Chances of winning in 6 out of 49 numbers lottery?

What are the chances of winning in a lottery if it's a draw of 6 numbers out of a total of 49 numbers and i bet on 23 numbers.





From my calculae it is a chance of 1 in 138.5 chance of getting all 6 numbers in my 23.





Please tell me if this is corrrectChances of winning in 6 out of 49 numbers lottery?
What you need is a permutation: you are going to choose 6 numbers out of a possible 49, without duplication of any number set.


So:nPk=(49!)/(49-6)!=10,068,347,520 permutations





This is all the possibles, and to account for unique combinations, use (49!)/[ (49 - 6)! x (6)! ]= 13,983,816





So this makes it 23/[10,068,347,520-13,983,816]


or a total chance of actually picking a winner of 1 : 2.284x10^9 or about one in 2 billionChances of winning in 6 out of 49 numbers lottery?
OK, any one combination has a likelihood of 360 chances in 49! (that's 49 factorial) ways of getting it right. That's a number considerably higher than 1 in 138.





Each of your 23 choices (assuming that you do not duplicate any ticket) has an equal chance.





Now for the part you missed - you add (not multiply) these 23 chances together. So, you now have 23 chances out of a very, very large number. I'd keep the money and spend it on fast cars.