What you have calculated is the odds of getting the right numbers in one specific order, but order doesn’t matter for the lottery. Here’s an easy way to calculate the odds:

for each ball, the factor is (# total balls left/# of correct balls), so:

(49/6) * (48/5) * (47/4) * (46/3) * (45/2) * (44/1) = 13.98 million

There must be something specific about the way the numbers are drawn to say its 14/1. Otherwise u are right and it shoud be in the billions.

answer 1 has it upside down.6/49*5/48*4/47*3/46*2/45*1/44
=
1/13983816
or
13983815:1

Answer 1 wasn’t calculating the probability, but the size of the sample space. Since there is only 1 winning combination, the prob is 1/his answer.

But, still, since order doesn’t matter, the easiest way to calculate it is to count the number of ways you can choose 49 objects 6 at a time:

49 choose 6 =

49!
—————-
(49-6)!*6!

=

49*48*47*46*45*44
—————————
6*5*4*3*2*1

different ways that the balls could be drawn from the hopper.

So, the chances of winning are 1 in that many

Oh, and you could think of it like this:

The original questioner was almost right in his way of counting it. You just have to divide by 6! to make it order independent.

The difference is between permutations and combinations.

If you calculate a permutation, then the calculation is
49×48x47×46x45×44
But this means that the person who has the ticket containing numbers 1, 2, 3, 4, 5, and 6 has a different ticket than the person whose ticket contains the numbers 2, 1, 4, 3, 6, and 5. And if the drawing result is 6, 5, 4, 3, 2, and 1 then both of them would lose.

The point is that with a permutation, it isn’t just the numbers that are important, but the order in which the numbers appear. That is not how the lottery works.

Instead, the lottery works on seeking a combination of numbers. If a person had a ticket with the numbers 1, 2, 3, 4, 5, and 6 and the balls actually drawn in the lottery are 6, 5, 4, 3, 2, and 1 (in that order), then the guy has one–he has the correct combination of numbers.

The way this is calculated is as follows
n! / ((n – r)! x r!)
Where n is the number of balls from which you are drawing and r is the number of balls being drawn. In this example, n is 49 and r is 6 (since you are drawing 6 balls out of 49. So
49! / ((49 – 6)! x 6!)
= 49! / (43! x 6!)
= (49 x 48 x 47 x 46 x 45 x 44 x 43!) / (43! x 6!)
= (49 x 48 x 47 x 46 x 45 x 44) / (6 x 5 x 4 x 3 x 2 x 1)
= 13,983,816
And 1 of those is the correct combination. So the probability of winning is 1/13,983,816 and the odds against winning are 13,983,815 to 1.