Eleven moves suffice.
François Brunault commented that the maker can get two moves to start on some hexagonal lattice (a lattice generated by unit vectors with an angle of $60$ degrees). In fact, by the fourth move, the maker can get three moves to start in a hexagonal lattice, and can choose these to be the vertices of an equilateral triangle of side $1$ so that the breaker has not played in this lattice yet.
Proof: Let the maker's first play be the origin $A$. Rotate the coordinates after the breaker's first move so that this play be ignored. Let the maker's second move be $B = (2x,0)$ where $x$ is a transcendental smaller than $1$. There are $4$ hexagonal lattices $H_{A,C}, H_{A,D}, H_{B,C}, H_{B,D}$ containing one element of $\lbrace (0,0), (2x,0)\rbrace$ and one element of $\lbrace C=(x,\sqrt{1-x^2}),D=(x,-\sqrt{1-x^2}) \rbrace$, the points of distance $1$ from the maker's first two moves. These hexagonal lattices have the property that the breaker hasn't played in any of them yet, and their pairwise intersections are empty or a point in $\lbrace A,B,C,D\rbrace.$ So, either the second play of the breaker misses both $H_{A,C}$ and $H_{B,C}$ or both $H_{A,D}$ and $H_{B,D}$. By symmetry, we can assume the second play of the breaker misses $H_{A,C}$ and $H_{B,C}$. Let the third play of the maker be $C$. This gives the maker an unopposed pair of points in both $H_{A,C}$ and $H_{B,C}$. The third play of the breaker can only be in one of these lattices. On the fourth move of the maker, the maker can play in the other to make an unopposed equilateral triangle of side length $1$. $\blacksquare$
Next, even if the maker is constrained to play in this lattice, the maker can force $5$ in a row by move $11$. Note that if the maker has an "open $3$" of $3$ points in row with two open spaces on either side, $- - \circ \circ \circ - -$, then the breaker has to respond immediately either just to one side or the other, or else the maker can make $5$ in a row in $2$ more moves. To avoid an explosion of cases, we'll let the breaker play both sides of an open $3$. Perhaps without this, the maker could force $5$ in a row in fewer moves.
We'll show that whatever the breaker's fourth move is, the maker can still force $5$ in a row by the eleventh move. By symmetry, we can assume the breaker's fourth move is in a sixth of the lattice between two of the perpendicular bisectors of the triangle's sides. We'll consider two possible moves within this sixth individually, and then all others.
o o x
o
5
o o x
o
x
5
o o x
o
x
x
6 5
o o x
o
x
x x
6 5
o o x
o
x x
x x
6 7 5
o o x
o
x x
x x
x 6 7 5 x
o o x
o
x x
x x
x 6 7 5 x
o o x
8 o
x x
x x x
x 6 7 5 x
o o x
8 o
x x x
x x x
x 6 7 5 x
o o x
8 o 9
x x x
This last play makes two open $3$s, with $7$ and with $8$, so with this $4$th move by the breaker, the maker can construct $5$ in a row by move $11$.
In the next sequence, I'll show the breaker's response immediately, again letting the breaker block both sides of open $3$s.
o o
o x
x
o o
o x
5
x
x
x 6 o o x
o x
5
x
x x
x 6 o o x
7 o x
5
x x
x x x
x 6 o o x
7 o x
5 8
x x x
x x x
x 6 o o x
7 o x
9 5 8
x x x
Again, the maker's 9th move creates two open $3$s, through the $7$ and through the $5$ and $8$, so with that choice of $4$th move by the breaker, the maker can construct $5$ in a row by move $11$.
Next we let the breaker's $4$th move block every other possibility in that sixth of the lattice, which will cover all of those possibilities simultaneously.
x x x x x
o o . x x x x x
o . x x x x x
. . . x x x x x
. . . . x x x
x x x x x
5 o o . x x x x x
o . x x x x x
. . . x x x x x
. . . . x x x
This play technically does not make an open $3$ since only one space to the right is open. So, the breaker does not have to respond to either side immediately. The other possibility is $2$ to the left. We will let the breaker play in all $3$ positions simultaneously.
x x x x x
x x 5 o o x x x x x x
o . x x x x x
. . . x x x x x
. . . . x x x
x x x x x x
x x 5 o o x x x x x x
o . x x x x x
6 . . x x x x x
x . . . . x x x
x x x x x x x
x x 5 o o x x x x x x
7 o . x x x x x
6 . . x x x x x
x x . . . x x x
x x x x x x x x
x x 5 o o x x x x x x
7 o . x x x x x
8 6 . . x x x x x
x x x . . . x x x
x x x x x x x x
x x 5 o o x x x x x x
7 o . x x x x x
8 6 9 . x x x x x
x x x . . . x x x
Again, the $9$th move creates two open $3$s, so the maker can construct $5$ in a row by move $11$.
