A draws a card from a pack of n cards marked 1,2, ..., n. The card is replaced in the pack and B draws a card. Then the probability that A draws a higher card than B is

If A draws card higher than B, then the number of favorable cases is (n - 1) + (n - 2) +... + 3 + 2 + 1 (as when B draws card number 1, then A can draw any card from 2 to n and so on). Therefore, the required probability is

$\frac{\frac{\mathrm{n}(\mathrm{n}-1)}{2}}{{\mathrm{n}}^2}=\frac{\mathrm{n}-1}{2\mathrm{n}}$