Odds of Lottery with Supp Numbers - Oz Lotto I'm calculating the probability of each division in Oz Lotto, where the
divisions
are as follows.
_
Division 1: 7 winning ..................................... (45,379,620:1)
Division 2: 6 winning, 1 or 2 supps ............... (3,241,401:1)
Division 3: 6 winning .................................... (180,078:1)
Division 4: 5 winning, 1 or 2 supps ...... (29,602:1)
Division 5: 5 winning .................................... (3,430:1)
Division 6: 4 winning .................................... (154:1)
Division 7: 3 winning, 1 or 2 supps ...... (87:1)
Oz Lotto is played with 45 numbers, 7 drawn winning numbers, and 2 drawn supplementary numbers.
_
The website lists the odds of each, as shown above, so the goal is to find the calculations that leads to those odds. Division 1, 3, 5, and 6 have been solved already using $( 7Cx * (45-7)C(7-x) ) / 45C7$ where x is winning balls in the division (Eg. in division 3, x = 6).
By trial and error, the calculation for Division 2 was found using $(7C6 * 2C1) / 45C7$, which equals 14/45379620, where the odds are 3,241,401:1.
I've tried to apply the same or similar formulas to division 4 and division 7 to reach the intended odds, but I've never found the correct calculation.
What are the correct calculations for Division 4 and 7 to achieve the same odds as listed above?
 A: Those numbers look more like reciprocals of probabilities rather than odds to me  
In general, the number of ways of getting $w$ winning numbers and $s$ supplementary numbers and $7-w-s$ losing numbers is ${7 \choose w}{2 \choose s}{36 \choose 7-w-s}$.  So 


*

*For division 2, if you have $6$ winning numbers, you cannot have more than $1$ supplementary number  so the calculation is $\dfrac{{7 \choose 6}{2 \choose 1}{36 \choose 0}}{45 \choose 7} = \dfrac{14}{45379620} \approx \dfrac{1}{3241401.4}$

*For division 4, if you have $5$ winning numbers, you can have $1$ or $2$ supplementary numbers so the calculation is $\dfrac{{7 \choose 5}{2 \choose 1}{36 \choose 1}+{7 \choose 5}{2 \choose 2}{36 \choose 0}}{45 \choose 7} = \dfrac{1533}{45379620} \approx \dfrac{1}{29601.84}$

*For division 7, if you have $3$ winning numbers, you can have $1$ or $2$ supplementary numbers so the calculation is $\dfrac{{7 \choose 3}{2 \choose 1}{36 \choose 3}+{7 \choose 3}{2 \choose 2}{36 \choose 2}}{45 \choose 7} = \dfrac{521850}{45379620} \approx \dfrac{1}{86.96}$
