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handya | 02:53 Thu 07th Jul 2005 | Business & Finance
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Can anybody tell me the odds of winning �10 on the lottery

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The odds of selecting any three numbers depends on where within the six they are positioned:

for the first three numbers drawn it is.. 1 in 921.2
for the last three numbers it is........... 1 in 15180
for the 2nd, 4th and 5th it is................
1 in 3312

the first and second examples being the best and worst odds.

There are 18424 ways of selecting 3 numbers from 49. when you buy a ticket you get 6 numbers, there are 20 ways of selecting 3 numbers from 6 so you cover 20 out of the 18424, so the probability of getting �10 is 18424/20 = 921.2 to 1

Not quite sure what Kempie is getting at, although he has the 921.2 right so I'm sure he'll enlighten me!

I was trying to factor in the odds of selecting the 3 other 'wrong' numbers and have totally discombobulated myself.

Apparently to select 3 of 6 from 49 is actually 1 in 56.66

http://icarus.mcmaster.ca/fred/Lotto/

That's great link Kempie, I think the point it is making is that in order to win �10 you have to get 3 in the 6 that are pick AND 3 in the 43 that are not picked. The 921.2 would be true if they only picked 3 numbers rather than 6.

I think therefore Handya that the answer you are looking for is 55.7 to 1

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