[pzhon]

[ QUOTE ]
If I take batches of 9 random numbers between 1 and 100, what will the average of the highest number in each batch be? I'd love to see some calculations so I can do it for 8 etc, it would help me to stop asking silly questions.


[/ QUOTE ]
Let X100 be 1 if some number of the 9 is 100, 0 otherwise.
Let X99 be 1 if some number of the 9 is 99+, 0 otherwise.
Let X98 be 1 if some number of the 9 is 98+, 0 otherwise.
...


The maximum of the 9 is X1+X2+X3+...+X100.

By the linearity of expected value (regardless of dependence),

E(X1+X2+X3+...+X100)=E(X1)+E(X2)+E(X3)+...+E(X100) .

E(Xn)
= Probability at least one number is in the range n to 100
= 1- Probability all 9 numbers are from 1 to n-1
= 1-((n-1)/100)^9.

So, the expected value of the maximum is

1+(1-1/100^9)+(1-(2/100)^9)+...+(1-(99/100)^9)
~ 90.4925

For batches of 8 numbers, the average maximum is about 89.3822.

For batches of 100 numbers, the average maximum is about 99.4279 ~ 100-(1/e + 1/e^2 + 1/e^3 + ...) = 100-1/(e-1). For batches of n*100 numbers, the average maximum is about 100-1/(e^n-1).