A run of exactly four dogs at a round table

Twelve animals, six dogs and six cats, are seated in a uniformly random arrangement around a circular table. What is the probability that somewhere around the table there is a block of exactly four consecutive dogs (flanked by cats on both sides)?

$\tfrac{15}{77}$. Two such blocks would need 8 dogs, so count arrangements containing one pattern cat, 4 dogs, cat. Choose the flanking cats ($6\cdot5$) and the ordered dogs ($6\cdot5\cdot4\cdot3$), then arrange this unit with the other 6 animals around the circle ($6!$ ways). Dividing by $11!$ gives $\tfrac{15}{77}$.