Expected Income Question An urn contains equal number of green and red balls. Suppose you are playing the following game. You draw one ball at random from the urn and note its colour. The ball is then placed back in the urn, and the selection process is repeated. Each time a green ball is picked you get 1 Rupee. The first time you pick a red ball, you pay 1 Rupee and the game ends. Your expected income from this game is..
The answer given is 0 but shouldn't it be positive as the as the income rises if you consecutively draw the green ball? Please help me clarify this doubt
 A: The expected value of our income $X$ is given by $$E(X)=\sum_i x_ip_i$$ over all outcomes $i$, where $x_i$ is the income and $p_i$ the probability of that outcome.
Let the $i^\text{th}$ outcome be drawing $i$ times. Then the first $i-1$ drawings will have been green, and the last drawing will be red, so we get income $x_i = i-2$.
The probability of the $i^\text{th}$ outcome is $P(\text{first } i-1 \text{ balls are green})*P(\text{last ball is red})$. Since we replace the ball, these events are independent, so $p_i = \left(\frac{1}{2}\right)^{i-1}*\frac{1}{2}=\left(\frac{1}{2}\right)^{i}$.
Now we can sum over all outcomes to find the expected value:
$$E(X) = \sum_{i=1}^{\infty}\left[(i-2)\left(\frac{1}{2}\right)^{i}\right]=\sum_{i=1}^{\infty}\frac{i-2}{2^{i}}$$
This converges by the ratio test, so we can do some manipulation:
$$2E(X) = \sum_{i=1}^{\infty}\frac{i-2}{2^{i-1}}=\sum_{i=0}^{\infty}\frac{i-1}{2^i}=-1+\sum_{i=1}^{\infty}\frac{i-1}{2^i}$$
And then subtract:
\begin{align}
2E(X)-E(X)&=\left(-1+\sum_{i=1}^{\infty}\frac{i-1}{2^i}\right)-\left(\sum_{i=1}^{\infty}\frac{i-2}{2^i}\right) \\
&=-1+\sum_{i=1}^{\infty}\frac{(i-1)-(i-2)}{2^i} \\
&=-1+\sum_{i=1}^{\infty}\frac{1}{2^i} \\
&=-1+1=0=E(X)\end{align}
A: Or you can simply do is express the expansion in the form of (1-x)^-2.
We are getting,
1/2*(-1)+1/21/20+(1/2)^21/21+(1/2)^31/22+.....
Which can be written as,
-1/2+(1/2)^3[1+21/2+3(1/2)^2+....]
The equation in the [] is the expansion of (1-x)^-2 where x=1/2
Thus,
-1/2+(1/2)^3[1-1/2]^-2
or, -1/2 + (1/2)^3*[1/2]^-2
or, -1/2 + 1/2
or, 0.
The method above is also correct but since these types of question is very common in some Indian competitive exams so the way I've done could be more helpful.
Thanks.
