Lotto Question

Hi All

I'm completely new to this sort of thing so please bear with me. I have a question regarding lotto draws.

We have shops here in South Africa that offer a different way to bet on lotto draws. Instead of the traditional pick 6 numbers to win Jackpot etc.
you can pick up to 4 numbers for a fixed return.

The draw itself involves 6 numbers and a bonus ball (7 total). the six numbers are 1-49 and the bonus is 1-49.

If I pick one number, that number must appear in any of the 7 numbers in the draw for me to win
If I pick two numbers, both numbers must appear in the 7 drawn for me to win.( if only 1 of the 2 is correct lose)
If I pick 3 numbers, all three must appear to win etc.

The odds offered are 5.5 for 1 number, 55 for 2 numbers, 450 for 3 numbers and 2275 for 4 numbers.

What is the probability of correctly picking 1,2,3 and 4 numbers with this format ?

Thanks.

There are two ways to win - either all of your numbers are in the first six, or all but one are in the first six and the other one is in the seventh.

There are (49 x 48 x 47 x 46 x 45 x 44) / 720 = 13,983,816 sets of 6 numbers that can be drawn from 49.

If you pick one number:
There are (48 x 47 x 46 x 45 x 44) / 120 = 1712304 sets of 6 numbers that include your chosen number
There are (48 x 47 x 46 x 45 x 44 x 43) / 720 = 12271512 sets that do not
The probability of winning = (1712304 / 13983816) + (12271512 / 13983816) x (1/49) = 1 / 7.1246

If you pick two numbers:
There are (47 x 46 x 45 x 44) / 24 = 178365 sets of 6 numbers that include both of your chosen numbers
There are 2 x (47 x 46 x 45 x 44 x 43) / 120 = 3067878 sets of 6 numbers that include one but not both of them
The probability of winning = (178365 / 13983816) + (3067878 / 13983816) x (1/49) = 1 / 58.0302.

If you pick three numbers:
There are (46 x 45 x 44) / 6 = 15180 sets of 6 numbers that include all three of your chosen numbers
There are 3 x (46 x 45 x 44 x 43) / 24 = 489555 sets of 6 numbers that include two but not all three of them
The probability of winning = (15180 / 13983816) + (489555 / 13983816) x (1/49) = 1 / 555.5545

If you pick four numbers:
There are (45 x 44) / 2 = 990 sets of 6 numbers that include all four of your chosen numbers
There are 4 x (45 x 44 x 43) / 6 = 56760 sets of 6 numbers that include three but not all four of them
The probability of winning = (990 / 13983816) + (56760 / 13983816) x (1/49) = 1 / 6509.0433

Hmmm...something doesn't look right. It doesn't make much sense that the 2-number bet has an expected return of 94.78% but the 4-number bet has an expected return of only 34.95%.

Thanks for the reply. Not sure either.

Does the 94.78% mean the house make a very small expected return, but the four numbers a larger expected return ?

Looking at the probabilities you calculated, if I look at it correctly the odds don't leave much room for the house to profit, except four numbers, unless it has something to do with the volume of tickets they can sell.

Are the odds quiet favourable for the punter ?

Sorry I'm a rookie at this stuff, just trying to make sense of it

A return of 94.78% means the house "should" keep 5.22% of whatever you bet.

A 5.22% house edge seems very low for a lottery game - are you sure it's 55-1 for two numbers?

(Also, are you sure the seventh number is 1-49? South Africa's Lottery site says the Powerball is 1-43.)

Quote: ThatDonGuy

A return of 94.78% means the house "should" keep 5.22% of whatever you bet.

A 5.22% house edge seems very low for a lottery game - are you sure it's 55-1 for two numbers?

(Also, are you sure the seventh number is 1-49? South Africa's Lottery site says the Powerball is 1-43.)

They have another Lottery game which is 49 numbers and you can pay extra for the last draw.

Are you playing the official game or is this a game offered by someone else which uses the draw from the official lottery?

I did a little more digging, and it appears that the "bonus number" is drawn from the same 49 balls as the first six. This makes the problem a little easier.

If you select N numbers, then treat it as if the lottery has N red balls and (49 - N) white balls, and you win if all N of the red balls are in the 7 balls drawn.

There are (49 x 48 x 47 x 46 x 45 x 44 x 43) / 5040 = 85,900,584 combinations of drawing 7 balls out of 49.

If you chose 1 number, you need to draw the 1 red ball and one of the (48 x 47 x 46 x 45 x 44 x 43) / 720 = 12,271,512 combinations of 6 white balls out of 48.
The probability of winning = 12,271,512 / 85,900,584 = 1 / 7.
For a payout of 5.5, the house edge is 21.4286%

If you chose 2 numbers, you need to draw the 2 red balls and one of the (47 x 46 x 45 x 44 x 43) / 120 = 1,533,939 combinations of 5 white balls out of 47.
The probability of winning = 1,533,939 / 85,900,584 = 1 / 56.
If this really does pay 55, then the "house edge" is only 1.7857%, which seems very low for a lottery game

If you chose 3 numbers, you need to draw the 3 red balls and one of the (46 x 45 x 44 x 43) / 24 = 163,185 combinations of 4 white balls out of 46.
The probability of winning = 163,185 / 85,900,584 = 1 / 526.4.
For a payout of 450, the house edge is 14.5137%

If you chose 4 numbers, you need to draw the 4 red balls and one of the (45 x 44 x 43) / 6 = 14,190 combinations of 4 white balls out of 46.
The probability of winning = 14,190 / 85,900,584 = 1 / 6053.6.
For a payout of 2275, the house edge is 62.4191%

Wow that is really low, yes I am sure it is 55/1.

@ Crystal Math, it is another company which uses the draws of different lotto games around the world. This particular one is UK 49's.

@ThatDonGuy sorry for the incorrect info previously, I thought maybe the bonus was separate.