Probability game with three players winning plus losing doesn't equal 1

The probability of losing is not correct. In such case there are two possible cases about the remaining players:

1) First player draws a lower card (this is the part you forgot in the question) and the second one a higher card;

2) First player draws a higher card and the second one a lower card.

Thereby,

$\Pr[lose] = \Pr[2^{nd} \; player\; wins] + \Pr[1^{st}\; player\; wins] = \frac{8}{9} \times \frac{1}{8} + \frac{1}{9} \times \frac{8}{8} = \frac{16}{72}$

You add all up and it sums to $1$.