Sam and Jane play with indicators. There are $30$ red balls and $50$ white balls. Sam and Jane take turns drawing balls until they have drawn them all. Sam goes first. Let $N$ be the number of times Jane draws the same color ball as Sam. Find $E[N].$
I have been proceeding with indicators...
$$  I_{j} =
\begin{cases}
1,  & \text{if the $i^{th}$ pick is the same as the $(i-1)^{th}$ pick.} \\
0, & \text{otherwise}
\end{cases}
$$
But I am having problems with this. Because when I proceed, I am find all the instances where the same ball is drawn on every draw, not just Jane's. I know the answer is $\frac{9}{19}$ but I can't get there. How do I change the indicators, or the probability?
 A: As you started, let $I_n=\begin{cases}1&\text{if Jane draws the same color on turn}~2n~\text{as Sam did on turn}~2n-1\\
0&\text{otherwise}\end{cases}$
We have $E[N]=E[\sum\limits_{n=1}^{40}I_n]$ which by linearity of expectation is $=\sum\limits_{n=1}^{40}E[I_n]$ and by symmetry is $40E[I_1]$
Why is $E[I_1]=E[I_2]=\dots$? Imagine that they pull out two balls and announce that they are starting in the middle of the count and that these were in fact the $(2n-1)^{st}$ and $(2n)^{th}$ balls respectively and then continue to the beginning of the count afterwards.  Clearly, the probabilities are the same regardless what label they give the turn numbers.
Let $S_r,S_w,J_r,J_w$ represent the events that Sam or Jane pulled a red or white ball on turn $2n-1$ or $2n$ respectively.
The probability that they matched color is then $Pr((S_r\cap J_r)\cup(S_w\cap J_w))=Pr(S_r)Pr(J_r|S_r)+Pr(S_w)Pr(J_w|S_w)=\frac{30\cdot 29+50\cdot 49}{80\cdot 79}$
The expected number of matches then will be $40$ times this number for a final total of:
$$\frac{1660}{79}\approx 21.01267$$
A: Hint:
Here's the simulated distribution (300,000 repetitions)

(let me know if you want the code, it's pretty straightforward though)
