Answer
$P(win~the~Jubilee~game~twice)=\frac{1}{497871360000}\approx0.000000000002009$
Work Step by Step
$P(win~the~Jubilee~game)=\frac{1}{705,600}$
$P(win~the~Jubilee~game~twice)=P(win~the~first~Jubilee~game~and~win~the~second~Jubilee~game)$
1) not a single event
2) AND
3) independent events
Use the Multiplication Rule for Independent Events (page 282):
$P(win~the~Jubilee~game~twice)=P(win~the~first~Jubilee~game~and~win~the~second~Jubilee~game)=P(win~the~first~Jubilee~game)\times P(win~the~second~Jubilee~game)=\frac{1}{705,600}\times\frac{1}{705,600}=\frac{1}{497871360000}\approx0.000000000002009$
