# Lottery second prize probability

In some lottery, there are $45$ numbers to enter, i.e. from $1$ to $45$, and $7$ winning numbers and $2$ supplementary numbers to draw. To win the first prize, one must have as same $7$-number in his/her ticket as the $7$-winning-number has been drawn. Order of the $7$ numbers doesn't matter. So the inverse of probability of winning the first prize is

$x_1=\dfrac{45\times 44\times 43\times 42\times 41\times 40\times 39}{7\times 6\times 5\times 4\times 3\times 2\times 1}=45379620$. Easy!

To win the second prize, one must have as same $6$ winning numbers in his/her ticket of the drawn-$7$-winning-number, plus one of the drawn-$2$-supplementary-number among his/her $7$-number-ticket. The inverse of probability of winning this prize is $x_2=3241401$. Not easy! I can't solve how this number, $x_2$, comes out. The results without step-by-step solution is here.

The result you have for $x_1$ is a straightforward 45-choose-7 combinations.
That still applies to the ticket selection for $x_2$, but for second prize there is more variety in what they can match. There are now more options to match against - in fact ${7\choose 6}{2\choose 1}=7\times 2 = 14$ options. Take the ratio of the ticket choice against the successful options and you're home.
To win second prize, you get to choose which of the 7 numbers you did not hit, then which of the 2 supplementary numbers you did hit. The odds are then $$7 \times 2 \times \frac{7\times 6 \times 5 \times 4 \times 3 \times 2 \times 1} {45\times 44 \times 43 \times 42 \times 41 \times 40 \times 39} = \frac{14}{45379620}=\frac{1}{3241401}$$