Why is the Average Vote per Users w/out Cameras...

Canon EF 70-200mm f/2.8L USM

02/02/2004 07:06:35 AM · #1

always an even number? That's virtually impossible for over 100 entries. Not on every single entry. Something is wrong here.

Edit: I'm talking about the paint with light. Just something I noteced on this one challenge.

Message edited by author 2004-02-02 07:07:47.

PaulMdx

Canon EOS-1D

02/02/2004 07:09:33 AM · #2

I guess because the number of voters without cameras is very small.

Canon EF 70-200mm f/2.8L USM

02/02/2004 07:17:20 AM · #3

Originally posted by PaulMdx:

I guess because the number of voters without cameras is very small.

If it was more than one, it would mean they all would have to vote the exact same score. Perhaps only one person without a camera voted? naaaaaaa hahahahaha

PaulMdx

Canon EOS-1D

02/02/2004 07:19:32 AM · #4

For the average score to be even??

6 + 8 = avg 7.0

4 + 6 = avg 5.0

Canon EF 70-200mm f/2.8L USM

02/02/2004 07:32:54 AM · #5

What are the odds in that on 120 entries all in the same challenge?

PaulMdx

Canon EOS-1D

02/02/2004 07:41:30 AM · #6

My point was, you said:

Originally posted by deafwolf:

If it was more than one, it would mean they all would have to vote the exact same score.

Which isn't the case, however small the possibility.

Gordon

RIM BlackBerry Curve

Canon EF 85mm f/1.8 USM

02/02/2004 08:02:02 AM · #7

Originally posted by deafwolf:

What are the odds in that on 120 entries all in the same challenge?

1 in 1^120 for 2 voters (unless it is too early here for my head to do stats)

Improbable events happen all the time, but in this case I think it is just one person.

dwoolridge

02/02/2004 08:54:39 AM · #8

Originally posted by Gordon:

1 in 1^120 for 2 voters (unless it is too early here for my head to do stats)

A sure thing in my books! Maybe you mean 1 in 2^120.

Gordon

RIM BlackBerry Curve

Canon EF 85mm f/1.8 USM

02/02/2004 08:56:45 AM · #9

Originally posted by dwoolridge:

Originally posted by Gordon:
1 in 1^120 for 2 voters (unless it is too early here for my head to do stats)
A sure thing in my books! Maybe you mean 1 in 2^120.

Actually I meant 10^120 or 1e120

At least I think I did. It's a bit later and I've had a coffee - but it isn't that much later!

dwoolridge

02/02/2004 09:10:17 AM · #10

Originally posted by Gordon:

Actually I meant 10^120 or 1e120

Oh, this answers the question: what's the probability 2 voters vote identically for 120 images?

I thought the more interesting problem was finding the probability the result will be integral for all images? Answer: 1 in 2^120.