Choosing die sizes when the sum must stay under 8

You and another player each independently choose a number of sides $k\in\{1,\dots,6\}$ for your own die, whose faces are then $1,\dots,k$, each equally likely. Both dice are rolled. If the two results add up to $7$ or less, each player is paid by a third party the value shown on their own die; if the total is $8$ or more, nobody is paid.

Both players choose optimally (an equilibrium). Let $s$ be the number of sides you choose and $p$ your expected payout. Find $s\cdot p$.

$\tfrac{49}{5}$. Compute the payoff table: against a $5$-sided die, your payouts for $k=4,5,6$ are $\tfrac{39}{20}$, $\tfrac{49}{25}$, $\tfrac{11}{6}$, so $5$ is a best reply to $5$ and $(5,5)$ is the symmetric equilibrium (smaller dice get undercut by a bigger one). Then $s=5$, $p=\tfrac{49}{25}$.