Hello, Stormey!
A few issues . . .
First, the number of possible outcomes is: _{37}C_{6} = 2,324,784.
. . The order of the numbers is not relevant.
Second, why would anyone buy two (or more) tickets with the same numbers?
Me and a friend of mine had into a little dispute about this issue, and I'd love if you guys could settle this.
The lottery in our country draws every weak 6 balls out of a pool of 37 balls (numbered 1,2,...37).
In each lottery ticket you need to guess 6 numbers.
For the sake of simplicity, let's assume that only a correct guess of 6 numbers will be considered as "winning the lottery" (even though in reality you'll also get a prize for guessing 5 or even 4 numbers).
The chance of winning the lottery, if you fill only one ticket, will then be 1 to 37x36x35x34x33x32. there's no argue here.
Now, he claims that any additional ticket I fill is going to improve my chances, so, if I will fill, say, n lottery tickets, my chances will then be n to 37x36x35x34x33x32 (adding 1/(37x36x35x34x33x32) n times).
I claim that this is not true, since this calculation ignores the fact that two (or more) tickets can have the same numbers in them.
For example, if I filled two tickets with exactly the same numbers, I didn't improve my chances whatsoever!
to this he replied "OK, so choosing different numbers in any ticket will give the above calculation..."
But then I thought about it a little and said that first of all, you can only produce 6 different lottery tickets (by the pigeonhole principle), so even if his observation is true, then for the chances will not be improved by as much as he said, and second, I'm not even sure that he's correct for n less then or equal to 6.
I know very little about probability and random variables, but is there a random variable (or a sequence of random variables) that describes the chance of winning the lottery with k lottery tickets? or is it to complex to determine?
Hello, Stormey!
A few issues . . .
First, the number of possible outcomes is: _{37}C_{6} = 2,324,784.
. . The order of the numbers is not relevant.
Second, why would anyone buy two (or more) tickets with the same numbers?
Hi Soroban!
I'm not saying anyone would do that, this was only given as an example that shows why the statement "every additional ticket will improve my chances, regardless of the numbers that was chosen", is incorrect.