Combinatorics: calculating Oz Lotto odds for divisions

AI Thread Summary
In Oz Lotto, players choose seven numbers from a pool of 45, with nine numbers drawn, including seven winning and two supplementary. The odds for winning in Division 4, which requires five winning numbers and one supplementary, are calculated as 1 in 29,602, while Division 7, needing three winning numbers and one supplementary, has odds of 1 in 87. A user expressed confusion over their calculations for Divisions 4 and 7, providing their formulas but questioning their accuracy. An alternative calculation method for Division 4 was also shared, emphasizing the complexity of combinatorial calculations. Accurate understanding of these odds is crucial for players aiming to win in Oz Lotto.
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Homework Statement
In Oz Lotto, balls are numbered 1 to 45. Nine are selected, seven of which are winning numbers and two being supplementary numbers. Players select seven numbers.

The odds of winning can be found here: https://www.lottoland.com.au/magazine/oz-lotto-everything-there-is-to-know.html

I tried calculating the odds, and get all of them right except for div 4 and 7. Could somebody please explain what I've done wrong?
Relevant Equations
[SUP]7[/SUP]C[SUB]5[/SUB] x [SUP]2[/SUP]C[SUB]1[/SUB] x [SUP]36[/SUP]C[SUB]1[/SUB]/ [SUP]45[/SUP]C[SUB]7[/SUB] =

1/30,012
In Oz Lotto, balls are numbered 1 to 45. Nine are selected, seven of which are winning numbers and two being supplementary numbers. Players select seven numbers.

The odds of winning can be found here: https://www.lottoland.com.au/magazine/oz-lotto-everything-there-is-to-know.html

I tried calculating the odds, and get all of them right except for Div 4 and 7. Could somebody please explain what I've done wrong?

Division 45 Winning Numbers + 1 Supplementary Number1 : 29,602

7C5 x 2C1 x 36C1/ 45C7 =

1/30,012
Division 73 Winning Numbers + 1 Supplementary1 : 87

7C3 x 2C1 x 36C3/ 45C7 = 1/91

Thanks
 
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I think I agree with your answers!
 
Last edited:
PS alternative calculation for division 4:
$$p = 7 \cdot 6 \cdot P(WWWWWSX) = 42 \cdot \frac{7}{45} \cdot \frac{6}{44} \dots \frac{2}{40} \cdot \frac{36}{39}$$
 
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