# Variance of taking balls from a bag question

A bag contains $$2$$ white balls and $$2$$ black balls. Each instance a ball is taken from the bag, if it's white it's returned to the bag and if it's black it's replaced by a white ball. The game ends when no black balls remain in the bag. Let $$Y$$ be the number of instances of the game; calculate $$\operatorname{var}(Y).$$ Answer in the book: $$14.$$

I am not sure how to write the probability function of $$Y;$$ please help and thanks in advance!

Let $$Y_1$$ be the random variable who value is the number of draws until the first black ball is drawn. This is a geometrically-distributed random variable with probability of success $$\frac12$$.
Similarly, let $$Y_2$$ be the random variable who value is the number of draws until the second black ball is drawn, after the first black ball is replaced by a white ball. This is a geometrically-distributed random variable with probability of success $$\frac14$$.
Then $$Y=Y_1+Y_2$$.
Let $$Y$$ be the number of black balls taken. The probability distribution table: $$\begin{array}{c|l} Y&P(Y)\\ \hline 2&\frac1{2^3}\\ 3&\frac1{2^4}+\frac3{2^5}\\ 4&\frac1{2^5}+\frac3{2^6}+\frac{3^2}{2^7}\\ 5&\frac1{2^6}+\frac3{2^7}+\frac{3^2}{2^8}+\frac{3^3}{2^9}\\ \vdots \end{array}$$ Note: $$P(Y=n)=\sum_{i=1}^{n-1} \frac{3^{i-1}}{2^{n+i}}=\frac13\cdot \frac1{2^n}\sum_{i=1}^{n-1}\left(\frac32\right)^i=\frac{2\cdot 3^n-3\cdot 2^n}{3\cdot 4^n}.$$ Hence: $$Var(Y)=\mathbb E(Y^2)-(\mathbb E(Y))^2=\\ \sum_{i=2}^\infty i^2\cdot \frac{2\cdot 3^i-3\cdot 2^i}{3\cdot 4^i}-\left(\sum_{i=2}^\infty i\cdot \frac{2\cdot 3^i-3\cdot 2^i}{3\cdot 4^i}\right)^2=\\ 50-36=14.$$ WA answers: 1, 2, 3.