# Heaps of Beans combinatorics problem

A game has four heaps of beans, contains $3,4,5,$ and $6$ beans. There are two players in this game, and they alternate moves. A move is either taking

1. one bean from a heap, provided at least three beans left in the heap, or

2. a complete heap of two or three beans.

The player who takes the last heap wins. I am trying to see if moving first or moving second will give you a winning strategy. I have a feeling that the first player could have a winning strategy, although I am still working on trying to find the winning strategy that awards this. If we start out with $$3,4,5,6$$ At this point, we want player $2$ to be left with a single stack of $4$ to deal with, a pair of stacks as $3,3$, or even left with $6$ as the lone stack. Thus, we need to find a strategy of player $1$'s that will lead to these particular stacks for player $2$. I am wondering, is there potentially a good form of backwards induction that ensures these situations will occur for player $1$?

There are only $10$ moves in the game whatsoever. This is because, each time you can take only $1$ ball till you reach $3$ heap, and then you must take the $3$ heap completely.
Thus, $3$ heap takes $1$ move
$4$ heap takes $2$ moves
$5$ heap takes $3$ moves
$6$ heap takes $4$ moves
Total $10$ moves. So, whatsoever, Player $2$ always wins.