Ten-door Monty Hall with three goats revealed

A game show has $10$ closed doors: one hides $1000 and the other nine hide goats (worth nothing). You pick a door at random. The host, who knows where the money is, then opens three of the doors you did not pick, each revealing a goat, and lets you either stay with your door or switch to one of the other unopened doors. Playing optimally, what is the expected value of the game in dollars?

$150. Staying wins with probability $\tfrac1{10}$. Switching wins only if your first pick was a goat ($\tfrac9{10}$) and you then pick the prize among the $6$ remaining closed doors ($\tfrac16$): $\tfrac{3}{20}$. So switch, for an expected $\tfrac3{20}\cdot1000=150$ dollars.