Best Answer Anza Power, 06 August 2013 - 07:23 PM

**1+n+n(n-1)/6**.

Ok so assuming the random cuts are made by selecting two random numbers in the range 0-2pi and cutting a line between these two points on the circumferential of the pizza.

First note that the probability of having 3 cuts intersect at the same point is 0, so we assume all intersections are between just two cuts.

Also note that all the pieces that result from the cuts are convex.

With these two things in mind, if there were n cuts and c intersections then the number of slices would be 1+n+c, you can convince yourself of that by looking at what happens when you slowly start making a new cut.

So what is c? well for two cuts what is the probability that they intersect? if the first cut was A-B and the second was C-D, then they'd intersect if and only if the values of A B C D were ordered in one of the following forms:

ACBD ADBC BCAD BDAC CADB CBDA DACB DBCA

And not the following forms:

ABCD ABDC BACD BADC CDAB CDBA DCAB DCBA

ACDB ADCB BCDA BDCA CABD CBAD DABC DBAC.

Or another way you can think of it is you choose 4 points on a circle and assign them labels ABCD with equal probability, what is the probability that AB are not neighbors?

So the probability for intersection is 1/3.

Let x_{i,j} be a random variable that gets value 1 if cuts i and j inttersect and 0 otherwise, it's obvious that E[x_{i,j}]=1/3.

It's obvious that c=sum_over_i_and_j(x_{i,j}), we can't calculate the probability for c because x_{i,j}'s are not independant, but we can calculate it's mean because the mean is linear doesn't matter if the variables are independant or not, so:

E[c] = E[sum(x_{i,j})] = sum(E[x_{i,j}]) = n(n-1)/2*E[x_{1,2}] = n(n-1)/2*1/3 = n(n-1)/6

So on average you will get 1+n+n(n-1)/6 slices