Answer
$P(win~both~games)=\frac{1}{3669120000000}\approx0.0000000000002725$
Work Step by Step
$P(win~the~first~game)=\frac{1}{5,200,000}$
$P(win~the~second~game)=\frac{1}{705,600}$
$P(win~both~games)=P(win~the~first~game~and~win~the~second~game)$
1) not a single event
2) AND
3) independent events
Use the Multiplication Rule for Independent Events (page 282):
$P(win~both~games)=P(win~the~first~game~and~win~the~second~game)=P(win~the~first~game)\times P(win~the~second~game)=\frac{1}{5,200,000}\times\frac{1}{705,600}=\frac{1}{3669120000000}\approx0.0000000000002725$
