1. ## Calculating lottery probabilities

The lottery i am talking about is a 6/49 game. You select 6 numbers from a pool of 49, and in the draw 6 numbers are taken. There is also an additional 'Bonus' number drawn from the remaining 43 balls which is used in some of the prize tiers.

I have managed to calculate the overall odds of winning in each tier as follows:

 Match Odds 6 1 in 13,983,816 5 + 1 BonusBall 1 in 2,330,636 5 1 in 55,491 4 + 1 BonusBall 1 in 22,197 4 1 in 1,083 3 + 1 BonusBall 1 in 812 3 1 in 61

Q1) How do i work out the probability for winning a prize in these divisions from these odds? Ive seen conflicting messages, but i assumed it was 1/odds, so the probability of matching 3 balls would be 1/61= 0.0164

In the lottery you can play via a system where you choose more than 6 numbers and enter all the combinations of those numbers. So if you did a system entry and chose 1,2,3,4,5,6,7 you would get lines of:
• 1+2+3+4+5+6
• 1+2+3+4+5+7
• 1+2+3+4+6+7
• 1+2+3+5+6+7
• 1+2+4+5+6+7
• 1+3+4+5+6+7
• 2+3+4+5+6+7

The lottery offers this all the way up to selecting all 49 numbers, which gives the number of different combinations as so:

 Numbers No. of combinations 6 1 7 7 8 28 9 84 10 210 11 462 12 924 13 1,716 14 3,003 15 5,005 16 8,008 17 12,376 18 18,564 19 27,132 20 38,760 21 54,264 22 74,613 23 100,947 24 134,596 25 177,100 26 230,230 27 296,010 28 376,740 29 475,020 30 593,775 31 736,281 32 906,192 33 1,107,568 34 1,344,904 35 1,623,160 36 1,947,792 37 2,324,784 38 2,760,681 39 3,262,623 40 3,838,380 41 4,496,388 42 5,245,786 43 6,096,454 44 7,059,052 45 8,145,060 46 9,366,819 47 10,737,573 48 12,271,512 49 13,983,816

Q2) How would i work out a) the odds of matching in each prize tier for each the combinations above? so the odds of matching three balls when i have chosen all the way from 6 - 49 balls for my system entry b) get the probability of matching in each prize tier from those odds?

I realise that at some point with the above combinations there will a point where its guaranteed that you will match the number and thus the probability is 1.0 but i dont know how to get to that point.

I had tried getting the probability of matching in each tier as 1/odds, then multiplying that value by the number of combinations, but that gave a probability way over 1.0 which i know isnt possible.

Any help in sending me the right way to calculate this is greatly appreciated!! Id like to learn the steps rather than just being given the answer

2. ## Re: Calculating lottery probabilities

Suppose you choose all combinations of $k$ numbers. Then, to get exactly $r$ balls (no bonus ball), you would have:
$$\dfrac{\dbinom{k}{r}\dbinom{49-k}{7-r}\dbinom{7-r}{1}}{\dbinom{49}{7}\dbinom{7}{1}}$$

To get exactly $r$ balls plus the bonus ball:
$$\dfrac{\dbinom{k}{r+1}\dbinom{r+1}{1}\dbinom{49-k}{6-r}}{\dbinom{49}{7}\dbinom{7}{1}}$$

In each of these formulas, the denominator is the total number of ways to choose 7 balls from 49, then choose one of them to be the bonus ball. The first numerator is the number of ways to choose the $r$ winning balls from the $k$ numbers you selected times the number of ways to choose the remaining balls from the balls you did not choose times the number of ways to choose a bonus ball among the balls that you did not choose.

The second numerator is the number of ways to choose the bonus ball from among your k numbers. Then, choose the $r$ winning balls from among your remaining numbers, and finally, choose the remaining winning numbers from the balls you did not choose.

3. ## Re: Calculating lottery probabilities

Originally Posted by SlipEternal
Suppose you choose all combinations of $k$ numbers. Then, to get exactly $r$ balls (no bonus ball), you would have:
$$\dfrac{\dbinom{k}{r}\dbinom{49-k}{7-r}\dbinom{7-r}{1}}{\dbinom{49}{7}\dbinom{7}{1}}$$

To get exactly $r$ balls plus the bonus ball:
$$\dfrac{\dbinom{k}{1}\dbinom{k-1}{r}\dbinom{49-k}{6-r}}{\dbinom{49}{7}\dbinom{7}{1}}$$

In each of these formulas, the denominator is the total number of ways to choose 7 balls from 49, then choose one of them to be the bonus ball. The first numerator is the number of ways to choose the $r$ winning balls from the $k$ numbers you selected times the number of ways to choose the remaining balls from the balls you did not choose times the number of ways to choose a bonus ball among the balls that you did not choose.

The second numerator is the number of ways to choose the bonus ball from among your k numbers. Then, choose the $r$ winning balls from among your remaining numbers, and finally, choose the remaining winning numbers from the balls you did not choose.
Thanks very much for that, doesnt look too complicated. Would you be able to do one as an example for me? Just so i know im putting the numbers in the right place

4. ## Re: Calculating lottery probabilities

Originally Posted by mikabrown
Thanks very much for that, doesnt look too complicated. Would you be able to do one as an example for me? Just so i know im putting the numbers in the right place
Sure. Let's say you want to buy tickets with all combinations of 9 numbers. Your odds of getting exactly 5 balls plus the bonus ball would be:

$$\dfrac{\dbinom{9}{5+1}\dbinom{5+1}{1}\dbinom{49-9}{6-5}}{\dbinom{49}{7}\dbinom{7}{1}} = \dfrac{120}{3,579,191} \approx \dfrac{1}{29,827}$$

I fixed it.