Question

A bag contains $\frac{n(n+1)}{2}$ counters, of which one is marked $1$, two are marked $4$, three are marked $9$, and so on; a person puts in his hand and draws out a counter at random, and is to receive as many shillings as the number marked upon it: find the value of his expectation.

Answer 1

Let $\left(\frac{n(n+1)}{2}\right)=N$

Expected value for counter $1$ $=\frac{1}{N}\times 1$

Expected value for counter $4$ $=\frac{2}{N}\times 4=\frac{2^3}{N}$

Expected value for counter marked $9$ $=\frac{3}{N}\times 9=\frac{3^3}{N}$

Similarly finding the expected value for each counter

Thus total expected value $=\frac{1^3}{N}+\frac{2^3}{N}+\frac{3^3}{N}................\frac{n^3}{N}$

$=\frac{1^3+2^3+3^3...........n^3}{N}=\frac{{\left(\frac{n(n+1)}{2}\right)}^2}{\left(\frac{n(n+1)}{2}\right)}=\left(\frac{n(n+1)}{2}\right)$