The probability that both marbles are red is given by: $$ P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red after 1st red is drawn}).$$ The probability that the first marble is red is $ \frac{4}{10} $. After drawing a red marble, there are 3 red marbles and 9 marbles total left in the bag, so the probability that the second marble is also red is $ \frac{3}{9} $. Therefore $$ P(\text{both red}) = \frac{4}{10}\times \frac{3}{9} = \frac{2}{15}.$$ Similarly, the probability that both marbles are blue is given by: $$ P(\text{both blue}) = P(\text{1st blue}) \times P(\text{2nd blue after 1st blue drawn}).$$ The probability that that the first marble is blue is $ \frac{6}{10} $. After drawing a blue marble, there are 5 blue marbles and 9 marbles total left in the bag, so the probability that the second marble is also blue is $ \frac{5}{9} $. Therefore $$ P(\text{both blue}) = \frac{6}{10}\times \frac{5}{9} = \frac{1}{3}.$$ Since drawing two red marbles and drawing two blue marbles are exclusive events, we add the individual probabilities to get the probability of one or the other occurring. Therefore: \begin{align*}P(\text{both same color}) &= P(\text{both red}) + P(\text{both blue}) \\ &= \frac{2}{15} + \frac{1}{3} = \boxed{ \frac{7}{15}}.\end{align*}