91Ó°ÊÓ

If n = $2$ and the result is r b r b, then we would win a total of $2$ units

Assume that deck of $2$n cards consists of equal number of red and black cards, n of each. Further, assume that the cards are shuffled and then turned over one at a time, whereby each time a red card is turned over, we win $1$unit if more red cards than black cards have been turned over by that time. Let X represents the amount that we win.

$\text{Notice that each card had a probability}\frac{n}{2n}=\frac{1}{2}\text{of being red or black.}$

If we define indicator variables $I_j$as:

$I_j=\{\begin{array}{ll}1,& \text{if}{E}_j\text{occurs}\\ 0,& \text{if}{E}_j\text{does not occur}\end{array}$

Whereby E_jdenote the event:

$E_j="\text{if we win}1\text{unit when}j\text{th red card is turned over ",}$

We have that

$X=\underset{j=1}{\overset{n}{∑}}{I}_j$

Therefore, the expected amount that we win is

$E\left[X\right]=E\left[\underset{j=1}{\overset{n}{∑}}{I}_j\right]=\underset{j=1}{\overset{n}{∑}}E\left[I_j\right]=\underset{j=1}{\overset{n}{∑}}P\left\{E_j\right\}$

So, let's find the probability P{E_j}. Since we win $1$unit if j th red card appears before j th black card and by symmetry, we have that

$P\left\{E_j\right\}=\frac{1}{2}⇒E\left[X\right]=\frac{n}{2}$

The expected amount that we win is$\frac{n}{2}$