Another balls game I have an infinite bag of balls. Balls can be black and white. 
Well, infinite, but I will get black ball from there with probability $p=0.6$ and white with $1-p=0.4$. 
I have two friends. One chooses black color and another one -- white color. With each turn of the game I'm giving balls from the bag of the chosen color to the corresponding person with corresponding probabilities. 
We decided that the person who gets $5$ more balls than another one wins this game. 
Well, the question is: how much time this game can long? (in terms of expected value of number of turns)
$------------------------------$
My attemt.
One wins with probability $p^{n}(1-p)^{n+5}$ at the $2n+5$-th turn of this game... so another one does that with $p^{n+5}(1-p)^{n}$ at the $2n+5$-th turn of this game.
Now, expected value: 
$$
M = \frac{\sum_{i=0}^{+\infty} (2i+5) \cdot \left( p^{i+5}(1-p)^{i} + p^{i}(1-p)^{i+5} \right)}{\sum_{i=0}^{+\infty} \left( p^{i+5}(1-p)^{i} + p^{i}(1-p)^{i+5} \right)}
$$
I feel like, I am messing something, because my answer is very close to $5$ turns, but intuitively it's clear that you aren't likely to end this game so fast. 
Edit:
$$
M = \frac{\sum_{i=0}^{+\infty} (2i+5) \cdot \left( p^{i+5}(1-p)^{i} + p^{i}(1-p)^{i+5} \right)}{\sum_{i=0}^{+\infty} \left( p^{i+5}(1-p)^{i} + p^{i}(1-p)^{i+5} \right)} = \frac{\sum_{i=0}^{+\infty} (2i + 5) \cdot p^{i}(1-p)^{i} }{\sum_{i=0}^{+\infty}  p^{i}(1-p)^{i} } = \frac{\sum_{i=0}^{+\infty} (2i + 5) \cdot p^{i}(1-p)^{i} }{\sum_{i=0}^{+\infty}  p^{i}(1-p)^{i} } =  \frac{ 2 \sum_{i=0}^{+\infty} i \cdot p^{i}(1-p)^{i} + 5\sum_{i=0}^{+\infty} p^{i}(1-p)^{i} }{\sum_{i=0}^{+\infty}  p^{i}(1-p)^{i} } = \frac{ 2 \sum_{i=0}^{+\infty} i \cdot p^{i}(1-p)^{i} }{\sum_{i=0}^{+\infty}  p^{i}(1-p)^{i} } + 5 = \frac{p(1-p)}{1-p(1-p)} + 5
$$
 A: Your calculation is correct, apart from the dropped factor of $2$ in the last line, but it is not finding the correct probability.
It is not true that on the $(2n+5)^{\text{th}}$ turn of the game, one player wins with probability $p^{n}(1-p)^{n+5}$. This does not take into account:


*

*the possibility that by then, the game may have already ended.

*more significantly, the different orders in which the balls may be drawn.


Also, if it were correct, you would not need the denominator.
The correct way to set up the problem is as a Markov chain with states $-5, -4, -3, \dots, 4, 5$ in which from state $k$ we go to state $k+1$ with probability $p$ and state $k-1$ with probability $1-p$. (The state $k$ represents a case in which the first player is ahead of the second by $k$ balls.)
Then we are looking for the hitting time from state $0$ to states $\{-5,5\}$, which can be obtained by solving the equations
\begin{align}
   h(k) &= 1 + p\cdot h(k+1) + (1-p) \cdot h(k-1) & -4\le k \le 4 \\
   h(-5) &= 0 \\
   h(5) &= 0
\end{align} 
for $h(0)$. When $p = \frac35$ I get an answer of $\frac{211}{11}$.
