Length giving one increasing run of six on average

Let $X_1,\dots,X_n$ be i.i.d. uniform on $(0,1)$. Call position $i$ the start of an increasing $6$-chain if $X_i<X_{i+1}<\dots<X_{i+5}$. For which $n$ is the expected number of positions starting such a chain equal to $1$?

$n=725$. A chain can start at positions $1,\dots,n-5$, and six i.i.d. continuous values are in increasing order with probability $\tfrac1{6!}=\tfrac1{720}$. By linearity the expected count is $\tfrac{n-5}{720}$, which equals 1 when $n=725$.