Hi All,(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Lottery Probabilities With Supplementary Numbers

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