Hi, folks.
This is a bit off topic, but may be of general interest to the folks on the
list. I'm posting here because this is the place I'd most likely expect to see
an answer!
I need a probability formula. The situation is as follows:
You have 7 objects owned by 7 people. If you were to randomly guess which
person owns which object, what is the probability that you would guess all 7
objects correctly? Guess 5 out of 7 correctly? Guess 4 out of 7 correctly?
Instead of just the answers, I'd prefer the actual probability formula. It's
been decades since I studied probability theory. A link to a web site with
probability formulas would be good, too (and would be on topic for a gaming
list! *grin*).
Any help would be appreciated!
> Allan Goodall wrote:
> Hi, folks.
Noooooo! Another probability thread!
Khaaaaaaaaaaaan!
Engaging filter now.
-Mike
Holy cow! Sarno Lives!!!!!!
:-)
> -----Original Message-----
> On Thu, Nov 17, 2005 at 08:45:07AM -0600, Allan Goodall wrote:
> I need a probability formula. The situation is as follows:
The number of ways of choosing k objects from a set of n objects, where
the order matters, is n!/(n-k)! (conventionally written as nPk). So if
you want to get 7 right, there are 7!/(7-7)! or 5040 ways you can guess,
of which one is correct.
It gets a bit trickier for other numbers. The short-cut is to construct
Pascal's triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
etc. where each number is the sum of the two diagonally above it. The relevant
row is n=7:
1 7 21 35 35 21 7 1
So if you want to know the probability of picking 5 out of 7 correctly:
(1) if you wanted to pick the _first_ 5 out of 7, that would be 7P5 or
2525. So your chance of doing that is 1/2525.
(2) But it's not just the first five; different ways of doing that. You might
get (1 for right, 0 for wrong):
1 1 1 1 1 0 0 1 1 1 1 0 0 1 1 1 1 0 0 1 1 1 1 1 0 1 0 1 etc.
See the triangle: counting up from 0, column 5 has 21 in it, which tells you
that there are 21 different variations without you having to work
them all out. So your actual chance is higher: 21/2525 or about 1/120.
http://mathworld.wolfram.com/Permutation.html
http://mathworld.wolfram.com/PascalsTriangle.html
I'm pretty rusty. Let's see.
Assuming the 7 objects are unique, I think the odds of getting all right are 1
in 7 factorial ("7 Permutations 7").
You can't get 6 right without 7, of course.
To get 5 right, multiply 1 in 7 factorial by the number of ways you can
pick 2 of the 7 ("7 choose 2" or 7 factorial divided by (7-2) factorial
times 2 factorial).
To get 4 right, multiply 1 in 7 factorial by the number of ways you can pick 3
of the 7 ("7 choose 3") then by the number of ways you can juggle 3 items in
three slots with none correct (Gut says 2, but I'm not sure of the underlying
math for larger numbers. For 4, I think it's 9.)
> -----Original Message-----