1. ## Probability Question

Hey,
Suppose that each of n sticks is broken into one long and one short part. The 2n parts are arranged into n pairs from which new sticks are formed. Find the probability (a) that the parts will be joined in the original order, (b)that all long parts are paired with short parts.
Thanks

2. Originally Posted by tbyou87
Hey,
Suppose that each of n sticks is broken into one long and one short part. The 2n parts are arranged into n pairs from which new sticks are formed. Find the probability (a) that the parts will be joined in the original order, (b)that all long parts are paired with short parts.
Thanks
Parts being joined in original order means that for each half a stick we have, we need one of the other sticks:

$\frac{1}{2}*\frac{n}{2n-1}2 = \frac{n}{2n-1}$

This needs to happen n times...

$P = (\frac{n}{2n-1})^n$

3. Originally Posted by tbyou87
Suppose that each of n sticks is broken into one long and one short part. The 2n parts are arranged into n pairs from which new sticks are formed. Find the probability (a) that the parts will be joined in the original order, (b)that all long parts are paired with short parts.
We must first dispose of possible ambiguities. We will assume that no two lengths are equal. That is, we have 2n different segments.
There are $\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^2 }}$ to partition that set into subsets of two elements each.
There is only one way for all the parts to match.

4. Originally Posted by Plato
We must first dispose of possible ambiguities. We will assume that no two lengths are equal. That is, we have 2n different segments.
There are $\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^2 }}$ to partition that set into subsets of two elements each.
There is only one way for all the parts to match.
This is the other option. I assumed that each stick was broken in the same manner.

5. Originally Posted by colby2152
This is the other option. I assumed that each stick was broken in the same manner.
That is a set of measure zero.

6. So in class today we went over this problem:
a) 1/((2n-1)(2n-3)...(1))
b) n!/((2n-1)(2n-3)...(1))

I'm not sure if these are equivalent to plato's response substituting in the denominator of (2n)!/(n!)^2.

Thanks

7. Originally Posted by tbyou87
So in class today we went over this problem:
a) 1/((2n-1)(2n-3)...(1))
I'm not sure if these are equivalent to plato's response substituting in the denominator of (2n)!/(n!)^2.
My apologies, but quite frankly what I posted is not correct. It should be $\frac{{\left( {2n} \right)!}}{{2^n \left( {n!} \right)}}$.
That is equivalent to what your class found.

LESSON: Do not work on two different but similar problems at the same time.
Cut and paste can be dangerous.

8. Is it possible to explain to me a bit where you got those values because i'm not really seeing it.
Thanks