Hi iq987,

You have had one correct solution so far; here is another way to solve the problem. (You can never have too many solution methods.)

Let $\displaystyle X_i = 1$ if card $\displaystyle i$ is the largest seen so far, i.e. $\displaystyle X_i > X_j \text{ for all } j < i$, and let $\displaystyle X_i = 0$ otherwise.

What is the probability that $\displaystyle X_i = 1$? There have been $\displaystyle i$ cards dealt so far, and each is equally likely to be the largest of $\displaystyle X_1, X_2, \dots ,X_i$; so

$\displaystyle Pr(X_i =1) = 1/i$

and

$\displaystyle E(X_i) = 1/i$

Hence

$\displaystyle E(X_1 + X_2 + \dots + X_N) = E(X_1) + E(X_2) + \dots + E(X_N) = 1 + 1/2 + \dots + 1/N$.

Here we have used the theorem that $\displaystyle E(X+Y) = E(X) + E(Y)$. This is true even if X and Y are not independent, which is essential to the solution because the $\displaystyle X_i$'s are not independent.