Breaking chocolate bars game
This is a combinatoric game. Using Conway notation, you can show that this game is always a number (cold game) and always an integer.
Let $A$ be the Left player and $B$ be the Right player.
Then show that the value $v$ of $(a,b)$ verifies :
$v(a,b)=-v(b,a)$
for $a\ge b$, let $|x|$ the number of bits in the binary representation of $x$ ($|0|=0$, $|1|=1$, $|2|=|3|=2$, |4|=|5|=|6|=|7|=3, etc). Then $$v(a,b)=\left\lfloor\frac{a}{2^{|b|-1}}\right\rfloor-1 $$
So if $A$ begins, he has a winning strategy as long as long as $a\ge 2^{|b|}$ (he needs $v>0$ to win)
If there are several pieces :
$$\mbox{A has a winning strategy} \Longleftrightarrow \left(\sum_i v(a_i,b_i)\right)>0$$
Exemple of values :
$$\tiny\begin{array}{r|cc} b_i\rightarrow& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16\\\hline a_i=1&0 &-1 &-2 &-3 &-4 &-5 &-6 &-7 &-8 &-9 &-10 &-11 &-12 &-13 &-14 &-15\\ 2&1 &0 &0 &-1 &-1 &-2 &-2 &-3 &-3 &-4 &-4 &-5 &-5 &-6 &-6 &-7 \\ 3&2 &0 &0 &-1 &-1 &-2 &-2 &-3 &-3 &-4 &-4 &-5 &-5 &-6 &-6 &-7 \\ 4&3 &1 &1 &0 &0 &0 &0 &-1 &-1 &-1 &-1 &-2 &-2 &-2 &-2 &-3 \\ 5&4 &1 &1 &0 &0 &0 &0 &-1 &-1 &-1 &-1 &-2 &-2 &-2 &-2 &-3 \\ 6&5 &2 &2 &0 &0 &0 &0 &-1 &-1 &-1 &-1 &-2 &-2 &-2 &-2 &-3 \\ 7&6 &2 &2 &0 &0 &0 &0 &-1 &-1 &-1 &-1 &-2 &-2 &-2 &-2 &-3 \\ 8&7 &3 &3 &1 &1 &1 &1 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 9&8 &3 &3 &1 &1 &1 &1 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 10&9 &4 &4 &1 &1 &1 &1 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 11&10 &4 &4 &1 &1 &1 &1 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 12&11 &5 &5 &2 &2 &2 &2 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 13&12 &5 &5 &2 &2 &2 &2 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 14&13 &6 &6 &2 &2 &2 &2 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 15&14 &6 &6 &2 &2 &2 &2 &0 &0 &0 &0 &0 &0 &0 &0 &-1 \\ 16&15 &7 &7 &3 &3 &3 &3 &1 &1 &1 &1 &1 &1 &1 &1 &0 \end{array} $$
EDIT (long answer to a comment) :
Yes, this is a cold game, so each time a player plays, the situation become worse for him. So skipping his turn is good.
All those (cold) games can be represented by a number (not necessarily an integer, it could be a real, or an ordinal). Here the game is simple, and we only get integers.
So the integer value of the game is how many turn a player can skip his turn and still lose. So if $v(a,b)=3$, $B$ can skip 3 times (assuming $B$ begins) and still lose (so this is a good game for $A$). The same for $v(a,b)=-3$ but you exchange $A$ and $B$. $v(a,b)=0$ means the first player will lose.
Usually, to compute the value of a situation $s$, we compute all values of all situations any player can go from $s$, and use the usual theory to infer the initial value (again "On numbers and games"). But you're right in your remarks and perhaps you can simply prove the values by reasoning on "how many times first player can cut at first without losing". But I guess you will just have to rebuild the theory in this particular case. You can't do without it on such games. Read the book !
This game is called Cutcake, and it is explicitly analyzed in both "On Numbers and Games" and "Winning Ways". (For some reason the spam-blocker thinks this answer is spam; not sure why.)