Prev: RE: [GZG] [OT] Probability formula needed | Next: RE: [GZG] [OT] Probability formula needed |

Date: Thu, 17 Nov 2005 15:05:33 +0000

Subject: Re: [GZG] [OT] Probability formula needed

```
On Thu, Nov 17, 2005 at 08:45:07AM -0600, Allan Goodall wrote:
>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?
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
Roger
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
```

Prev: RE: [GZG] [OT] Probability formula needed | Next: RE: [GZG] [OT] Probability formula needed |