Tough problem to write up. Answer is $c \geq \frac{1}{2}$.

Neat problem.

For c<1/2. Assume bob writes on the board 2^n ones and 2^n zeroes. The idea goes like this, if Alice can use X ones and Y zeroes to make a number M, bob can use X zeroes and Y ones to make a number 1-M. Also, the number of zeroes/ones decreases by at most one, unless Alice forges them both into a 1/2. So you only have to consider how can Bob take advantage of Alice spawning a lot of 1/2.

The answer is $c \geq \tfrac{1}{2}$. For $c \geq \tfrac{1}{2}$, Alice's strategy for $n$ even is as follows. By shifting and scaling assume WLOG that not all the integers are the same parity. Merge two opposite-parity integers to get a half-integer, and then merge this arbitrarily until she makes $\tfrac{1}{2}n$ moves. This leaves Bob with $\tfrac{1}{2}n-1$ "untouched" integers and one number $k$ whose $\nu_2$ equals $-\tfrac{1}{2}n$. But it is clear that Bob loses due to $\nu_2$ reasons, since without merging with $k$ we can never get to a $\nu_2$ at most $-\tfrac{1}{2}n$, but merging with $k$ always leaves us with a single number of $\nu_2$ more than $-\tfrac{1}{2}n$; in particular, when we end up with $2$ numbers we have one with $\nu_2$ at most $-\tfrac{1}{2}n$ and another with $\nu_2$ more than this. Now consider $c=\tfrac{1}{2}-r$ for some $r>0$. Bob employs the following strategy. He makes $\lfloor \tfrac{1}{2}n\rfloor$ numbers equal to $0$ and $\lceil \tfrac{1}{2}n\rceil$ numbers equal to $1$. After Alice makes her moves, Bob considers each number not equal to $0$ or $1$ formed by Alice. If a number $m$ was formed by merging $t>\tfrac{2}{r}+1$ zeroes and ones (hence $t-1$ merges are employed), then Bob directly forms $1-m$ by swapping the roles of $1$ and $0$. These two merges use a total of $t$ zeroes and $t$ ones. For the numbers formed from merging at most $\tfrac{2}{r}$ zeroes and ones, Bob merges numbers created with the same process and then create a single copy of $1-m$ for each instance of $m$. Because there are a finite number of numbers formable from at most $\tfrac{2}{r}$ zeroes and ones (independent from $n$), Bob uses a bounded number (say $C$) of zeroes and ones in this process. Alice can make at most $\tfrac{r}{4}n$ numbers from merging more than $\tfrac{2}{r}+1$ zeroes and ones, so between Alice's moves and Bob's responses to them at most $(\tfrac{1}{2}-r)n+\tfrac{r}{4}n+C$ zeroes and ones each are used. For sufficiently large $n$ this leaves at least $1$ zero and $1$ one unmerged (in particular, it is always possible for Bob to respond as above). Now, Bob can finish by merging each Alice-created $m$ with his own $1-m$, as well as merging all the remaining zeroes together, merging all the remaining ones together, and merging the resulting $0$ and $1$ together. This leaves us with every number equal to $\tfrac{1}{2}$ (and at least two numbers), from which Bob can clearly win. $\blacksquare$ Remark: "Naively" responding to formation of $m$ with formation of $1-m$ works for $c \leq \tfrac{1}{4}$, but Alice could form a bunch of $\tfrac{1}{2}$ from merging $0$ and $1$ (or, say, $\tfrac{1}{4}$ from merging $0$ and $1$, then $\tfrac{1}{2}$ and $0$) which is a problem for $c \leq \tfrac{1}{2}$, since to respond Bob needs too many numbers. However, in reality if Bob is trying to finish with two copies of $\tfrac{1}{2}$ you could have a milllion copies of $\tfrac{1}{2}$ and still respond to them with a single $0$ and $1$ by merging them beforehand, and refining this argument finishes.

The problem is quite easy for P3.

IAmTheHazard wrote: Why do we need to consider cases t>2/r+1 and t<=2/r?