Question

In a state lottery, a player selects ten different numbers between 1 and 60 inclusive. If those ten numbers are drawn in the lottery, the player wins millions of dollars. Nguyen is considering buying a ticket, and his friend Brad suggests, "Why don't you pick 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10?" Nguyen replies, "Are you kidding? Those will never come up!" Brad comes back, "They have just as good a chance as anything else!" Is Brad's claim true or false?

Answer 1

please help

Answer 2

i am stuck:0

Answer 3

Well, it seems to me that he might be right. There is as good a chance a number like 20 will be picked as there is for a number like 3.

Answer 4

I think it is true his response

Answer 5

thank you

Answer 6

But, picking them in that order would be near impossible. But there is a chance those numbers could be picked at random, just like there is a chance 11,12,13,14,15,16,17,18,19 and 20 could be picked.

Answer 7

and because it is in between 1-60

Answer 8

thank you sooooooo much I'm so bad at probability :D

Answer 9

It's okay. Tag me if you need more help, as I've done a little probability and like it.

Answer 10

Don't forget to medal, and then close this question so it's off the list, as now you no longer need help :)

Answer 11

Think about the problem like this. Think about ALL probability problems like this. Always, if you can, make them simpler so you can get some intuition.

Rather than having numbers 1-60, let's just have numbers 1-10.

Now if your friend says pick the number "1" out of the 10. Is this any less likely than say a "5" ?

Hopefully you'll see that every possibility is equally likely and one choice isn't more likely than another.

Now choose two numbers, say 1 and 2. Is this more likely than say 3 and 7 ?

The system is the same so we don't expect one pair of numbers to be more likely than another because of the symmetry of the problem.

When I say "symmetry" I mean, think of all combinations, for example how many pairs can we make with only 10 numbers?

12
21
14
38
39
....
and so on

There are
10*9=90 permutations, but half of these are duplicates, like 12 and 21, so really there are 90/2 = 45 combinations

There is a formula for this, if you are not aware:
$$
{{10}\choose{2}}=\cfrac{10!}{2!(10-2)!}=\cfrac{10!}{2!8!}=45
$$
http://www.wolframalpha.com/input/?i=10+choose+2

So the chance of say 14 being the winning lottery ticket is 1/45, the same as if you were to choose 12. So by "symmetry" I mean that each choice is, from a probability perspective, the same.

For your problem, the number of combinations for this lottery is

$$
{{60}\choose{10}}=75,394,027,566
$$
and the chance of winning is 1/75,394,027,566 or about \(1.3× 10^{-11}\) or 0.000000000013.

So combinations like 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 and 11,12,13,14,15,16,17,18,19 and 20 are each equally likely, by symmetry.

Hope this helps.