- #1
JamesV
- 4
- 0
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:
[itex]\binom {45}6 = 8145060[/itex]
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:
[itex]\binom {6}5 \times \binom {45-8}1= 222[/itex]
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:
[itex]\binom {6}4 \times \binom {45-8}2= 9990[/itex]
But I can get the right answer by:
[itex]\binom {6}4 \times \binom {45-6}2= 11115[/itex]
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
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:
[itex]\binom {45}6 = 8145060[/itex]
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:
[itex]\binom {6}5 \times \binom {45-8}1= 222[/itex]
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:
[itex]\binom {6}4 \times \binom {45-8}2= 9990[/itex]
But I can get the right answer by:
[itex]\binom {6}4 \times \binom {45-6}2= 11115[/itex]
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