Expected amount of parts in random cut If there is a square (or circle), we cut it $i$ times in uniform random location, the cut paths are straight and in uniform random angle. Not just horizontal and vertical paths.
What is the the expected amount of the parts in $i$ times?
For example:
if $i = 1$, it has $2$ parts, so the the expected amount is $2$.
if $i = 2$, it may have $3$ parts or $4$ parts.
Is there any theory can be the solution of this problem?
I'm sorry for my poor English, so I draw 3 images to describe what I mean. Hope this can clear my question. Thank you all.
http://i.stack.imgur.com/arTSH.png
 A: This is my interpretation of the problem: at every time you choose a random point of the figure (either a square or a circle or any convex set with positive measure) and draw a straight line by that point either horizontally or vertically with probability $\frac12$ for each case. Is this interpretation correct?
If so, at every time $i$ the probability to randomly choose a point lying in one of the previous lines is null. Therefore the lines up to time $i$ altogether form an $m\times n$ grid with $m+n=i+2$. Note that an $m\times n$ grid splits the figure in $mn=m(i+2-m)$ parts.
If $a_i$ denotes the number $m$ of rows in the grid, then
$$
\mathbb P(a_i=m) ~=~ \frac{\mathbb P(a_{i-1}={m-1}) ~+~ \mathbb P(a_{i-1}=m)}{2}
$$
Indeed, at time $i$ you have $m$ rows if and only if either at time $i-1$ you had $m-1$ rows and at time $i$ you drew a horizontal line, or at time $i-1$ you had $m$ rows and at time $i$ you drew a vertical line. Both these two cases have conditional probability $\frac12$.
You can easily check by induction that
$$
\mathbb P(a_i=m) = \frac{1}{2^i}\binom{i}{m-1}
$$
so, the expected value you are looking for at time $i$ is
$$
\sum_{j=1}^i
\sum_{m=1}^{j+1}
m(j+2-m)\frac{1}{2^j}\binom{j}{m-1}
$$
By suitably modifying and deriving the identity $\sum_{k=0}^n\binom nkx^k = (1+x)^n$ (and then evaluating it in $x=1$) you can end up with a closed form of the previous expression.
