Two artists want to paint all sand grains in 2 colors. They stop if one artist picks a grain already painted by the other. In mean how many picks? Let $N$ be the (big) number of sand grains.
Artist $B$ want to paint them blue. Artist $R$ want to paint them red.
We start with all sand grains unpainted.
They decide $B$ can start.
$B$ picks a random sand grain (all always chance $\frac{1}{N}$) and paint it blue. 
After this it's $R$ turn. $R$ picks a random sand grain and paints it red (if not already blue).
If an artist picks a sand grain which is:

*

*not painted yet the artists paints it with the own color and place it back afterwards. After this it's the other artist turn again.

*already colored in the own color the same artist will pick a new sand grain (this can repeat multiple times).

*already colored but not the own color the drawing session will end for both artists.

(That means they do fair share. They have either the same amount of colored sand grains or $R$ has just $1$ more.)
What is the expected/mean count of picks required for this experiment in relation to $N$?
We assume $N$ can grow as big we want (but not $\infty$). Does it converge to a equation $f(N)$?

The expected number of picks $n$ could be written as:
$$\mathbb{E}[n,N] = \sum_{i=2}^{\infty} i\cdot p_i$$
Can we approximate it for large $N$. Does it converge to something?
 A: This is a collision problem, so if it follows the pattern of other collision problems it seems reasonable to guess that the answer will be $O(\sqrt{N})$.
If there have been $2j$ turns without stopping then the conditional probability B paints the next grain blue is $\frac{N-2j}{N-j}$
while if there have been $2j+1$ turns without stopping then the conditional probability R paints the next grain red is $\frac{N-2j-1}{N-j}$
so the probability they have not stopped after $2j+1$ turns is $\frac NN\frac {N-1}{N} \frac {N-2}{N-1} \frac {N-3}{N-1} \cdots  \frac{N-2j}{N-j} =\frac{ (N-j)!(N-j-1)!}{(N-2j-1)! N!} =\frac{2N-2j-1 \choose N}{2N-2j-1 \choose N-j}$
and that  they have not stopped after $2j+2$ turns is $\frac NN\frac {N-1}{N} \frac {N-2}{N-1} \frac {N-3}{N-1} \cdots  \frac{N-2j}{N-j}\frac{N-2j-1}{N-j} =\frac{ ((N-j-1)!)^2}{(N-2j-2)! N!}=\frac{2N-2j-2 \choose N}{2N-2j-2 \choose N-j-1}$
To find the expected time when they stop, you have to sum these two sets of probabilities from $j=0$ upwards until they become $0$ (which they do after $N+1$ turns) plus the probability of $1$ they have not stopped by $0$ turns.  This is easy enough to program though I do not see an easy way to get a closed form.  Empirically, the expected number of turns seems to be close to $\sqrt{N\pi }$ and even closer to $\sqrt{N\pi } - \frac12$ for large $N$.
The $\sqrt{N}$ is not a surprise, but why this particular expression I do not know. For example using R:
expectedturns <- function(N){
  probstillplaying <- 1
  cumprobsum <- 1
  j <- 0
  while (probstillplaying  > 0){
    probstillplaying <- probstillplaying * (N - 2*j) / (N - j)
    cumprobsum <- cumprobsum + probstillplaying 
    probstillplaying <- probstillplaying * (N - 2*j - 1) / (N - j)
    cumprobsum <- cumprobsum + probstillplaying
    j <- j+1
    }
  return(cumprobsum)
  }

with example expectations
expectedturns(1) 
# 2
expectedturns(2)   
# 2.5
expectedturns(3)   
# 3
expectedturns(4)
# 3.416667
expectedturns(10^2)
# 17.30196
expectedturns(10^4)
# 176.7531
expectedturns(10^6)
# 1771.9546

Note that that $\sqrt{1000000\pi }-\frac12 \approx 1771.95385$ so very close.  The error in the approximation seems to be  $O\left(\frac1{\sqrt{N}}\right)$.
