Expected Value of 3 numbers I don’t understand expected value, can somebody tell me and help me do this problem?
Generalization: what is the expected value of the biggest number if we choose $k$ numbers from $1,2,...n$.

$$
\begin{aligned}
E&amp;=\frac{1}{\binom{n}{k}}\sum_{i=k}^{n}{i\binom{i-1}{k-1}}\\
&amp;=\frac{1}{\binom{n}{k}}\sum_{i=k}^{n}{k\binom{i}{k}}\\
&amp;=\frac{k}{\binom{n}{k}}\sum_{i=k}^{n}{\binom{i}{k}}\\
&amp;=\frac{k}{\binom{n}{k}}\binom{n+1}{k+1}\\
&amp;=\frac{k}{\binom{n}{k}}\frac{n+1}{k+1}\binom{n}{k}\\
&amp;=\frac{k}{k+1}(n+1)
\end{aligned}
$$

in Your case, substitute $n=10$ and $k=3$
The expected value of a variable is the average of possible values weighted by their probabilities. It will not necessarily be an whole number even if only whole numbers are possible values.

The largest of the three numbers picked cannot be 1 or 2 because there are insufficient smaller numbers available.

If the largest number is 3 then there is only 1 possible combination for the other numbers (1 2).

If the largest number is 4 then there are 1+2=3 possible combinations for the other numbers (1 2) &amp; (1 3)(2 3).

If the largest number is 5 then there are 1+2+3=6 possible combinations for the other numbers (1 2) &amp; (1 3)(2 3) &amp; (1 4)(2 4)(3 4).

Following the same pattern, if the largest number is 6 then there are 1+2+3+4=10 possible combinations for the other numbers. This pattern is the triangular numbers. The next are 15, 21, 28, 36.

The expected value of the largest number picked is therefore 
$$E(X)=\frac{3*1+4*3+5*6+6*10+7*15+8*21+9*28+10*36}{1+3+6+10+15+21+28+36}=\frac{990}{120}=8.25$$