Chance calculation

[Kedas]

If you have an ordered list of numbers and you randomly change them what is the chance that at least one number will stay on its previous position?

 

Or more practically if you give everyone a number and you let them blindly pick a number. What is the chance that at least one persons picks his/her own number again.

[Tartaglia]

1-(1/n)

[Kedas]

It looks right for n=2 and n=3

but n=1 should be 1.

and if n is very large than the change that at least one stays at its current position is very high.

 

So if I have a cube of water and give it a good (random) shake.

Then I can say it is very likely that at least one molecule arrived on it's previous position.

That sounds a bit weird?

[Tartaglia]

The working is

 

Prob(>= 1) = 1 -P(0) = 1 - (n-1!/n!) = 1 - (1/n) for n> 1

 

In the case of n = 1, P(>= 1) = 1 - 0/1 = 1. The failure comes from definition of 0! = 1

[Kedas]

Thanks for the explanation

1-((n-1)!/n!) looks more right because it doesn't look too simple

[uncool]

Not quite - P(0) is not (n-1)!/n!. For example, P(0) for 4 is 3/8, not 1/4. The actual answer is:

 

1/1! - 1/2! + ... + (-1)^(n - 1)/n! (iirc). See

http://en.wikipedia.org/wiki/Derangement

=Uncool-

[Tartaglia]

Quite right uncool - I do apologise. I didn't consider that there are two types of numbers after one has been chosen.Those whose positions are taken and those whose positions aren't

[Kedas]

Thanks uncool.

My weird 'feeling' example feels much more right now