As Karl has remarked, the order in which the cards are distributed are irrelevant, so strange as it might seem, we might as well distribute $6$ cards to the "dog" first as the easiest way to compute

P(at least one oudler in the dog) $=1-\Large\dfrac{\binom{72}{6}}{\binom{78}{6}}\;\approx 0.3017$

As Karl has remarked, the order in which the cards are distributed are irrelevant, so strange as it might seem, we might as well distribute $6$ cards to the "dog" first as the easiest way to compute

P(at least one oudler in the dog) = $1-$ P(no oudler in the dog) $=1-\Large\dfrac{\binom{72}{6}}{\binom{78}{6}}\;\approx 0.3917$

As Karl has remarked, the order in which the cards are distributed are irrelevant, so strange as it might seem, we might as well distribute $6$ cards to the "dog" first as the easiest way to compute

P(no oudler in the dog) $=1-\Large\dfrac{\binom{72}{6}}{\binom{78}{6}}\;\approx 0.3017$

As Karl has remarked, the order in which the cards are distributed are irrelevant, so strange as it might seem, we might as well distribute $6$ cards to the "dog" first as the easiest way to compute

P(at least one oudler in the dog) $=1-\Large\dfrac{\binom{72}{6}}{\binom{78}{6}}\;\approx 0.3017$