Stay or switch urns after drawing a one-dollar chip

Two identical-looking urns each hold ten chips. Urn A has four $\text{\char36}1$ chips and six $\text{\char36}10$ chips; urn B has three $\text{\char36}1$ chips and seven $\text{\char36}10$ chips. You choose an urn at random, draw one chip, and it is a $\text{\char36}1$ chip (it is not replaced). You may now draw a second chip either from the same urn or from the other urn, and you are paid the value of this second chip. With the best choice, what is your expected payout?

$52/7$ dollars (stay). Seeing a $\text{\char36}1$ chip makes urn A more likely, with posterior $4/7$ versus $3/7$. Staying pays $7$ from A and $8$ from B on average, giving $\tfrac47\cdot7+\tfrac37\cdot8=\tfrac{52}{7}\approx 7.43$, while switching gives only $\tfrac47\cdot7.3+\tfrac37\cdot6.4\approx 6.91$.