WARNING: This is wrong! 3^2 = 9, so we can't encode it in 8 different symbols, but left here for history.

The 15 bit encoding could be visualized in a different way, too, that is using 3 bits to represent two successive cells state:

    000 "__"
    001 "X_"
    010 "_X"
    011 "XX"
    100 "XO"
    101 "OX"
    110 "O_"
    111 "_O"

There are 5 successive cells (4.5 actually, 9/2, but we need to be discrete here, and discard the last), so 5*3 = 15.

However this encoding shows that if you want to represent multiple boards one after the other you are actually just using 13.5 bits for each board, as it should be in theory.

Because to represent 18 total cells (two boards) you need 9*3 bits = 27/2 = 13.5 bits per board.