# Is Nim a (strong) positional game?

A positional game is a kind of a combinatorial game described by:

• $$X$$ a finite set of elements. (Often $$X$$ is called the board and its elements are called positions.)
• $$F$$ a family of subsets of $$X$$. (These subsets are usually called the winning-sets.)
• A criterion for winning the game.

For $$n$$-pile Nim, wouldn't $$X$$ be the union of disjoint posets, e.g. $$\{(1,2,...,k_1), (1,2,...,k_2),..., (1,2,...,k_n)\}$$? Then a move consists of choosing a point in one of the posets and removing that point and everything above it in said poset.

Then $$F$$ would be $$\{\{{(1,2,...,j),(),()}\}, \{{(),(1,2,...,j),()}\}, \{{(),(),(1,2,...,j)}\}\}$$

And the winning criterion is being the first one to "receive" a position in $$F$$?

This seems right to me, but it appears that player two can have a winning strategy, which contradicts the notion of Nim being strong positional.

No. For a positional game, the legal moves are to take any element of $$X$$ which hasn't already been taken. In this setting of Nim, taking an element of one of the posets can mean that several other elements of $$X$$ stop being legal moves.
Also, I don't think your $$F$$ makes sense. $$F$$ should be the sets such that if you can take all of one set in $$F$$, you win the game. That's not what you have (your $$F$$ would mean that the winner is the first to take the last $$j$$ elements of one of the posets).