Marbles in 3 buckets

Given 3 buckets, each containing $n$ marbles. Alice and Bob are playing a game, where on each turn, the player may remove 1, 2, or 3 marbles from a single bucket. Alice goes first. The player who removes the last marble wins. Determine all $n$ such that Alice has a winning strategy.

Solution

If $n$ is divisible by 4 then Bob has a winning strategy, and if $n$ is not divisible by 4 then Alice has a winning strategy.

First note that $(4m,k,k)$ is a losing position for any $k \geq 0,m > 0$. Because if you remove $p$ marbles from the first bucket, your opponent may remove $4-p$ marbles to go back to $(4(m-1), k k)$. And if you remove from any of the second or third pile, your opponent may mirror that to go back to $(4m, k-p, k-p)$. Thus, no matter what you do, your opponent always has a legal move to make.

Therefore if $n$ is divisible by 4, assuming Bob plays optimally, Alice loses the game.

But if $n$ is not divisible by 4, say $n = 4m+a, a=1,2,3$. Then Alice may remove $a$ from the first bucket, to arrive at $(4m,n,n)$ which is a losing position for Bob.

Generalization

If there are $n$ buckets of $n$ marbles, we can employ the same strategy. The goal is to force your opponent into a "symmetric" situation where no matter what he/she does, you can always mirror it.

If $n$ is divisible by 4, then $(n,n,\dots,n)$ is a losing position because no matter what you do, your opponent can always mirror it. In general, the situation where there are an even number of buckets left and each pair of buckets contains the same number of marbles is a losing position.

If $n=4m+1,4m+3$, then you can employ the same strategy as above. Alice transforms the configuration into $(4m, a,a, b,b, \dots, z,z)$ and from there no matter what Bob does, Alice can always mirror it.

Variation

Now, suppose there's only $n$ marbles total and it is to be distributed among all 3 buckets, with each bucket containing at least one marble. Alice determines the distribution, and Bob makes the first "regular" turn. Determine all numbers $n$ such that there exists a way for Alice to distribute in such a way that Bob always loses.

Solution

At this point, we have to characterize the conditions in which configuration $(a,b,c)$ is a losing position. It's clear to see that $(a,0,0)$ is a losing position if and only if $a \equiv 0 \mod 4$.

Now consider the configuration $(a,b,0)$. If $a \equiv b \mod 4$ then this is a losing position, because the opponent can always mirror you move to keep the condition true. Therefore, if $a \neq b \mod 4$, it is a winning position because we can turn it into the above-mentioned losing position for our opponent.

Now consider the configuration where there are three non-zero buckets left. As we saw above, $(4a,b,b)$ is a losing position because your opponent can always mirror your steps, and restoring to original configuration, and consequently, $(a,b,b)$ is a winning position if $a \neq 0 \mod 4$ because you can turn it into the form $(4a,b,b)$.

We can take this reasoning one step further by claiming, the configuration $(a,b,c)$ where $b \equiv c \mod 4$ is a losing position if and only if $a \equiv 0 mod 4$. Indeed, if $a$ is divisible by 4, then any move you make in the first bucket will be countered in the first bucket, and any move you make in the second or third buckets will also be countered to keep $b \equiv c \mod 4$. Thus either you will bring $a$ below 4 and your opponent can empty the first bucket, leaving you to $(0,b,c)$ which is a losing position, or you will empty one of the second or third buckets first. In the latter case, then your opponent can still mirror it to arrive at $(a,0,c)$ where both $a,c$ are divisible by 4, which is a losing position. Conversely, if $a$ is not divisible by 4, you can make it divisible by 4 to be a losing position for your opponent.

Finally we prove that $(a,b,c) \equiv (1,2,3) \mod 4$ is a losing position, because any move that is made from here will result in a winning position. If we take from first bucket, we now get $(2/3/0, 2,3) \mod 4$ all of which are winning positions. If you take from second bucket, we get $(1, 3/0/1, 3)$ also winning positions, similarly if we do from third bucket. The point is that there is no way to avoid $(a,a,b)$ form where $b $ is not divisible by 4, or $(0,a,b)$ form.

Thus, if $n > 4$ is even, Alice has a winning strategy as follows. If $n = 4m$ then Alice distributes the marbles as $(4(m-1),2,2)$. If $n=4m+2$ she distributes it as $(4m,1,1)$. Both are losing positions Bob. But if $n=4$, then Bob has a winning strategy. Alice has no choice but to distribute the marbles as $(1,1,2)$ initially, and Bob can turn it into $(1,1,0)$ which is a losing position for Alice.

If $n$ is odd, Bob has a winning strategy no matter how Alice distributes the marbles. If all the three buckets contain odd number of marbles, then two of them will be congruent modulo 4, and the third one will not be divisible by 4, and that is a winning position for Bob. If two buckets contain even numbers and the third one odd, we have two choices. Either the two even buckets are both $\equiv 2 mod 4$, which means it's still a winning position for Bob, or at least one of them is divisible by 4, which is also a winning position for Bob.

Bottom line, Alice has a way to distribute the marbles into a losing position for Bob if and only if $n$ is even and $n > 4$.