Originally Posted by

**mattstan** Sorry for asking another lottery question, I did a search and there seem to have been a lot, but none answer my question.

I'm trying to work out the probability of getting 5 out of 6 numbers correct in the British National Lottery, actually I know the probability - I'm trying to work out how to work it out (so to speak).

Here's how the lottery works:

1) 6 unique numbers between 1 and 49 are selected by each player on their ticket.

2) There are 49 balls with the integers 1 through to 49 on them.

3) 6 'Winning' balls are chosen during the lottery draw and a further 'Bonus Ball' is drawn at the end, making 7 balls drawn in total.

There are lots of winning tickets but the large cash prizes are for:

a) Matching all 6 winning numbers.

b) Matching 5 of the 6 winning numbers and matching the 'Bonus Ball' number.

c) Matching 5 of the 6 winning numbers.

a) is pretty easy to work out:

Either:

(49 * 48 * 47 * 46 * 45 * 44) / 720 = 13,983,816

Divided by 720 because there are 720 ways that the 6 balls can be drawn and the order in which the numbers are drawn does not matter.

6 * 5 * 4 * 3 * 2 = 720

Or you can work it out like this:

1 / ( (6/49) * (5/48) * (4/47) * (3/46) * (2/45) * (1/44) ) = 13,983,816

Now is where I am going wrong, trying to calculate c) - matching 5 of the 6 winning numbers and with the number that is not matched also not matching the 'Bonus Ball' number.

1 / ( (6/49) * (5/48) * (4/47) * (3/46) * (2/45) * (42/44) ) = 332,948

I got the 42/44 because there are 44 remaining balls, one of them is the 6th winning number and one of them is the 'Bonus Ball' number, so only 42 of the 44 remaining numbers is a 'not matched number'.

However the advertised odds of matching 5 of the 6 winning numbers (and also not matching the bonus ball) is 55,492. That's quite a long way from the 332,948 that I calculated.

Can someone please explain to me where I am going wrong and how I should have calculated it?

Many thanks.

PS. I guess I should have written "13,983,816 to 1", etc., since for easy readability I chose not the express the probability as a value between 0 and 1.