Lottery Probabilities With Supplementary Numbers

[JamesV]

Hi All,

I'm trying to figure out the probability of winning the lotto. 8 numbers are drawn between 1 and 45. The first six are 'winning' numbers, the last two are the 'supplementary' numbers. To win division 1, you need to get all six winning numbers right:

\binom {45}6 = 8145060

Hence, the probability of winning division 1 is the inverse of that, or 1.23E-7.

To win division 3, you need to get 5 of the six winning numbers right. Now, from their website I know the odds of winning division 5 is 1/36689, which is the same as 222 / 8145060. I can come up with the 222 by:

\binom {6}5 \times \binom {45-8}1= 222


Now, that gives me the right answer, but I can't really work out why. If I apply the same formula to division 4, which requires 4 of the six winning numbers, I get the wrong answer:

\binom {6}4 \times \binom {45-8}2= 9990

But I can get the right answer by:

\binom {6}4 \times \binom {45-6}2= 11115

I know what the probabilities are (from the website), but cannot understand how they came up with them, or why they're inconsistent between division 3 and 4. Can anyone help?

Cheers,
James

 

[phinds]

I'm trying to figure out the probability of winning the lotto.

Small. Really, really, small.

 

[JamesV]

True, but hardly satisfies my mathematical curiosity.

 

[micromass]

The chance on winning the lottery is the same as saying that you need to find WannabeNewton in New York City. You go to New York City and you go to a random house and you knock on the door, and he answers.

 

[JamesV]

So no good ideas then?

 

[Office_Shredder]

The website mentions some supplementary numbers but for the life of me I can't find a basic description of what they are, so I assume that changes the odds somewhat.

If those weren't around, then the number of ways to pick 5 winning numbers should be (6 choose 5), pick 5 winning numbers to select, *( 45-6 choose 1), pick one non-winning number to select. The number of ways to pick 4 winning numbers would be (6 choose 4)*(45-6 choose 2)

If the supplementary numbers are just two extra numbers they draw, then to pick 5 winning numbers but no supplementary numbers you would have (6 choose 5) ways of picking your winning numbers, then (45-8 choose 1) ways of picking your non winning, non supplementary number. Since there's no distinction between 4 winning numbers and 4 winning numbers plus supplementary numbers in the rules, the number of ways to pick 4 winning numbers is still (6 choose 4)*(45-6 choose 2)

 

[JamesV]

I think they're just two extra numbers they pull out afterwards from the same pool of numbers.

Why, in the first case, is it (45-6 choose 1)? I understand the 45 is the number of options each ball could be, but what's the 6?

 

[Office_Shredder]

6 is the winning numbers, which you can't pick if you are picking a non-winning number.

 

[haruspex]

The reason for the apparent inconsistency is here:
2 5 winning numbers + 1 or 2 supplementary numbers
3 5 winning numbers
4 4 winning numbers
With 5 winning numbers and either supplementary you would score div 2, so the correct definition of div 3 is 5 winning and no supplementary. There isn't a division for 4 winning plus some supplementary.