Stopping on three consecutive face-down cards

Three face-down cards carry the values $n$, $n+1$ and $n+2$ in random order, where $n$ is unknown to you. You turn them over one at a time. After seeing the first card you may stop and be paid its value, or continue; after the second you may stop and take the second card's value, or else you must take the third card's value. Under an optimal strategy the expected payout is $n+c$. Find $c$.

$\tfrac43$. Shift so the cards are $0,1,2$. Stopping at the first card gives $1$ on average, so look at the second card: if it exceeds the first, keep it; otherwise take the third. Over the six equally likely orders this pays $1,2,2,2,1,0$ for a mean of $\tfrac86=\tfrac43$, and checking every rule based on the observed difference shows nothing does better.