Colored ball problem with $2$ colors, but each ball is always replaced with $1$ color I'm trying to find a solution to this problem but I'm having a hard time thinking through how to solve it:
You have one bag with $N$ balls, each being red or white.
You draw some number of balls, each time replacing the ball with a red ball. IE both red and white balls are always replaced with red balls regardless of the color drawn.
How do you solve for an expected number of red/white balls?
Example:
$100$ balls, $70$ red and $30$ white. You draw $15$ balls. How many red balls do you expect to draw?
It would be pretty straightforward if the balls were replace with their own color, but I'm stumped here.
Ideally I'd love to have something I could punch into a cell formula in Google Sheets, but if that's not possible then I'd be happy to have any kind of explanation.
 A: Let's say you have $N$ balls, $W$ of which are white, and you draw (and replace with red) $m$ times. You want to know the expected number of white balls drawn.
This is the sum over $1 \leq k \leq m$ of the expected number of white balls drawn in the $k$-th round, namely the probability that the $k$-th draw is a white ball.
Now let us imagine the modified procedure. When you draw a red ball, you put it back in the bag. When you draw a white ball, you paint it in red and then put it back in the bag.
Thus in the $k$-th round, you get a white ball if and only if the ball is originally white and has never been drawn out in the previous $k - 1$ rounds.
The probability that the ball is originally white is $\frac WN$. The probability that this ball has never been drawn before is $(1 - \frac1N)^{k - 1}$.
Thus the probability that you draw a white ball in the $k$-th round is equal to $\frac W N \cdot (1 - \frac 1N)^{k - 1}$.
Taking the sum over $1 \leq k \leq m$, we get that the expected number of white balls drawn in $m$ rounds is equal to $W(1 - (1 - \frac 1N)^m)$.
A: Suppose there are $r$ red balls and $w$ white balls, and you draw $k$ balls. For each white ball, the probability it gets picked is $1-(1-\frac1{r+w})^k$. Using linearity of expectation, the expected number of white balls you pick is
$$
\mathbb E[\text{# white balls picked}] = w\times\left[1-\left(1-\frac1{r+w}\right)^k\right]
$$
Since $
(\text{# white balls picked})+(\text{# red balls picked})=k
$, the expected number of red balls picked is $k$ minus the above expression.
A: One strategy to solving such a problem, is to think of what you are interested to count. For example, we are interested to count the number of red balls drawn in a finite number of trials (and it's average). Let $X$ be the number of red balls drawn in $M$ trials.
Can you think of expressing $X$ as a sum of random variables?
Let $I_j$ be an indicator function that lights up, when a red ball is drawn in the $j$th draw. Then, we can write:
\begin{align*}
X &= I_1 + I_2 + \ldots + I_M
\end{align*}
Taking expectations on both sides and by the linearity of expectations, we have:
\begin{align*}
E(X) &= E(I_1) + E(I_2) + \ldots + E(I_M)
\end{align*}
Note that, the events "$\text{A red ball is drawn in trial }i$" and "$\text{A red ball is drawn in trial }j$" are not independent. However, linearity of expectations is applicable, regardless of whether the random variables are independent or not!
Now, the expectation of an indicator function $I_A$, $E(I_A)$ equals the probability of the occurrence of the event $A$, $P(A)$. That's because, $I_A$ is a Bernoulli random variable with success probability $p$, and the expectation of a Bernoulli random variable is just $p$, so $E(I_A) = P(A)$.
The unconditional probability of drawing a red ball in the $j$th draw, without any knowledge of the previous draws is $(70/100)$. So, $E(I_j) = P(\text{red ball is drawn in the }j\text{th trial })=70/100$.
So,
\begin{align*}
E(X) &= \sum_{j=1}^{M}E(I_j)=\frac{70}{100}\times M
\end{align*}
