Consider a collection of $N$ cards numbered $1,2,...,N,$ where $n\geq 3$. Two cards are drawn at random and set aside. Suppose $n$ cards are selected at random from the remaining $(N-2)$ cards using SRSWR and their numbers are notes as $y_1,y_2,...,y_n$. If $S=\sum_{i=1}^{n}y_i$ then $E(S)=\frac{N+1}{2}$.
I am almost clueless in this problem. I know the definition of expectation but not able to prove it using only definition. Kindly tell if I need to know something more (may be some theorems)
I am new to Statistics.
(Assuming that your $n=N$).
Suppose that the values of the two cards drawn aside are $X_1$ and $X_2$. Then when you draw a card at random from the remaining $N-2$ cards, its value has a uniform distribution over the set $\{1,2,\dots,N\}\setminus \{X_1,X_2\}$. So the expectation of $Y_i$ is given by $$ \mathbb{E}(Y_i\mid X_1,X_2)=\left[\frac1{N-2}\cdot 1+\frac1{N-2}\cdot 2+\dots+\frac1{N-2}\cdot N\right] - \left[\frac1{N-2}\cdot X_1+\frac1{N-2}\cdot X_2\right] =\frac1{N-2}\cdot \left[\frac{N(N+1)}2-X_1-X_2\right]. $$ Now, $$ \mathbb{E}(X_1)=\frac1{N}\cdot 1+\frac1{N}\cdot 2+\dots+\frac1{N}\cdot N=\frac{1+2+\dots+N}{N}=\frac{N+1}2 $$ $$ \mathbb{E}(X_2\mid X_1)=\left[\frac1{N-1}\cdot 1+\frac1{N-1}\cdot 2+\dots+\frac1{N-1}\cdot N\right] - \frac1{N-1}\cdot X_1 $$ so $$ \mathbb{E}(X_2)=\mathbb{E}\mathbb{E}(X_2\mid X_1)=\left[\frac1{N-1}\cdot 1+\frac1{N-1}\cdot 2+\dots+\frac1{N-1}\cdot N\right] - \frac1{N-1}\cdot\mathbb{E} X_1 =\frac{1+2+\dots+N-\frac{N+1}2}{N-1}=\frac{N+1}2 $$ which you can alternatively get by symmetry (it does not matter which of the two was drawn first).
Finally, Ssnce $Y_1,\dots,Y_n$ are i.i.d., $$ \mathbb{E}(Y_i)=\mathbb{E}\mathbb{E}(Y_i\mid X_1,X_2)=\frac1{N-2}\left[\frac{N(N+1)}2-(N+1)\right]=\frac{N+1}2 $$ (not surprisingly again, because of symmetry).