Two greedy chocolate eaters play the following game involving $n$ pieces of chocolate and an additional parameter $\alpha$ with initial value $1$: Each player eats either $\alpha$ pieces of chocolate or he increments $\alpha$ by $2$ (replacing thus $\alpha$ by $\alpha+2$) and eats then $\alpha$ pieces of chocolate (he eats thus $2$ pieces more than his opponent in the preceding move). The game stops if less than $\alpha$ pieces remain and is won by the player having eaten more. There is thus the possibility of a draw in the case of equality.

My computer seems to pretend that the first player can always win this game if the initial number of pieces is odd (at least for all odd numbers $\leq 451$).

Is there an easy reason for this?

(In the case of an even number of pieces, the game seems to be more or less balanced but there are many cases of a draw when both players use optimal strategies.)

Update: I have accepted Douglas Zare's answer containing an elegant explanation for the existence of a winning strategy for the first player. (Un)fortunately, this explanation does not give much information on the winning strategy and a part of the mystery remains.

**Added 4/25/13**: The above game is an example of a simple combinatorial game where
the first player has a non-obvious (for most odd values of $n$) winning strategy.

Douglas Zare's proof works also for the following variations (variation (1) is an even simpler game with less possibilities, for the remaining ones there are in general more possible moves):

(1) (in honor of Fibonacci): If a player has augmented (by $2$) the number of pieces, then the next player has to take the same amount (the value of $\alpha$ cannot be augmented during two consecutive moves).

(2) After a move with $\alpha$ pieces eaten, the next move can eat $\alpha-2,\alpha$ or $\alpha+2$ pieces ($\alpha-2$ has of course to be positive).

(3) Combine (1) and (2): $\alpha-2$ (if positive) or $\alpha$ pieces can always be eaten, $\alpha+2$ pieces only if $\alpha$ did not increase during the preceding move.

(4) More generally, one can choose for each odd number $\alpha\geq 1$ a set $\mathcal P(\alpha)$ of possible odd numbers for the next value of $\alpha$. Douglas Zare's proof works if $\mathcal P(1)=\lbrace 1 ,3\rbrace$. A perhaps interesting choice is $\mathcal P(\alpha)=\lbrace 1,\alpha,\alpha+2\rbrace$ which allows a sort of "reinitialization" of the game.

(5) Variation (4) can be combined with (2): Possible moves depend not only on the value of $\alpha=\alpha_1$ eaten by the preceding player but also by the values $\alpha_2,\alpha_3,\dots,\alpha_k$ eaten two moves, three moves etc, ago. D. Zare's proof applies if $\mathcal P(1,\emptyset,\dots)=\lbrace 1,3\rbrace$ and $\mathcal P(\alpha_1,\dots,\alpha_i,3,1,\emptyset,\dots,\emptyset)\supset\mathcal P(\alpha_1,\dots,\alpha_i,3,\emptyset,\emptyset,\dots,\emptyset)$ for all choices of $i\leq k-1$ odd numbers $\alpha_1,\alpha_2,\dots,\alpha_i$ .

Are there instances with interesting winning strategies?