  We consider a classic rendezvous game where two players try to meet each
other on a set of $n$ locations. In each round, every player visits one of the
locations and the game finishes when the players meet at the same location. The
goal is to devise strategies for both players that minimize the expected
waiting time till the rendezvous.
  In the asymmetric case, when the strategies of the players may differ, it is
known that the optimum expected waiting time of $\frac{n+1}{2}$ is achieved by
the wait-for-mommy pair of strategies, where one of the players stays at one
location for $n$ rounds, while the other player searches through all the $n$
locations in a random order. However, if we insist that the players are
symmetric --- they are expected to follow the same strategy --- then the best
known strategy, proposed by Anderson and Weber, achieves an asymptotic expected
waiting time of $0.829 n$.
  We show that the symmetry requirement indeed implies that the expected
waiting time needs to be asymptotically larger than in the asymmetric case.
Precisely, we prove that for every $n\geqslant 2$, if the players need to
employ the same strategy, then the expected waiting time is at least
$\frac{n+1}{2}+\varepsilon n$, where $\varepsilon=2^{-36}$.
