How to Calculate Probability for Lottery Numbers?

[member 529879]

1. The buyer of a lottery ticket chooses four numbers from the numbers 1 to 32. Repetition is not allowed.
   
   How can this be solved? I thought it would be solved by multiplying the probability of each number being picked. 1/32 x 1/31 x 1/30 x 1/29. Though this seemed to give me the wrong answer.

 

[phinds]

1. The buyer of a lottery ticket chooses four numbers from the numbers 1 to 32. Repetition is not allowed.
   
   How can this be solved? I thought it would be solved by multiplying the probability of each number being picked. 1/32 x 1/31 x 1/30 x 1/29. Though this seemed to give me the wrong answer.
   

I think you're looking at it the way you look at a problem involving drawing marbles out of a box, in which case "no repetition" is expressed as "marbles don't get put back". That's not the way to look at this problem. Try again.

 

[member 529879]

I found the formula to use in this situation and found that I had to divide by 4! To get 35,960, but why is this not the same as drawing marbles out of a box?

 

[phinds]

... why is this not the same as drawing marbles out of a box?

Depending on formulas is a bad idea. You need to depend on your understanding of the various kinds of problems, else you end up just applying formulas blindly, which doesn't always work.

Why would the chance of the 3rd number, just as an example, being 14, have anything to do with the first number being 3 (for example)? That is, what are the odds of the 3rd number being 14? What are the odds of the 1st number being 3?

EDIT: Hm ... I think *I* may be the one not thinking this through. I was thinking the odds of anyone number being some specific value is 1/32 but that would allow for duplicates.

Nuts. I think your original thought (1/32 x 1/31 x 1/30 x 1/29) should be right.

EDIT: EDIT: Well, nuts, nuts. I'm back to thinking I was right in the first place and your first thought was wrong. What I REALLY think, now, is that I should go to bed.

 

[Trevor Vadas]

Your missing that if the winnings numbers were 1,2,3,4. You could win with 1,2,3,4 or 2,3,1,4 or ... The way you originally calculated it was that you needed to get the number in order.

 

[insightful]

Just to finish this off, the correct probability is: 4/32 x 3/31 x 2/30 x 1/29 = 1/35960.

 

[WWGD]

You find all choices , which is $32C4$ and just one of these is the winning combination. If order does not matter , then use $32P4$

 

[statdad]

"4/32 x 3/31 x 2/30 x 1/29 = 1/35960."

True, but it doesn't give an indication why.

Nothing in your statement says order is important, so combinations can be used to count. The number of ways to select 4 numbers is
$$C^{32}_4 = \dfrac{32!}{4! \cdot 28!} = \dfrac{32 \cdot 31 \cdot 30 \cdot 29}{4 \cdot 3 \cdot 2 \cdot 1}$$

Since the only way to win is to have exactly the four numbers selected by the authorities the probability is
$$\dfrac 1 {C^{32}_4} = \dfrac{4 \cdot 3 \cdot 2 \cdot 1}{32 \cdot 31 \cdot 30 \cdot 29}$$

which is the expression you give.

"If order does not matter , then use 32P4"
You mean to say ``If order does matter, ...'' here.

 

[WWGD]

I meant if different orderings of the same 4-ple are considered equal. That is my parsing of "if order does not matter".

 

[statdad]

My point was that you said "if order does not matter use 32P4" (use permutations when orders does not matter) was a typo on your part