# combinatorics – Can you win the monochromatic urn game?

In the (monochromatic) urn depletion game, you are given $$n$$ vases, each containing some number of balls $$a_1,ldots, a_n geq 0$$. You win the game if you can remove all of the balls from the vases; you must draw them one at a time, and the only rule is that you cannot draw from the same vase twice in a row.

The problem is to decide, given the occupancy numbers $$a_1, ldots, a_n$$, whether the game is winnable.

Example: The game (AAA, A) (three in one vase; one in another) is unwinnable.

I’ve already got an efficient algorithm for winning the game: at each step, draw from the vase with the largest number of balls $$a_i$$ (among vases you can legally choose). If the game is winnable at all, this algorithm will win it.

So instead of an algorithm, I am looking for a property of the numbers $$a_1,ldots, a_n$$ which would enable someone to calculate whether the game is winnable. Evidently there’s a formula implicit in the algorithm above, but I wonder if it’s possible to find an explicit and simple one.

I’ve tried establishing the result for small $$n$$: if $$n=1$$, $$a_1$$ must be 0 or 1. If $$n=2$$, then $$|a_1-a_2|$$ must be 0 or 1. If $$n=3$$, the condition is slightly more complicated but might be expressible in terms of the differences $$|a_i-a_j|$$.

(I previously asked a related question about the multicolor version of this game)