[bigpooch]

Basically, pzhon answered it all but he wasn't explicit
about 4):

<font color="white">
A very rough approximation for the EV of the game for
general n is sqrt(pi*n) - 8 (including the $7 fee for the
privilege of playing) and the expected number of correct
guesses above n/2 is very roughly 1/2[sqrt(pi*n)-1].

It is a mere check to see that the game is +EV for n&gt;=21
and -EV for n&lt;=20.
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