Assume the contrary. Each lucky guest has at least two friends. Note that a lucky guest having exactly 2 friends can be made unlucky in two different ways: the hat colours of her friends should be changed from $(a, b)$ to either $(a, a)$ or $(b, b)$. A lucky guest having at least 3 friends can be made unlucky in just one way: the hat colour of her uniquely determined friend should be changed to uniquely determined colour. There are $2 \cdot 29$ possibilities to change hat colour of one guest. If each of these changes produces some unlucky guest then each person should have exactly two friends. Consider a graph $G$ where vertices correspond guests and two vertices are adjacent if and only if guest corresponding to these vertices are friends. Let us colour vertices according to hat colours of corresponding guests. Then $G$ will be decomposed into cycles of the form $a, b, c, ..., a, b, c$. Since $3 \nmid 29$ we get a contradiction. Done.