# Statistical probabilities to win the lottery

I'm trying to figure out what is the statistical probabilities to win the lottery.

Let's assume that we pick 6 numbers out of 40 so: $$\frac1{40×39×38×37×36×35}=\frac1{2763633600}$$ Up until this point it makes sense to me but here is the thing that I don't understand :

"So, at this point, your odds of winning are 1 in 2763633600. But, since you can choose your winning numbers in any order, your chances of winning are somewhat better than this. Your chance betters by the number of different ways that a sequence of 6 numbers can be written down, which for 6 numbers is 6! (6 factorial) or 720. Divide 2763633600 by 720 to account for this, to get 3838380.

In other words, there are 720 different ways that the 6 numbers you choose can be filled out on your lottery ticket--if you choose your 6 numbers correctly, any of these ways will make a winning ticket."

I don't get it because I could choose the winning numbers in any order on the first equation why do i need to divide by 720 ?

The problem is to find the total number of outcomes. We pick six numbers from the forty to get $40\cdot39\cdot38\cdot37\cdot36\cdot35$. However, this counts, for example, $1,2,3,4,5,6$ and $6,5,4,3,2,1$ as two different tickets, when in fact they are the same.
For any given ticket, there are $6!=720$ ways to arrange those numbers, so each ticket which should have been counted once was counted $720$ times. Thus we divide our original count by $720$ to get the number of distinct tickets.
This is the same as $40$ choose $6$, or $40\choose 6$ by the way.
Your first count was count of all ordered $6$-tuples. And in lottery it does not matter in which order do numbers come out; 1 13 is same as 13 1. Beacuse order does not matter, you look at combinations. So the answer to your question is how many different $6$-combinations can you pull from 40? And the answer is $\displaystyle\binom{40}{6}$. So the probability of winning a lottery with one combination is $\displaystyle\frac{1}{\displaystyle\binom{40}{6}}$.