jmc

There are 16 cards in a standard deck which are either diamonds or aces. The probability that neither card chosen is a diamond or an ace is ${\left(\frac{36}{52}\right)}^2={\left(\frac{9}{13}\right)}^2=\frac{81}{169}$. Therefore, the probability that at least one of the cards chosen was a diamond or an ace is $1-\frac{81}{169}=\boxed{\frac{88}{169}}$.