The formulation of the question is very unclear, so I'll make some guesses about what was meant. First, I assume that all vertices have the same sort of data available, namely how many winning positions (not how many sequences of moves leading to those positions) there are for one or the other player after any number of subsequent moves. (In particular, the statement "'C' vertex contains different data" means only that the numbers are different for C, not that a different sort of information is given there.) Second, I'll assume that the *condition "if the game is not finished" means merely that the indicated number of moves should make sense. (I'll comment below on the alternative interpretation that, at a vertex corresponding to a finished game, we are not told who won.) Third, I'll assume that the absence of that *condition in the "3 moves" line of the example is not significant, mainly because I can't think of anything general to infer from that absence.
With all these assumptions, I'd find optimal moves for Player 1 by the following standard method for analyzing game trees. First, discard the information at most of the vertices; keep only the information at the leaves of the tree, telling who won the finished plays. Second, work backward from the leaves toward the root, labeling each vertex with the following information: Who has a winning strategy from that vertex? How many moves will it take to win (against the opponent's best delaying strategy)? If the player who has the winning strategy from this vertex is the one who is to move, then what is the first move of that strategy? It is easy to assemble all this information for a vertex given the corresponding information for all its children, so we can work backward from the leaves and fill in the information throughout the game tree. (Note that there is nothing probabilistic about the solution, or, for that matter, about the problem.)
In the (intuitively very strange) case that the *condition was intended to say that we don't know who wins at which terminal position, then, in general, there is not enough information to design an optimal (or even a reasonable) strategy. Perhaps in this situation, probability can become relevant if we assume that the opponent also doesn't know the winning conditions of the game and therefore uses some mixed strategy. In that case, the theory of equilibrium mixed strategies might be relevant, but I think this interpretation of the question is too improbable to be worth any more comment.
Best Answer
The answer depends upon, of course, if the players make their moves intelligently or stupidly. In either case, build an alpha-beta tree of move sequences and search either depth-first or breadth-first, stopping and backtracking as winning configurations are found. However, there are some rigorous points you have not defined about this question, including whether perhaps it came to you as a homework problem. Hoping that this is not the case, here's a take on it:
Yes, if $n=2$, then the first player to move obviously wins by picking any one of the four open spots, and the game is over. If $n=3$ (which you don't define for odd $n$, but lets say for odd $n$, $(n+1)/2$ spaces wins, then the first player to move wins automatically also, because no matter which space they take in the $3 \times 3$ grid, the next player can only block one space, and in the 2nd round, first player easily wins.
For $n=4$, the first player wins, with the same strategy as shown for $n=3$. $A$ picks any square in the $4 \times 4$ grid, $B$ picks any other square but can only block $A$ in one of the two dimensions in which this grid lies, therefore on the first step of the second round, $A$ wins by picking a square adjacent to his first square.
Are you sure that you've thought this out for the smaller grid games?
So for the $4 \times 4$ configuration, assuming that $A$ plays intelligently, there are
$4 \times 4 = 16$ first moves for $A$
$16 - 1 = 15$ first moves for $B$, if $B$ plays stupidly, or either $2, 3,$ or $4$ moves if $B$ plays intelligently
$4 \times 1 + 8 \times 2 + 4 \times 3 = 4+16+12=32$ second moves for $A$ which win the game and end the game ($A$ could have picked one of the $4$ corners, $B$ blocks one way leaving $1$ way for $A$ to play, if $A$ picks one of the non-corner edge pieces (8 ways to do that), then $B$ blocks one way leaving two ways for $A$'s second winning move, and if $A$'s first move is a center piece, $B$ can block one of the $4$ adjacent squares, leaving $3$ adjacent squares for $A$ to pick and win.
So if $A$ and $B$ both play intelligently, the sum of these are the possible games for a $4 \times 4$ grid:
$4 \cdot 2 \cdot 1 + 8 \cdot 3 \cdot 2 + 4 \cdot 4 \cdot 3 = 8 + 48 + 48 = 104$
If $A$ and $B$ do not play intelligently, then the number of possible games is much larger, and the number of "winning configurations" or "end positions" is much larger.