better: Spoiler can win in two rounds. Namely, play $b_1$ and $b_3$; however Duplicator responds, the vertices of $A$ they choose will be connected (or equal).

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    Well played, as they say! In the definition I'm used to, though, Duplicator cannot choose equal vertices in response to distinct vertices. – 2017-02-12
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    @FabioSomenzi The definition I learned says that they can, but they lose immediately if they do. – 2017-02-12
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    @FabioSomenzi Prohibiting Duplicator from choosing equal vertices in response to distinct vertices seems to be just about the same as prohibiting Duplicator from simply giving up. Choosing equal vertices in response to distinct vertices (or vice versa) loses the game instantly, since partial isomorphisms have to be one-to-one functions. – 2017-02-12
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    Never mind, I just misread Noah's parenthetical expression. Of course, partial isomorphisms have to be one-to-one. – 2017-02-12
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    @NoahSchweber, my definition also allows choose equal vertices but it is pointless for player. Thanks/ – 2017-02-12
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Spoiler wins in three rounds by picking $a_1$, $a_2$, $a_3$. Duplicator must then pick three of the $b_i$'s, but there's no way to get a partial isomorphism between structures $A$ and $B$ that way.

Since Spoiler's three picks are distinct, the three $b_i$ must be distinct. But then it's easy to check that at least one pair of $b_i$'s is not in the edge relation.

Since Spoiler wins in three rounds, Duplicator doesn't have a strategy for three rounds.