Probability: Random stick breaking vs non random stick breaking If I broke a stick into n parts randomly, and broke another same length stick into n parts, with each part being equal.
If I pick a stick from both piles which is expected to be longer.
I couldn't figure out the basic to start, I can say on a average the length of a part on first stick is (1/N), and answer seems to be both will be same but I just don't feel right about it 
 A: 
If I broke a stick into n parts randomly

Start from there. What is the distribution function of the length of the sticks? What you need is some formal way to express what is the probability that each part of the stick has some length.
The answer is: It depends on how you split the stick.

EDIT: I made a mistake in one of the calculations. The answer is still "it depends" but the two examples I picked happen to yield the same result (shame on me). I'll cross out the parts that are invalid and add a third case where the results are actually different.
The possibilities are countless; here are my favorite two:


*

*You pick $N-1$ points on the stick independently with uniform probability. The case where any two are the same has probability 0, so I'll assume they are distinct. This determines $N$ parts.

*You pick $N$ real numbers to represent the length of each part (that their sum is the length of the stick) so all parts are treated equally. To formalize this we get a $(N-1)$-simplex (think the equivalent of a triangle in $N-1$ dimensions) as the space of possibilities. Since this is a bounded figure, it is possible to take a uniform density function on it.

*(New!) Start by splitting the stick in a random point picked with uniform probability. Set apart the first piece and keep going $N-1$ times.


Case 1 looks way nicer, doesn't it? Unfortunately it has the unfortunate effect that the length of the $N$ pieces do not have the same distribution: the ones in the middle are more likely to be shorter than the ones at the edges. EDIT: It so happens that this approach brings the same result in terms of probablity distributions for the length of the pieces as Case 2. The proof is not trivial though.
To visualize that think at $N=3$. Your choices for the two cuts can be represented with points on a square (each coordinate tells you where one cut is made). Compare the areas of the regions on that square where each individual piece is longer that $\frac13$ and multiply that for the probability of you picking that piece (I'll assume that is uniform). You will notice that the analysis for the two edge pieces will be completely symmetrical while the middle piece requires a special treatment.
In the end the probability that the stick you pick is longer than $\frac13$ is:
$$\frac13*\frac49+\frac13*\frac49+\frac13*\frac49=\frac49$$
Case 2 has the added complexity of higher-dimensional geometry but it is symmetrical and it is thus easier to obtain a general formula. A bit of geometry will tell you that the probability of any piece to be longer than $\frac1N$ is: $$\left(\frac{N-1}N\right)^{N-1}$$
I do not include the computations because they are very easy to make for those that know high-dimension geometry (just use similitude) but too long to fit in an answer otherwise.
Case 3 is totally different. The first piece is likelier to be longer since it is taken out of the whole stick (average length $\frac12$) and subsequent parts are expected to keep getting shorter as they are taken out of a progressively smaller part (average length halves each step).

Naturally you can split the stick in lots of other ways. You could say you pick apart one piece at a time for $N$ times but split the longer half on the next step. You can be trying to keep it "fair" by being "close" to $\frac1N$? How hard are you trying to be "fair"? How good are you at it? The possibilities are endless.
A: For simplicity, assume the length of the stick is 1.  We assume splitting a stick of length 1 randomly into $n$ parts means that we generate $n-1$ points independently from a Uniform(0,1) distribution, which points then divide the stick into $n$ intervals.
If $n=2$, the probability that a randomly selected interval is greater then $1/2$ is $1/2$.  For $n >2$, the probability that a randomly selected interval is greater than $1/n$ is less than $1/2$, so the "deterministic" stick is likely to have the longer piece.
We claim that all intervals on the random stick have the same distribution, and that the probability that an interval has length greater than $t$ is $(1-t)^{n-1}$.  To see this, suppose we generate $n$ points $X_1, X_2, \dots , X_n$ from a Uniform(0,1) distribution and use them to partition a circle of unit length into $n$ intervals.  By symmetry, the lengths of all the intervals have the same distribution.  Suppose we now cut the circle at $X_n$ to produce a line of length 1 partitioned randomly by the points $X_1, X_2, \dots , X_{n-1}$ into $n$ intervals, which all have the same distribution.  To see that $P(L > t) = (1-t)^{n-1}$ where $L$ is the length of an interval, consider the left-most interval.  Its length is greater than $t$ when all the $n-1$ points are greater then $t$, which happens with probability $(1-t)^{n-1}$, and all the other intervals have the same distribution.
So for any interval chosen on the "random stick", the probability that its length is greater than $1/n$ is 
$$P(L > 1/n) = (1-1/n)^{n-1}$$
The function $(1-1/x)^{x-1}$ is strictly decreasing for $x > 1$ and is equal to $1/2$ at $x=2$, so $P(L > 1/n)$ is $1/2$ for $n=2$ and is less than $1/2$ for $n>2$.
A: Just to confirm the solution by @awkward; consider the simplest non-trivial case, $N=3$, and let's assume that the stick length is equal to $1$.
We cut the stick at 2 points, $X_1$ and $X_2$, independently and uniformly chosen from $[0,1]$; $F_{X_1}(x)=F_{X_2}(x)=x, \ 0<x<1$
The lengths of the 3 parts are


*

*$Y_1=min(X_1,X_2)$

*$Y_2=|X_1-X_2|$

*$Y_3=1-max(X_1,X_2)$
Let's find the distribution of $Y_1$
$$P(Y_1>x)=P(X_1>x,X_2>x)=P(X_1>x)P(X_2>x)$$
so
$$1-F_{Y_1}(x)=(1-F_{X_1}(x))(1-F_{X_2}(x))=(1-x)^2$$
and
$$F_{Y_1}(x)=1-(1-x)^2, \ 0<x<1$$
It can be checked that $Y_2$ and $Y_3$ have the same distribution.
Now the probability that the length of one part is greater than $1/3$ is
$$P\left(L>\frac{1}{3}\right)=1-F_{Y_1}\left(\frac{1}{3}\right)=\left(1-\frac{1}{3}\right)^2=\frac{4}{9}<\frac{1}{2}$$
