Coin bias making a tie as likely as one player winning

Brad and Chad each flip their own coin, both with heads probability $p$, until they see a first head. Let $B$ be the event that Brad needs strictly fewer flips than Chad, $C$ the event that Chad needs strictly fewer flips than Brad, and $E$ the event that they need the same number. For which $p$ is $\mathbb{P}[E]=\mathbb{P}[C]$?

$p=1/2$. By symmetry $\mathbb{P}[B]=\mathbb{P}[C]$, so all three events must have probability $\tfrac13$. A tie has probability $\sum_{k\ge1}\big(pq^{k-1}\big)^2=\tfrac{p^2}{1-q^2}=\tfrac{p}{2-p}$ with $q=1-p$, and $\tfrac{p}{2-p}=\tfrac13$ gives $p=\tfrac12$.