Can you win the monochromatic urn game?

The game is unwinnable iff the largest number is greater than or equal to the sum of all the others, plus 2. If the largest number is this big, then there are too few balls in the other vases to separate all the balls from this vase. If there are fewer balls than this in the largest vase we use induction to prove it is winnable.

Firstly if there is only 1 ball the game is trivially winnable,and if there are 2 balls they are in different vases so the games is again winnable. Suppose it is winnable when there are $n$ balls. If there are $n+1$ balls then remove a ball from the biggest number, and a ball from any other vase. Note that if a different vase now has the largest number it can have at most 1 more than the previous largest. The largest number still satisfies the condition and the smaller game is winnable.

The intuitive answer looks like

Let $s=\sum\limits_{i=1}^n a_i$ and $a_{(n)}=\max\limits_{1\le i\le n} a_i$.

Then the game is winnable iff $a_{(n)} \le s - a_{(n)} +1$, i.e. iff $a_{(n)} \le \left\lceil\frac{s}2\right\rceil $

** Not an answer, but a suggestion **

Suppose you generalize the problem to this: A "game" is a sequence of natural numbers $$ a_1, \ldots, a_n $$ together with a number $k$ between $1$ and $n$. A move in a game consists of picking any number $i$ from $1$ to $n$ with $a_i \ne 0$ and $i \ne k$; such a move changes the game to $$ (a_1, ..., a_i - 1, \ldots, ...a_n; i) $$ i.e., the $i$th slot is reduced by $i$ and the special index is changed to $i$.

A game is bad if all $a_i = 0$ except for $a_k$, which is nonzero; a game is won if all $a_i = 0$.

You can now recursively define a function $$ W(a_1, \ldots, a_n, k) $$ that's "True" if either the current game is won, or there's an $i$ such that $$ W(a_1, \ldots, a_i - 1, \ldots n, i) = True $$ and False otherwise.

Finally, you can see the the numbers $(a_1, \ldots, a_n)$ have your special property exactly if $$ W(a_1, \ldots, a_n, 0; n+1) = True $$ i.e., if you set up a game where there's an extra vase (the $n+1$th one) that's just been emptied, and all the other vase-fullnesses are as specified.

For $n = 3$, for instance, this tells you that a 3-vase game is winnable is you can reduce it to a 2-vase game where the two vase amounts differ by $1$. Alternatively, you can construct all winnable 3-vase games by sprinkling balls into vases, without ever dropping two balls into the same vase in sequence. So because $$ (3, 4, 0) $$ is a winnable 2-vase game (expressed as a 3-vase game with an empty vase), so is $$ (93, 94, 4) $$ because you can get there by adding to each of 3, 2, 1 in sequence, four times, and then adding to each of 2,1, in sequence, 86 times.