Expectation of the sum of two cards without replacement = Expectation of the sum of two cards with replacement

Question

We have $10$ cards numbered from $1$ to $10$. We pick two cards among them. What is the expected value of the sum of these two cards ?

I have solved this question the hard way using the law of total expectation (conditioning on the first draw) and I have found that the answer is $11$ (which I checked is right).

But $11=5.5\cdot 2$ so it is equal to the expectation of the sum if we picked the the two cards with replacement (because then the mean of each card would be $5.5$)

So I wonder if there is a hidden argument to get this answer faster ? Thanks !

Answer 1

If $X_1,X_2$ are the numbers on the two cards drawn then:$$\mathbb E(X_1+X_2)=\mathbb EX_1+\mathbb EX_2=\frac{11}2+\frac{11}2=11$$

Application of linearity of expectation.

Answer 2

The expectation of the sum of two random variables is the sum of their expectations.

$$\mathsf E(X_1+X_2)=\mathsf E(X_1)+\mathsf E(X_2)$$

This is called the Linearity of Expectation.   It works whether the random variables are independent or not.

It is a very useful thing to know.

So whether drawing with or without replacement the expected value is the same.   $\tfrac {11}2+\tfrac {11}2=11$.

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From first principles:

$$\begin{align}\mathsf E(X_1+X_2) =&~\begin{cases}\sum_{i=1}^{10}\Big(\tfrac i{10}+\tfrac 1{100}\sum_{j=1}^{10}j\Big) & : \text{with replacement}\\\sum_{i=1}^{10}\Big(\tfrac i{10}+\tfrac 1{90}\mathop{\cdot\sum_{j=1}^{10}}\limits_{j\neq i} j\Big) & : \text{without replacement}\end{cases} \\[1ex]=&~\begin{cases}\tfrac 1{10}\sum_{i=1}^{10}\Big(i+\tfrac 1{10}\frac{110}{2}\Big) & : \text{with replacement}\\\tfrac 1{10}\sum_{i=1}^{10}\Big(i+\tfrac 19(\frac{110}{2}-i)\Big) & : \text{without replacement}\end{cases} \\[1ex]=&~\begin{cases}\tfrac 1{10}\Big(\frac{110}{2}+\frac{110}{2}\Big) & : \text{with replacement}\\\tfrac 1{10}\Big(\tfrac{8}{9}\frac{110}{2}+\tfrac {10}9(\frac{110}{2})\Big) & : \text{without replacement}\end{cases} \\[1ex]=&~\begin{cases}11 & : \text{with replacement}\\11 & : \text{without replacement}\end{cases} \end{align}$$